on operators in relative homotopy groups

Annals of Mathematics

On Operators in Relative Homotopy GroupsAuthor(s): J. H. C. WhiteheadSource: Annals of Mathematics, Second Series, Vol. 49, No. 3 (Jul., 1948), pp. 610-640Published by: Annals of MathematicsStable URL: http://www.jstor.org/stable/1969048 .

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ANNALS OF MATHEMATICS

Vol. 49, No. 3, July, 1948

ON OPERATORS IN RELATIVE HOMOTOPY GROUPS

By J. H. C. WHITEHEAD

(Received June 17, 1947)

1. Introduction Let Q be an arcwise connected topological space and P C Q an arcwise con-

nected sub-space. Let et e 7rn(Q, P) be represented by a map1

(1.1) o: ,(I, I, ao) -(Q. PI po),

where In is the hypercube in R', given by o _ t, * * , t 1, I' is its boundary, ao = (0, , 0) and ao and po e P are base-points for the relative homotopy group 7r,(Q, P). Let

ft (I n I n) _>(Q) p)

be a deformation in which ft(ao) may wander, provided fi(ao) = fo(ao) -po, and let ar e 7r, (P) be the element represented by the map 0: (I, 0, 1) -+ (P, Po, Po) (I = I), where 0(1 - t) = ft(ao). Let2 ar be the element of 7n(Q, P), which is represented by the map fi. The transformation a -* ra, for a fixed ar e 7r,(P) and variable a e 7rw(Q, P), is an automorphism, Ta, of 7rn(Q, P). Also 0f + T-

is a homomorphism of 7r,(P) in the group of automorphisms of 7rn(Q, P). Thus 7rn(Q, P) is a group with operators, the operators being in 7rj(P).

If P = po, so that 7r,(Q, P) is the absolute homotopy group, rn(Q), the above process only produces the identity. In this case an automorphism a -* aa, with a e Xrn.(Q)a ? r1i(P), is defined' by means of a deformation ft: In -* Q such that ft(In) 0(1 - t), where 0: (1, 0, 1) --+Q, po, po) represents a- e 7ri(Q).

Thus we have two distinct processes for defining operators. The purpose of this note is to give a generalization of the second process from absolute to relative homotopy groups. The method is to deform the identity map i: P -+ P by means of a homotopy it: P -* Q, such that4 {l = i and 4o is a map of the form 4o: (P, po) -* (P, po). Let fo be given by (1.1), let go = fo In and let gt = 6_,go . Then the homotopy gt may be extended throughout In to give a deformation ft: In -* Q and the analogue of 0a e 7rn(Q) is the element of 7rn(Q, P), which is represented by the map fi .

I If Ai , Bi are subsets of A1, B1 (i = 1 , - *, m) the symbol f:(Al, * * * , AM)-.. (B1 , , - Bi) will denote a map of AI in B1, such that f(Ai) C B,. A homotopy

ft: (AI, *, Am) - (B1, - , Bin) will mean one such that ft(Ai) C Bi for each i. In practise we shall always have A, c Ai1, B, C B;i1.

2 Cf. J. H. C. WHITEHEAD, Proc. Lond. Math. Soc., 48 (1944), p. 283. 3 S. EILENBERG, Fund. Math., 32 (1939), 167-175, and J. H. C. WHITEHEAD, Proc. Lond.

Math. Soc., 45 (1939), pp. 278 et seq. 4Note that tj(P) need not be in P when 0 < t < 1.

610



RELATIVE HOMOTOPY GROUPS 611

2. The operators

Let QP be the function space of maps f: P -* Q, with what R. H. Fox5 has called the "compact-open" topology. WTe require P to be such that the continuity in the composite variable (p, t) of a map h: P X I -> Q, is equivalent to the continuity of the map h*: I -- QP, where h*(t): P -* Q is the map defined by h*(t)(p) = h(p, t). As proved by Fox this is the case, if, for example, P is regular and locally compact, or if P satisfies the first countability axiom. Let A C QP be the arc component of the identity map i: P -- P and let B be the sub-set of A, which consists of maps of the form f: (P, po) (P, po):

By a path, I, in A we shall mean a map, : I -* A, and t(O), t(1) will be called its end points. Let yi(A, B) denote the aggregate of classes of paths in A, whose end points are in B, two paths to, t, being in the same class if, and only if, there is a deformation Us (O _ s < 1) such that is(O), is(l) C B. We shall denote elements of yi(A, B) by u, x, y, x1 etc. If x e yi(A, B) and t e X we shall denote the arc components of B, which contain t(O) and t(1), by x(O) and x(1) and shall call them the extremities of x. We denote by e the arc component of B which contains i, and y1(A, B, e) will stand for the set of elements x e yl(A, B) such that x(l) = e. Our operators will be the elements of -y1(A, B, e), rather than individual paths, and the purpose of this section is to associate an endomorphism, a -- xa, of rn(Q, P) with each element x e -y1(A, B, e). In general a -c xa will not be an automorphism. For example, if Q = In, p = Pn, po = ao and if x e -yi(A, B, e) is the element represented by a path which joins the constant map In

-* ao to i, then it will be apparent that xa = 0 for each a e 7rn(Q P). In a later section we shall exhibit a sub-set of these endomorphisms, which is a group of automorphisms of 7rn(Q, P).

If two paths, t and -q, have the same end points and if one is deformable into the other with the end points held fixed, then, and only under these conditions, we shall write t-7. If t and tq are twvo paths in A such that t(1) = 71(O), then 4 will stand for the path which consists of i, followed by 77 and is given by

(071)(t) =(2t) if O < 2t < 1

= (2t-1) if I < 2t < 2.

More generally, if , , im are paths in A such that

(X)= s+1 (O) (i = 1, * ,m- 1)

then 4i *. im will denote the path which is given by

(i1 ... im)(t) = j(mt-j + 1) if j- 1 < mt ?g.

If m = 3 it is easily verified that (6122) % 2 and a similar result holds for ti, * , G with m > 3 and any system of brackets.

R. H. Fox, Bull. Amer. Math. Soc., 51(1945), 429-32. In this topology a neighborhood of a map fo: P -* Q is the aggregate of maps f: (P, C) - (Q, U), for an arbitrary compact set C C P and open set U C Q, such that fo(C) C U.



612 J. H. C. WHITEHEAD

If t is any path, we denote by t-' the path which is given by t-'(t) = t(1-t). Clearly t-q-1 _, t1 _ , for any paths X, t, v such that t(1) = n(0)

= t(0). Let a e 7rn(Q, P), let fo e a be given by (1.1) and let t be a given path in A,

which joins a point t(O) e B to t(1) = i. Let g0 = fo I I' and let gt be the homotopy, which is given by gt = (l - t)go. Let ft be an extension of gt to I'. Since t(O) e B it is a map of the form t(O): (P, po) -- (P. po) and it follows that f1(P) C P. fi(ao) = po. Therefore fi represents an element of 7rn(Q, P) which we denote by ta. Notice that, if t is a path in B, which termin- ates at t (1) = i, then f t (i) C P. ft (ao) = po , whence (a = a.

LEMMA 1. The elements ta is independent of a) the choice of fo from the maps in the class a and b) the way in which gt is extended to give ft . Also c) if t=- , then ta = -qa. The proof of this lemma is of a more or less standard type and will be postponed

till ?11. LEMMA 2. If i, t/ are paths in A such that t(0) e B, t(1) = n(0) = -(1) = i, then

t(r/q) = (trq)afor each a E 7rn(Q2 P). Let fo e a be given by (1.1) and let go = fo I In. Then fi e M~a, where ft is an

extension to In of the deformation gt = -(1 -t)go . Since n(O) = i we have gi = go. Also f2 e t (ra), where fist is an extension to In of the deformation g1+t = - t)gl = -t)go. Letf' = f2t(O < t < 1). Thenfi = f2 e a) and ft is an extension to In of gt = g2t . But

g2c = q(l-2t)gO if 0 2t < 1

= g1+(2g-1) if 1 _ 2t _ 2

= t(2 - 2t)go.

Therefore

qt= =(1-2t)go if 0 < 2t < 1

(2.2.) -(2- 2t)go if 1 < 2t ? 2,

and it follows from (2.1.) that gt = (e)(1 - t)o. Therefore fi e (tn~x, whence t(7a) = (t77)a.

LEMMA 3. Let X be a path in B and t a path in A such that (0) = X(1), t(1) = i. Then (X) a = {a for every a e 7nr(Q, P).

Let a E 7rn(Q, P), let fo e a be given by (1.1) and let g0 = fo I In. Let gt = (l- t)go and let gt be given by

4t = t(1 2t)go if 0 < 2t < 1

= X(2-2t)go if 1 < 2t < 2.

Let f be an extension of g' throughout I' (fo = fo). Then fi (XX)a. Since X is in B we have X(t) (P) C P. X(t) (po) = po, whence At) C P. tf(aO) = Po if 1 _ 2t < 2. Therefore fl/2 also represents the element (XW)a e 7rn(Q, P). But ft = ft/2




(O ? t < 1) is an extension of gt = t(1 - t)go to In Therefore fi e ta, whence =a (XW)a. Let t be a path in A such that t(O) e B, t(1) E e. Let tq and 12 be any paths in e

such that 711(0) = 172(0) = t(1), X1(1) = X2(1) = i. LEMMA 4. (t-i)a = (t72)a for any a e irn(Q, P). Since Da = D'a if v-D'we have

(Q71i)a = q2X2 Xq)

I {(t2)o I}a (no = X2 71)

(tM2) (fla) by Lemma 2.

But 71o is a path in e. Therefore floa = a, whence (t7ql)a = (t72)a. Let a e 7rn(Q, P), x e ]y1(A, B, e), let t e x and let -q be a path in e such that

-q(0) == t(1), 71(1) = i. Let xa denote (tn)a. THEOREM 1. The element xa e 7rn(Q, P) does not depend on the particular choice

of the path t e x or of the path -q, in e, which joins t(1) to i. The fact that xa does not depend on the choice of the path - follows from

Lemma 4. Let to, ti C x and let t, be a deformation of to into ti such that t,(0) E (O)

t8(1) e x(l). Let Xo and X1 be the paths described by t,(O) and t,(1). That is to say let Xi(s) = t8(i) (j = 0, 1). Then to is deformed in to the path XotiXj, with its end points held fixed, by means of the deformation G8, which is given by

=s(t) X0(3t) = t3g(0) if 0 < 3t < s

3 -s if s?< 3t < 3-s

= 1(3 -3t) = t.3_3t(l) if 3 - s ! 3t < 3.

Therefore to X0otXT'. It follows that

(toi7)a (X4iXj11 n)a

= { Xo(4171') a (n'= X-11 )

=(1%'q')a by Lemma 3.

Therefore xa does not depend on the choice of t e x and the theorem is established.

THEOREM 2. The transformation a -- xa, for afixed x e -yi(A, B, e) and a variable a e 7rn(Q, P), is an endomorphisrn of 7rn(Q, P).

The proof of this, which is essentially the same as that of the corresponding theorem for the operators discussed in ?1, will also be postponed till a later section (?12).

3. A multiplication in -y,(A,B) In this section we define a multiplication between certain pairs of elements in

y1(A, B). Let t and v be paths in A such that t(1) and t(0) are in the same arc




component, c, of B and t(1) e B. Let 77i, 712 be any two paths in c such that 71o(0) - 712(0) = (1), 71 (1) = 772(1) = (0)

LEMMA 5. If either6 a) every closed curve in c is contractible in A, or b) if c contains a map, f, which is a homeomorphism,

then t712t 7 where -q' is a path in B, such that 71'(0) = We have

(3.1) -

where 7qo = 711712. In case (a) we simply shrink the closed curve qo into the point t(0), holding the latter fixed, and thus obtain

where 1' is the constant path, given by 71'(t) -(1). This establishes the lemma in case (a).

In case (b) we may suppose that t(0) = f. For if not let r be a path in c, which joins f to (0). Then txqj_ t(, j(r-1)(r-)cj _ 1, 2) and we replace r by -r and apj by 13T. To simplify the notation we assume that t(0) = f in the first place. Then 10(0) = 1o(1) = f, where 710 means the same as in (3.1). We shall show that F'7q1oD is deformable, rel. 0 U 1, into the path q', which is given by 1(t) = t(I)f-17o(t). In fact such a deformation is achieved by means of the 2-parameter family of maps P(1 - s)f-1?7o(t). The desired homotopy, O5(Os = 0 61 = 7'), is defined explicity by

08(t) = ( (- 3t) if 0 _ 3t < 1- s

(3.2) = (s)f1o 1 (3t-1 + s) if 1-s 3t < 2 +

= -(3t-2) if 2 + s < 3t < 3.

Therefore F'171ot =7', whence t712t -t71? and the lemma is established. Let x, z C yi(A, B) and let x(1) = z(0). Let t Ex, v Ez and let 1 be a path in

x(1), such that 71(0) = t(1), 71(1) = t(0). Let xz 6 -yi(A, B) be the element which is represented by the path .

THEOREM 3. If the arc component x(1) of B satisfies either condition (a) or (b) of Lemma 5, then xz depends only on x and z and not on the choice of the paths r ex,

e z or of the path q, which joins (1) to (0) in x(1). The fact that xz does not depend on the choice of v follows from Lemma 5.

Before proceeding further we note the obvious fact that, if t is a path in A, with its end points in B, and if X, Au are paths in B, such that X(1) = t(0), ,u(0) = then t and Xtj/ represent the same element of y,(A, B).

6 I do not know what supplementary conditions, if any, are necessary for the Lemma.



RELATIVE HOMOTOPY GROUT'S 615

Now let to, ti C x and oI, t C z and let is, I. be homotopies such that (s(j), ?8(j) C B(j = 0 1). Let Xj(s) = t8(j), jj(s) = t8(j) and let qj be a path in x(l) which joins {j(l) to Dj(O). Then, as in the proof of Theorem 1, we have

67ox1o0 (Xo6X1\1 )77o(ytlo~11 1)

-?(t]X0tl~l1 ( )-1 -o %7 1 A7 = Xi 77oi'o)

--Xo(6i71) 77'M71l by Lemma 5,

where 1(t) e B. But Xo and t7'jul are paths in B. Therefore to?7oho and 017lD represent the same element, xz e y1(A, B), and the theorem is established.

4. A family of groups contained in 'yi(A,B) We define a multiplication between the arc components of B, which we denote

by a, b, c etc. If fe QP, 9 e PP, then fg EQP is defined. If fo, go C B thenfogo(P) C P, fogo(po) = po and if ft, gt are deformations of fo, go into fi = gi = i, then ftgo, followed by gl, is a deformation of fogo into i. Therefore fogo e B. Iffo , fi C a and go, gi C b, where a and b are arc components of B, and if ft, gt are homotopies of the form f,, qt: (P, po) > (P, po), then the homotopy fogt determines a path in B, which joins fogo to fJg,. Therefore fogo and fig, are in the same arc component of B, which we denote by ab. Clearly a(bc) = (ab)c and ea = ae = a, where c is the arc component containing i.

Let t be a path representing a given element x e 7y1(A, B) and let f e a, where a is an arc component of B. Then a path if, joining t(O)f to t(l)f, is defined by (if) (t) = (t)f. Clearly the element of y1(A, B), which contains if, is independent of the choice of t ex and f e a. We shall denote it by xa. It is also clear that x(ab) = (xa)b and we shall write it as xab. If a is an arc component of B and y e -y1(A, B), then ay is, in general, undefined. Therefore a product of the form xay, or xaby, will always mean (xa)y, or (xab)y. The extremities of xa are the arc components x(O)a, x(1)a.

Let x, y C -y1(A, B), let x(1) = y(O) and let x(l) satisfy either the condition (a) or (b) of Lemma 5. Let a be any arc component of B such that the arc component x(l)a = y(O)a also satisfies either condition (a) or (b) of Lemma 5. Then xy and (xa) (ya) are both defined and I say that

(xy)a = (xa)(ya).

For let f e a and let t e x, q e y be such that t(1) = -(O). Since {(t)fI(t) - (77)(t)f it follows from (2.1) that (trq)f = (cf) (nif), whence (xy)a = (xa) (ya).

Let maps f, f' C B be such that ff, f'f C e. That is to say, each of ff', f'f -1 -1 - is homotopic to the identity, rel. po. Then aa1 = a a = e, where a, a- are the

arc components of B, which contain f, f' respectively. Under these conditions we shall describe f and ff' as homotopy inverses, rel. po, of each other and shall describe a-' as the inverse of a and a as the inverse of a-'. It follows from this definition that any maps fi e a, fl e a-' are homotopy inverses, rel. po, of each other.




Let G be the aggregate of arc components of B, which consists of elements with homotopy inverses, rel. po. Since (ab)c = a(bc), ea = ae = a, aal = a a = e if a e G, it follows that, with this multiplication, G is a group. We now make the assumption that P, Q are such that Lemma 5 is valid provided c e G, and hence that Theorem 3 is valid if x(1) = z(O) e G. This is certainly true if, for example, each arc component of B is simply connected7. For then the condition (a), in Lemma 5, is verified. So far as I know, this assumption may not imply any further restriction on P and Q. If it does, the following results are valid, without this further restriction, provided G is replaced by the sub- group, whose elements are arc components containing homeomorphisms.

Let -y1(A, G) C '1(A, B) be the sub-set which consists of elements whose extremities are elements of G. Our assumption concerning P and Q implies that, if x, y C -y1(A, G) and if x(1) = y(O) then a unique product xy is defined, as in Theorem 3. Also xa e -y1(A, G) if x e -y(A, G), a e G.

We now define the product

(4.1) xy = x{x(1)-1y(O)y

of any ordered pair of elements x, y C '1 (A, G). If x(1) = y(O), then xy = xy. The extremities of x-y are (x-y)(O) = x(O){x(1)'-1y(O) and (x-y)(1) = y(1). Let x, y, z C yi(A, G) and let aj = x(j), bj = y(j), c; = z(j)(j = 0,1). Re- membering that (xy)a = (xa) (ya), we have

(x *y) .z = (xal'boy) z

= (xaT'boy)bYlcoz

= (xalT'bobT 'co) (ybT'1co)z and

x * (y *z) = x * (yb11coz)

= (xalT'bobT'lco) (ybj'1co)z

(4.2) = (x.y).z

Therefore the multiplication defined by (4.1) is associative. We now write x * y as xy for any elements x, y C -yj(A, G). Notice next that, if x, y C 'yi(A, G), b e G, then, since (xb)(1) = x(1)b, we have

xby = xbb-1x(lf-y(O)y = xx()-1y(O)y = xy.

Let H. C 'y1(A, G) be the sub-set consisting of all the elements, x, such that x(1) = a, for a given a e G.

THEOREM 4. With the multiplication defined by (4.1), Ha is a proup. Let u e Ha be the element which is represented by a path in the arc component

a (e.g. by a constant path). Then obviously xu = ux = x for every x e Ha.

7e.g. if P is a linear graph. For then 7r2(P) = 0 and it is easily verified that each arc component of B is simply connected.




Because of (4.2) it only remains to show that, to each x e Ha corresponds an element x-' e Ha, such that x-'x = u. Since x'bx = x'x, for any x' e -yi(A, G), b e G, and since x'b E Ha if b = x'(1)Y'a, we have only to find an element x 'yE'i(A, G) such that x'x = u. Let t e x and let x' ey'1(A, B) be the element which is represented by the path t-'. Then x' e -y1(A, G), since x'(O), x'(1) - x(1), x(O) C G, and obviously x'x = u. Thus the theorem is established.

Since (xy)c = (xc) (yc) if x, y C -yi(A, G), c c, G it is easily verified that x -* xa-'b is an isomorphism of Ha onto Jib for any b E G.

5. A group of automorphisms of 7rn(Q, P)

Let T. be the endomorphism, a -* xa, of wrr(Q, P) for any x e He. THEOREM 5. Tx is an automorphism of wrn (Q, P) for each x e He. The map

x -> Tx is a homomorphism of H. in the group of automorphisms of 7rn(Q, P). Let x, y C He and a e ir, (Q, P) be given and let x t X, X e y be such that t(1)

= y(l) = i. Then it follows from the argument used in proving Lemma 2, with (2.2) replaced by

t = (1-2t)gO if O < 2t < 1

= (2 -2t)-q(O)go if 1 < 2t < 2,

that x(ya) = (xy)a. Clearly ua = a if u is the unit element of He, represented by the constant map on i. Therefore xa = 0 implies 0 = x-'(xa) = (x-l x)a = a, whence T. is an isomorphism into. Also x(x-'a) = (xx-l'a = a for a given a e 7rn(Q, P), whence Tx is onto. Therefore Tx is an automorphism of 7r2n(Q, P). Finally x(ya) = (xy)a means that TxT, = TX,, or that x -* Tx is a homomorphism of He in the group of automorphisms of irn(Q, P). Thus the theorem is established.

6. Decomposition of H, Let (1) II C H. be the aggregate of elements in He, which are represented by deforma-

tions of P over itself; that is to say, by paths, i, such that t(t)P C P (in general t(t)(po) # po unless t = 0 or 1), and

(2) let r C He be the aggregate of elements in He, which are represented by deformations of P, rel. Po; that is to say by paths, i, such that (t) (Po) = po. Clearly II and r are sub-groups of He. If x e He let 0(x) by the element of 7ri(Q), which is represented by the track of Po in the deformation t(t): P - Q. where t e x. That is to say, 0(x) is represented by the map f:I J- Q, which is given by f(t) = (t)po . Since is(O) (po) = s(l) (Po) = Po if is(O), (1) C B it follows that 0(x) depends only on the element x e He and not on the choice of the path ex. Also0(xy) = 0(x)0(y) since {f(t)ri(O)}(po) = (t){n(O)(po)} = W(t)(po) if 1(O) e B. Therefore 0H HIe * 7r1(Q) is a homomorphism.

Let t be a path, which joins a point t(O) E B to t(1) = i and is such that {(t) (P) C P for each t e I, though possibly t(t) (Po) 5 Po unless t = 0 or 1.



618 T. H. C. WHITEHEAD

LEMMA 6. If P is an8 A.N.R. (absolute neighborhood retract) then the map t(O) has a homotopy inverse, rel. po. That is to say the arc component of B, which contains t(O), is an element of G.

Let ft: pO P be the deformation of the map fo = i I po, which is given by f'(po) = t(t)(po) (N.B. t(O)(po) = po, since t(O) e B). Since P is an A.N.R. the deformation f' can be extended throughout P to give a deformation ft P - + P. with fo = i. Let n (t) = fi-, . Then n1(t)(po) -fi-t(po) = t(1 -t)(po). Let gt: P P be defined by

9t = t(2t.)rn() if 0 _ 2t _ 1

= r) (2t -1) if 1 _ 2t < 2.

Then go = t(O>)1(O), g, = q(1) = i and

gt(po) = t(2t)7h(0)(po) = t(2t)(po) if 0 _ 2t ? 1

- r(2t - l)(po) = t(2 - 2t)(po) if 1 ? 2t ? 2.

Therefore there is a homotopy p.: I -- P given by

p8(t) = {2(1 - s)t} (po) if 0 ? 2t ? 1

= {2(1 - s)(1 - t)}(po) if 1 < 2t _ 2,

such that po(t) = 9t(po) I pi(t) = t(O) (po) = Po. Let9 R = (P X 0) U (po X I) U (P X 1) and let X: R -P be the homotopy,

which is given by

X8(pj) = gj(p) (p e Pt j = 0, 1)

Then Xo(p, t) = gt(p) for any (p, t) e R and Xj(p, 0) = go(p) = t(O)7(O)(p) for any p E P. Also Xj(po , t) = pi(t) = po . Since P is an A.N.R., the deformation X8 can be extended to a deformation, j, P X I -* p, of the map Ao , which is given by Mo(p, t) = gt(p). Let t: P -* P be the deformation, which is given by bt(p) = Mj(p, t). Then ao(p) = AL(p, 0) = Xi(p, 0) = t(O)7(O)(p), S5(P) = AL(p, 1) = Xj(p, 1) = g9(p) = p, St(po) = Mi(po , t) = Xj(po, t) = Po. Therefore t(O>)(O) C i rel po. Similarly t(O)t(O) i, rel. po, and the lemma is established.

THEOREM 6. (a) II is invariant in He. (b) If either P is a polyhedron or Q is an A.N.R., then r = 0-1(1), and is there-

fore invariant in H..

8 K. BORSUK, Fund. Math., 19 (1932), p. 222. As is nowadays usual, we do not require an A.N.R. to be compact (cf. W. HUREWICZ and H. WALLMAN, Dimension Theory, Princeton (1941), pp. 82, 86). It is, of course, trivial that t(O) has a homotopy inverse, e.g. the identity. The point is to show that it has a homotopy inverse rel. po.

9 In proving this lemma we reproduce an argument used by Eilenberg (loc. cit., p. 169), which will be convenient for future reference.




(c) if either P is a polyhedron or both P and Q are A.N.R.'s, and if the natural homomorphism1 q5: 7ri(P) -i 7r,(Q) is onto, then" He = iir.

Let x e H and y e He be represented by paths t e x and 7X e y, such that {(t)(P) C P, 2(1) = = i. The element y-1 is represented by the path 7'= t'9, where g is a homotopy inverse2, rel. Po , of n(0). Also n'(0) = Xo = 77(1)g = g. It follows from (4.1) that yxy -, written as (yx)y-1, is represented by the path

{nt(O)EO'(O)n' = {nt(O)n'(O)} {tx'(O)} {tF'n'(O) I. This path is deformable into the path' n(0)tq'(0), with its end points fixed, by means of the 2-parameter family of maps -(s) (t) '(0). More explicitly, the homotopy in question, Pa, is given by

p8(t) = q(3t)t(0)q'(0) if 0 _ 3t _ 1 - s

(l7 ) (3t-1 +s)- (0) if 1-s < 3t < 2 + s

= (3-3t) '(0) if 2 + s _ 3t ? 3.

Since t(t)(P) C P, n(0) n'(0) e e, it follows that tq(0))q'(O) represents an element of H. Therefore yxy'1 e II, which disposes of (a).

Now let either P be a polyhedron'4 or Q an A.N.R. Obviously 0(x) = 1 if x e r, whence r C 0-1(1). Conversely, let 0(x) = 1 and let to e x. Then the circuit described by Po in the deformation to(t): P - Q is contractible. In (6.1) let po(t) = (t) (Po) and let p8 be a homotopy, rel. 0 U 1, of p0 into the constant map P1: I -* Po. Also let the map gj P -> P in (6.1), be replaced by o0(j) (j = 0, 1). Then 0., given by t.(t)(p) = (p, t), where A8 is defined as in the argument which follows (6.1), is a homotopy, rel. 0 U 1, of to into a path ti, such that 6,(t)(po) = Po. Therefore x e r, whence 0-1(1) C r. Therefore r = 0-1(1), which disposes of (b).

Now let either P be a polyhedron or P and Q both A.N.R.'s and let the homomorphism 4: 7r,(P) -* 7r,(Q) be onto. Then it follows from an extension argument of the kind used in the last paragraph that any element x e He has a representative path, I, such that t(t) (Po) e P. As in the proof of Lemma 6, let Xq be a

10 4, is the homomorphism in which corresponding elements are represented by the same map. This condition means that every element of wr (Q) has a representative circuit in P.

11 It is tempting to argue that II n r = 1, and hence that H. = II X r, since any path, which satisfies both of the above conditions (1) and (2), for II and r, lies in e. However an element x $ 1, of H., may be represented by one path, which satisfies (1), and another which satisfies (2).

12We are retaining the condition on P and Q, concerning Lemma 5, which was imposed in ?4.

13 Notice that ,(O)E(t)X,'(O) is defined since t(t) (P) C P. This does not alter the fact that, in the preceding formulae, iq(O)t means bi(O) It.

14 In this case we use the extension theorem given in P. ALEXANDROFF UND H. HOPF, Topologie, Berlin (1935), p. 501 (Hilfsatz 1 a).



620 J. HI. C. WHITEHEAD

path such that tq(0) e B, t/(l) = i, t/(t)(P) C P, r/(t)(po) = - t)(po). Then it follows from Lemma 6 that the map t(0) has a homotopy inverse, rel. Po, and hence that the arc component of B, which contains n(O), is an element of G. Therefore Xq represents an element y e II. Since -(t) (Po) = - (1-) (po) we have 0(y) = 0(x-'). Therefore 0(yx) = 1, whence yx = z e r. Therefore x = y- z, with y-' e II, z e r, and the proof is complete.

We now assume that either P is a polyhedron or P and Q both A.N.R.'s. Then we have

THEOREM 7. The group II is the image of 7r,(P) in a homomorphism

J : 7ri(P) > H, which is onto and is such that

a) X = 4), where 0: He -* 7r,(Q) and 4 7r: (P) -*> 7r(Q) mean the same as before, b) 4t'(a)a = aa, for each a e 7ri(P), a e irn(Q, P), where ca means the same as in ?1, c) if ;(a) = 1, thena*3 = ffor any f e 7r.r(P) (m 1 = Bif1 m > 1,

and = Crf3 Vifm = 1). Let z e HI and let ?o e z be such that Po(t)(P) C P. I say that, if the circuit

described by Po in the deformation pc(t): P -- P is contractible in P, then z = 1. For it follows from an extension argument of the kind used in Lemma 6 that there is a homotopy, id, rel. 0 U 1, such that Ps(t)(P) C P, P1(t)(po) = Po. Then P, e z and ?, is a path in e. Therefore z = 1.

Let a e 7ri(P) and let f: (I, 0, 1) -* (P, Po, po) be a map representing C. As in the proof of Lemma 6, there is a path, I, such that {(1) = i, t(t) (P) C P, W(t) (Po) = f(t). Then t(O) (Po) = f(O) = Po and it follows from Lemma 6 that t represents an element x e I. We write x = 4u(a). Let f': (I, 0, 1) > (P, Po, po) be another representative of a and let I' be a path such that {'(I) = i, {'(t) (P) C P. ('(t)(po) = f'(t). Let x' e H be the element represented by {'. Then x-x' is represented by the path v = (-1q) t', where g is a map lying in the arc component x(0Y)-x'(O), of B, and 77 is a path in this arc component joining (t-'g) (1) = t(O)g to 1'(0). Since t(t)(P), g(P), 7(t)(P), ('(t)(P) C P we have t(t)(P) C P. Also, since g(po) = n (t)po = Po , we have

t(O)(po) = (l- 3t)g(po)

= (l- 3t)(po) = f(1 - 3t) ifO < 3t < 1

=0 (3-1) (po) = Po if 1 3 3t < 2

-'(3t -2)(po) = f'(3t -2) if 2 < 3t ? 3.

But f - f' in P, rel. 0 U 1, since f, f' C a. Therefore the circuit described by PO in the deformation P(t) is contractible in P. Therefore x-x' = 1, or x' = x. Therefore 4i(u) is a single-valued function of a e 7r1(P). Thus A,(a) is uniquely defined as the element of H, which is represented by any path I, such that

15 Here AB -* O8 is the automorphism of the absolute homotopy group 7rm(P), which was discussed in ?1.




,(1) e e, $(t) (P) C P and the deformation t(t) carries po round a circuit representing a.

Let a-, o2 C xri(P) and let 41(o-r) (r = 1, 2) be represented by a path r , such that t,(t) (P) C P and the circuit described by po in the deformation {r(t) represents o, . We may also assume that 41(1) = i. Then 1 (u1) 1(u2) is represented by the path 71 = U122(O) 2, and the deformation -q(t) carries po round a circuit representing U102. Moreover -(t)(P) C P. Therefore -q also represents the element #,(0102). Hence 4f(uiu2) = #,(0u1)1t(u2), or 4, is a homomorphism.

The homomorphism 41 is onto. For if x e II is represented by a path I, such that t(t) (P) C P, then x = ?*(a), where a is represented by the track of po in the deformation t(t). Obviously @(a) = +(a) for any a e xri(P). It is also obvious from the definition of A,(a) that 4,(uo)a = uoa for any a e 7rn(Q, P). This disposes of (a) and (b).

Let a e 7ri(P) be given and let 4(o) be represented by a path, I, such that t(1) =i, t(t) (P) C P and h e a, where h: (I,0, 1) -* (P, po , po) is given by h(t) = (1- t) (po). Let ,3 e irm(P) be represented byfo: (Im',') -* (P, po) and let ft = t(1 - t) fo . Then ft(Im) = (1 - t)(po) = h(l - t). Therefore fi e Am3. But fi = t(O)fo, whence /3 -* u is the automorphism induced by the map t(O): (P, po) -i (P, po). If A(a) = 1, then 4/(u)(0) = e and t(O) is also joined to t(1) = i by a path, I', in e; that is to say by a path, I', such that ('(t)(P) C P, ('(t)(po) = po . Thereforef, is also obtained from fo by the deformation ('(1 - t)fo: (Im, Jm) (P, po). Therefore a- : = /3 and the proof is complete.

Any two maps fo , f: (P, po) -* (P, po), in the same arc component, a, of B are connected by a homotopy ft: (P, po) (P, po). They therefore determine the same endomorphism of 7rm(P) for any m ? 1. We shall denote this endomorphism by 1 --* a(13), where 13 e 7rm(P). If a and b are two arc components of B and if f e a, g e b, thenfg e ab and it follows that (ab)(fl) = a{b(fI)}. Also e(f) = ,B and it follows that f --* a(8) is an automorphism if a e G; also, writing Sa for 13 -> a(G), that a -> Sa is a homomorphism of G in the group of automorphisms of rrm(P).

THEOREM 8. If a e 71(P), x = 41(a) e H, y e He, then yxy1 = y(0)(0) Let t e x and t7 e y, where t(t) (P) C P, and let a be the element represented

by the map g: I -> P, which is given by g(t) = (t)(po). It was shown in the proof of Theorem 6(a) that -q(O)47q'(O) e yxy-, where -q'(0) is a homotopy inverse, rel. Po, of -q(O). Since -q'(O) (po) = po the track of po in the deformation 7(1) (t)7'(0) is the map g*: I - P, which is given by

g*(t) = q(0)0)n'(0)(p0)

= 7(O) (t)(po)

= 7(0)g(t).

Therefore g* represents the element y(O)(o) e Xri(P), and the theorem follows. A possible application of these operators is to the case where Q is a complex

and P is its (n - 1)-dimensional skeleton. With this in mind we shall study



622 J. H. C. WW-HITEHEAD

the case in which P consists of k (n - 1)-spheres with a single common point, po, no two meeting each other anywhere else. We shall give an algebraic1s description of the group r, assuming that the homomorphism 71r2(P) -+ 7r2(Q), in which corresponding elements are represented by the same map, is onto if n > 2. If n > 2 and if 7r,(Q) = 1, then it follows from Theorem 6(c) and Theorem 7 that r = He, since 7r,(P) = 1. If n = 2 it appears that the automorphisms of 7r2(Q, P), which correspond to the elements of I f r, are precisely the inner automorphisms. Hence, modulo the inner automorphisms, the image of Hr in the group of automorphisms of 7r2(Q, P), or the image of He if 4): 7r,(P) -- ir,(Q) is onto, is the direct product of two groups, corresponding to IT/iI nr and r/il n r, each of which is amenable to algebraic analysis"7. For this we shall need a purely algebraic section.

7. Algebraic digression'8

Let L and A be arbitrary multiplicative groups. We shall denote the elements of L by small Roman letters and those of A by small Greek letters. Neither L nor A need be commutative. Let'9 a >-+ Pa be a homomorphism of L in the group of automorphisms of A. Thus 4)a :A- A is an automorphism and 4)ab 4 'acb. We shall write

b b) c =Ccb (C -1) b ( b) -1 -b qkb(Ca) = ab (ab) = ac (aot)b (of)=

Let a: A -* L be a fixed homomorphism, which is an operator homomorphism in the sense that20

(7.1) a8ab= baab-'

and let

(7.2) afa-1 Ma

for every pair of elements a, j3 C A. Notice that (7.2), with ,3 = a, implies ba a aj. A crossed homomorphism, f: L -- A, is a transformation such that

(7.3) f(ab) = f(a)f(b)a

16 i.e. algebraic in terms of 7r,1(Q, P). 17 For an algebraic account of the operators of the kind discussed in ?1, which correspond

the elements of II, see J. H. C. WHITEHEAD, Annals of Math., 42 (1941), pp. 421 et seq.; also Annals of Math., 47 (1946), 806-10.

18 This is closely related to the theory of group extensions. See R. BAER, Auto- morphismen von Erweiterungsgruppen, Paris (1935); also S. EILENBERG and S. MACLANE, Annals of Math., 48 (1947), p. 55 and M. Hall, Annals of Math., 39 (1938), p. 222.

19 In our application of this we shall take L = 7r,-(P), A = r.(Q, P) for n > 2. We start with the algebra which is appropriate to the case n = 2. In this case 4. is the operator corresponding to a -ri(P). If n> 2, then 4o is taken to be the identity for every aern_4(P).

20 Notice that Oab and Oact can only mean (aa)b and a(qO), respectively.



RELATIVE H3MOTOPY GROUPS 623

for any pair of elements a, b C L. From (7.3), first with a = 1 and then with b = a', we deduce

(7.4) f(l) = If(a-l)a = f(a)-.

Let C be the totality of crossed homomorphisms of L into A. We shalldefine a multiplication, f. g, of any elements f, g C C and shall show that, with this multiplication, a certain sub-set of the elements in C is a group F. In our special case a factor group of F is the algebraic counterpart of F C He .

In order to define f. g we associate with each f e C the transformations Sf: L -* L and f A -* A, which are given by

(7.5) Sf(a) = af(a)a, f (a) = f(a)a.

Notice that

Sf(Oa) = af(la)ala = atf(&a)aj = 61f(a) or

(7.6) Sfa = a~f . Notice also that

E (Oa) = f(aoa) Oa

- f(aa3a-l)oa by (7.1)

- f(a)f(aO)af(a-l)"azf by (7.3)

- f(a)f(COI)aIaf(a-l)a by (7.2)

(7.7) - f(a)Ef(O)af(a)- by (7.4)

I say that Sf and Jf are endomorphisms of L and A. For

Sf(ab) = af(ab)ab

= a {f(a)f(b)a } ab by (7.3)

= af(a)aaf(b)a-lab by (7.1)

= df(a)aaf(b)b

(7.8) = Sf(a)Sf(b)

and Ef (a0-) = M~ao-) aio

= f(daalR3)aCR3

-f(= a)f(da)io by (7.3)

= f(OaO)aMfGdf)1 by (7.2)

(7.9) = Ef 4(C')7_' (0 4{




The endomorphism Ef is to be compared with the endomorphism a -* xa, of Theorems 1 and 2, and Sf with the corresponding endomorphism, , -* x(O) (p3), of 7rwi(P), which was discussed in the passage preceding Theorem 8.

With every ordered pair of elements f, g C C we now associate the map (f * g) : L -* A, which is given by

(7.10) (f *) (a) = f { S, (a) } g (a).

We have (f g) (a) = f {dg (a) a } g (a)

= fdg(a)f(a) 1

alg(a) by (7.3)

(7.11) = fdg(a)g(a)f(a) by (7.2)

whence

(7.12) (f g)(a) = ZI{g(a)}f(a).

I say that f g e C. For

(f g) (ab) = {qg (ab) }f(ab)

= Ej {g(a)g(b)a }f(a)f(b)a

= f {g(a) },f {g(b)a}f(a)f(b)a

= Jg{g(a) }f(a)Ef{g(b) }af(b)a by (7.7).

= (f g) (a) { (f * g) (b) Ia.

Therefore f g e C. I say that

(7.13) Sf.= SfSg , ,f.g = , , .

For Sp.g(a) = a(fg)(a)a

= a[fI S(a) }g(a)]a = df{S(a)}dg(a)a = f SfS(a) } S(a) = Sf{Sg(a)}.

Also EfZ.(a) = (f g) (ae)a

=Ef I g (ca) }f(dlae)a = fI Oa) I J) f (ae)

= I {gOa)aC'I} by (7.9)

w Efhesl(ai)( which establishes (7.13).




I say that

(7.14) (f.g).h = f.(g.h)

for any f, g, h C C. For

{(f.g) .h} (a) = (fi.g){Sh(a)}h(a)

= f I SgSh(a) }g{Sh(a) }h(a)

= f{I SgSh(a) } (g * h) (a)

= f{Sg-h(a)}(g.h)(a) by (7.13)

= {If.(g.h)}(a).

Let j: L -* A be the map given by j(L) = 1, where 1 denotes the unit element in A. Obviously j E C and

(a) [S,= 1,>2 = 1 (7.15) (b) 'Si 12fEj

(b) tjf =f j = f

where 1, in (7.15 a), denotes the identical automorphism of both L and A. Let F C C be the sub-set consisting of all those elements f e C such that Sf

is an automorphism. I say that F, with the multiplication f. q, is a group. For it follows from (7.15 a) that j e F and, in the presence of (7.14) and (7.15 b), it only remains to prove that, to each g e F corresponds a g' e F such that g' * = j. Given g e F, let g': L -* A be the map, which is defined by

g'(a) = {gS-1(a) }'.

I say that g' E C. For let a' = So, '(a) for any a e L. Then (ab)'= ab', g'(a) = g(a")V and ag(a')a' = S,(a') = a. Therefore

q'(ab) = g(aV)

= {g(a')g(b) })I

I gq(b')aq(a')a (a,)}1

- {g(b~)ag(a) }-

- g(a)- g(b)a

= go(a)g (b) .

Therefore g' e C. It follows from (7.10), withf = 9' that

(g'.g)(a) = g'{Sg(a)}g(a) = {jgS'Sg(a) } 1g(a)

= (a)-1 (a)

= 1.




Therefore g'.g = j. It follows from (7.13), with f = 9' and from (7.15 a) that S,, S, = 1. Therefore S0, = Si', whence g' e F. We have proved that g' *g = j and it follows that F is a group. Moreover it follows from (7.13) that f -+ S, is a homomorphism of F in the group of automorphisms of L.

Let F' C C consist of all those elements f e C such that Ef is an automorphism. Then F' is a group. For we have only to find an f e F' such that fief' = j, for a given f e F'. Let f': L- A be defined by

f'(a) = En f(a)-1 }. Then f' e C. For >f2{f'(a)I = f(a)-' and

Ef {f'(a)f'(b)'} = Ef{f'(a) }I f If'(b)aI

= f(a) f(a){I f f(b)} af(a) by (7.7) = f(b)Yf(a)'

= f(ab)X'

= E}f '(ab) 1. Since OJf is an automorphism it follows that

f'(ab) = f'(a)f'(b)a, or that f' e C. Also

(f-f')(a) = Z,{f'(a)}f(a) = 1,

whence f f' = j. It follows from (7.13) and (7.15a) that f = Ef7, whence f/ e F'. Therefore F' is a group. It follows from (7.13) that f -f is a homomorphism of F' in the group of automorphisms of A.

I say that F' = F. For let f e F' and let f' be the inverse of f, so that f' .f = fif' = j. Then it follows from (7.13) that Sf Sf = Sp Sf = 1. Therefore Sf is an automorphism. Therefore f e F, whence F' C F. Similarly F C F' and it follows that F' = F.

Let E be the set of all maps of the form ea: L -* A where ea is given by ea(x) = aa,

for a fixed a e A and variable x e L. We have ea(xy) = aaco = aacaza-XV

= aoa-x(aa-W)c

= e .(x) e.(y) -

Therefore ea e C. Further

Sea(x) = aea(x)x

= a(aaX)x (7.16) = aa(xaa-13 -7)x

= (Oa) x (a)-'.




This is an (inner) automorphism, whence ea e F. Also

(7.17) = aaL

= CaY1.

Thus Lea is an inner automorphism of A. Moreover a given inner automorphism, - ata 1, of A is the automorphism E., . Therefore the automorphisms EY,

for all elements a e A, are precisely the inner automorphisms of A. It follows from (7.12) and (7.17) that

(ea.en)(x) = Zea.j{ef(x)}ea(x)

= a 3-Zala-1 -

(7.18) =

= e,(x).

Clearly el = j where 1 denotes the unit element in A. Therefore it follows from (7.18), with (3 = a-1 that the inverse of ea is ed,-i . Hence, and from (7.18), E is a sub-group of F. Also a -* ea is a homomorphism of A on E.

If f E C we have (f e) (x) = Ef{ea(x)}f(x)

= Zf(aa)f(x)

= Ef(a)Zf(a)T(x) by (7.7)

= En(a)f(x) Ef (ca) f(x)-1f(x)

= E(a)f(x) Ef(a). Also

(e: .f) (x) = 1X) I efx)

(7.19) = bf(x)131313 by (7.17)

= Af(x) 3'.

Therefore f et = e, -f with ,3 = Zf(a). Hence E is an invariant sub-group of F.

Notice two special cases of the above. First let L be a free group, which is freely generated by a1, , ak . Then a map f e C is uniquely determined by the elementsf(ai) = ai together with the rule (7.3). If a = a" a (eK = A 1),

then it follows from (7.3), (7.4) and induction on p that

(7.20) f(a)= albl .pbw

where be = a,1 ... a9- if ) ,

=a * ad if EX = -1.




Let us writef = (ai, l , ak), where ai = f(ai). If g = (A . , ok) we have f-g = (yi, * * * , y) where

= f (fg)(ai)

= f { g (ai) } g (ai)f(ai) by (7. 1 1)

= f(OIi)/3,ai. Secondly let L and A be commutative and let ab = a for every a e A, b e L.

Then C is the group of homomorphisms f: L -* A, with the composition law given by (fg-1) (a) = f(a)g(a)F1 or (f - g) (a) = f(a) - g(a) if A is written additively. If L and A are written additively we have

(7.21) Sf = af + 1; E, = fa + 1

(7.22) fg = fag + f + g,

where 1 in (7.21) stands for the identity automorphism both of L and of A. Also E = 0. If L is a free Abelian group, freely generated by al, * *, as, then we may writef = (ai, , *, ak), where a; = f(ai). If g(bi) = 0i, for any g e C, thenf Dg = (yl, . Y ,7k), where

k

es = as + 3i + > nijai, i=1

if ads = E lc a ni1ai.

8. A special case of 7r2(Q, P)

We now continue from where we left off at the end of ?6. We first take n = 2 and P = 81 U ... U So, where Si, * * , Sk are 1-spheres with a single common point, po , no two of which meet anywhere else. Since 1r2(P) = 0 it follows without difficulty that each arc component of B is simply connected. Therefore Lemma 2 is valid if c is any arc component of B. It follows that the multiplication xay, which is given by (4.1), is defined for any x, y C yd(A, B), provided only that x(1) e G. In particular it is defined for any pair of elements x, y C 'y1(A, B, e). We shall apply the results of ?7 to this situation, taking L = -ir(P), A = r2(Q, P), 0: A -- L to be the boundary homorphism and the operators a __+a to be those discussed21 in ?1. The purpose of this section is to prove the following theorem.

THEOREM 9. There is a homomorphism, h: F -4 r, of the group F, defined in ?7, on the group r, defined in ?6, such that

a) ThL(f) = >f for each f E F, where Tx means the same as in ?5, b) h1(H fn r) = E, h(ea) = ct(0a) e i fn I for any ea e E, where i1: 'ri(P) II

means the same as in Theorem 7,

21 The relation cig = O(a) a is satisfied if a,B C ir2(Q, P) without any restriction on P and Q. For a suitably chosen map representing a,$ can be deformed into one representing f#a by a deformation in which the base point describes a circuit representing 8a. Writing ,3 = 1 we have a6a = a and hence (7.2).




c) h-1(1) is the group Eo C E, which consists of all the elements ea such that aa = 1.

To prove this we introduce a new group AP, which is defined in the same way as r, except that

1) we only admit paths such that (t) (po) = po for each t e I; and 2) we only admit deformations of paths which, in addition to (s(O), (,(1) C B,

satisfy the condition ,8(t) (po) = po for every s, t C I. It is easily verified that this restriction may be imposed throughout ??2 to 5 without invalidating any of the arguments.22 I say that 1 is isomorphic to F. For let t be a path, which represents an element x e A, and assume that t(1) = i. Let X: (I, 0, 1) -(P (P, po, po) be a map representing a given element a E L and let ,g: 12 Q be the map, which is given by ,(s, t) = ( - s)X(t). Then ,(s, 0) = ,4(s, 1) = (1-s)(po) = po ,(O, t) = X(t) eP, .(1, t) = (O)X(t) E P. There- fore A(I2) C P, ,u(O, 0) = Po, whence 4 represents an element f(a) E A. The element f(a) is obviously unaltered by a deformation, s, of the path 4o =

P. FIG. 1

such that t8(0), is(1) C B, {8(t)(po) = Po. Therefore the mapf: L -> A depends only on the element x E ( and not on the choice of t e x. The map f satisfies the conditions f(ab) = f(a)f(b)a for the reason23 indicated in Fig. 1. Thereforef is a crossed homomorphism. Clearly af(a) = r(a)a-', where a -- r(a) is the endomorphism of iri(P), which is induced by the map t(0): (P, Po) -+ (P, po). Therefore r = Sf,, where Sf is defined by (7.5). Clearly Sf is an automorphism if t(0) has a homotopy inverse rel. po. The converse follows from the special nature of P. Thereforef E F, since x e ( and accordingly x(0) E G. Hence, writing f = x(x), we have defined a transformation x: -* F.

22 All we have to do is to replace A by the arc component of the identity in the sub- space of QP, which consists of all maps of the form f: (P,po) -* (Q, po).

23 It is hoped that, by the time this is printed, or soon afterwards, there will have ap- peared one or more books on homotopy groups, which will give the detailed arguments im- plicit in this and similar statements. Some of the essential arguments, in rather crude form, are to be found in J. H. C. WHITEHEAD, Proc. Lond. Math. Soc., 45 (1939), pp. 281, 282 and Annals of Math., 42 (1941), 409-28.



630 J. H. C. WHTEHEAD

The transformation x is a homomorphism. For let x, y C cl and let t E , , E y be such that t(1) = -(1) = i. Then tr(O)q e xy and it is clear from Fig. 2 that x(xy)(a) = fIS,(a) }g(a) = (f g)(a), wheref = x(x), 9 = x (y). Therefore x(xy) = x(x) x(y), whence x is a homomorphism.

The homomorphism x is onto. For let f: 7ri(P) -> 7r2(Q, P) be a given element of F. Let free generators a,, , ak, of 7r1(P) be represented by maps X1, Nk, , where Xr is a map X,: (1, 0, 1) -* (S1 , , pO) such that Xr I (0, 1) is a homeomorphism of the open interval (0, 1) on S1- Po (r = 1, , k). Let f(ar) be represented by a map24 1r: 1I2 -+ Q, such that Aur(O, t) = Mr(t),

IAr(S, 0) -ir(S, 1) = Po- Let a(s): (P. po) -- (Q, Po) be the map defined by t(S)(p) = St -s, Xr T(p) } if p e S1, for each r = 1, *, 1. Since Ar(S, 0) = 7r(s, 1) Po this is a single-valued map, which is continuous2" in (p, s). Also

t(O)(P) = j1r.l1 Xjr(p)} e P, M(l)(p) = Ar{O, XrT(P)} = Xr{Xi(p)} = P(p 81r), t(s)(po) = gr(l - s, 0) U AL(l - S,1) = Po U Po = Po. Therefore the path t represents an element iny7l(A, B, e) and t(s)(po) = po. The map t(O): P -* P

FIG. 2

maps S1 on the circuit XA: I P, where Xr(t) = r(1, t). Also af(ar) = aa 1 where a' e iri(P) is the element represented by Xr . Therefore ?r = af(ar)ar =

S1(a,) and it follows that the map t(O) induces the endomorphism Sf. Since f e F it is an automorphism and it follows from the special nature of P that t(O) has a homotopy inverse rel. Po. Therefore the path t represents an element x e k. Obviously f = x(x), whence x is onto.

24 In general, if E1 is any segment on P2, then any element a E7r2(Q, P) can be represented by a map M:I2 Q, such that p I El is a given map u':E1 -* P, provided M'(0, 0) = pe if (0, 0) e E'.

2a Here and in similar definitions, we rely on the following theorem. Let O: X -+ Y be a map of a space X onto a space Y, which has the identification topology associated with the map 4,. That is to say, 4,(Xo) is closed if Xo C X is any 'saturated' closed set, i.e. one such that, if +-l(y) n Xo $ 0 then 4+-(y) C Xo for every y e Y. Then the theorem states that, given a mapf: X - Z, of X in any space Z, the transformationfo-1: Y-- Z is continuous if it is single-valued. This is a corollary of a theorem proved by N. BOURBAKI (Livre III Topologie G6n6rale, Chap. 1, p. 53, Theorem 1). If X and hence Y = +(X) are compact, Y being a Hausdorff space, then Y has the identification topology associated with the map o. For any closed set XO C X is compact. Therefore 4(Xo) C Y is compact, and hence closed, whether Xo is saturated or not.




I say that x is an isomorphism. For let x(x) (ar) = 1 for each r = 1, , A, and let Aur: 12 -> Q be given by iir(S, t) = o(1 - S)Xr(t), where 4o e x and Xr means the same as in the preceding paragraph. Then Our represents the element x(x) (ar) e 7r2(Q, P). Since Xr(X) (ar) 1 there is a homotopy Aur(O < U < 1), rel. 12 X of AUOr = Ar into a map .,Lr 12 - P. The deformation Qu , given by (u(s)(p) = 4ur I - s, Xr'(p)} if p e S, carries Pointo a path, 4j, in e. Also u is rel. 0 U 1 and tu(s) (Po) = Po since /cur is rel. 12 and 11r(S, 0) = Ar(S, 1) = Po.

Therefore tj e x, whence x = 1 and x is an isomorphism. There is a natural homomorphism h,: 1 -, r, in which corresponding elements

are represented by the same path. We define h: F -, r as the homomorphism h = hlx-'.

We now dispose of Theorem 9 (a). Let a given element a e 7r2(Q, P) be represented by a map go :j2 __ Q, such that go(O, t) = go(s, 0) = go(s, 1) = po . Let f e F be given and let h(f) be represented by a path i, such that t(1) = i, i(s) (po) = po. Let gu: 12 - Q be the homotopy, which is given by

(S t) go(2- ) if 0 < 2s < 2- u

==t(3-2s-u)go(1, t) if 2-u ? 2s < 2.

Since go(O, t) = go(s, 0) = go(s, 1) = pO, t(s)(po) = po we have

gu(0, t) = 90(0, t) =p = (1 -u)go(O, t)

gu(S, 0) = Po = g(1 -u)qo(s, 0)

qu(s, 1) = Po = t(1-u)go(s, 1)

g7u(II t) = -)90(11 0).

Therefore gu l2 is the deformation t(1 -u)g', where g' = go i2 In other words, gu is an extension to 12 of the deformation g' = t(1 - u)g'. Since t = h(f) it follows that the map gl represents Th(f) (a). But

gl(s, t) = go(2s, t) if 0 ? 2s ? 1

= t(2-2s)go(1, t) if 1 < 2s < 2.

Now the map X: I -> P, which is given by X(t) = go(l, t) represents aa, whence A: 12 Q, given by g(s, t) = t(1 - s)go(l, t), represents f(aa). Therefore gi represents f(a) = Ef(a) (see Fig. 3). Therefore Th(f) (a) = Ef(a), which disposes of (a).

Before starting the proof of (b) we note that, in defining the isomorphism x: - F, we need not have restricted ourselves to paths, t, such that t(1) = i. In proving (b) we shall consider paths, (, such that t(1) is a map in which a part, containing Po, of each Srl(r = 1, ***, k) is shrunk into Po. More precisely, let Xr: I -> mean the same as before, Xr I (0, 1) being a homeomorphism on S1 - po, and let a: I -I be given by a(t) = 0 if 0 4t < 1, a(t) = (4t - 1)/2 if 1 < 4t ? 3, a(t) = 1 if 3 < 4t _ 4. Then t(1) shall be the map p: P > P,




which is given by p(p) = XArX'r(p) if p e Sr . It is obvious that t(1) e e and also that ar e 7rj(P) is represented by the map XrcT: I -* P.

We now prove (b), first showing that h(eO) e II f r for any a T Or2(Q, P). Let a be represented by a map g: 12 Q such that g(O, t) = g(s, 0) = g(l, t) = po.

po o( ' f(boc)

FIG. 3

FIG. 4

Then, as indicated in Fig. 4, aa-ar is represented by any map, Ar: 12 Q, such that

,Yr(s, t) = g(s, 4t) if 0 < 4t ? 1

C P if 1 <4t<3 (8.1)

= g(s, 4-4t) if 3 4t < 4

.r((O t) = XrO(t) for each t E I.

Since ,Ar(S, 0) =Mr(S, 1) po a path, {, is defined by

(8.2) 0 (s)(P) -r s, X'(P)} if p e(

and o(1)(p) = ,.{OXT'(p)}= XrTXrT'(p) = p(p). Therefore o(l) = pee. Since /r(l, t) 6 P, Ar(l, 0) = ,r(l, 1) = po it follows that 4o(O) e B. Also to(s)(po) = Ar l - s, X-'(po)} = po, since lir(l - 8,0) = lir(l - s, 1) = po. Therefore to




represents an element x' e ). It follows from (8.2) that gr(s t) = W - S)Xr(t) and hence that ea = x(x'). Let x = hi(x') = h(ea). Then x also is represented by the path to. I say that a homotopy, tu, is defined by

u(S)(P) = - Lr{1 - s,(1 - Iu)X7'(p) + 1u} if p E S. For Xir (p) is single-valued except when p = Po , and XrT'(po) = 0 U 1. Therefore in order to prove that tu(s) (p) is single-valued, and hence continuous in (p, s, U), it is sufficient to prove that ur(l - s, u/4) = Ar(1 - S, 1 - u/4). But this follows at once from (8.1). Therefore (u is a homotopy of the path to. Since /Lr(I2) C P we have (u(O)(P), tu(1)(P) C P and we also have,

u(OP) (p)= u/4) U Mr(l, 1 - u/4) =po

u(lO) ()= ur(O u/4) U ArA(O, 1 - u/4) = po.

Therefore tj, as well as to, represents the element x = h(ea). If 0 < t < 1 then 1/4 ? t/2 + 1/4 < 3/4, whence

(8.3) ( (S)(p) = Ar - s, -Xr (p) + 1/4} e P if p E Sr Since 6 E h(e,) it follows that h(ea) e II. Therefore

h(E) c II nlr or E C h-'(lii n ).

On comparing (8.3) and (8.1) we see that

WS() (pO) = Mr(l - s, 1/4) U Yr(l - s, 3/4) = g(l - s, 1).

Since g(O, t) = g(s, 0) = g(l, t) = po it follows that Oa is represented by the map 9': I -> P, which is given by g'(s) = g(l - s, 1). Therefore aa is represented by the track of Po in the deformation j (s), whence h(e,) = (aOa), where 41 means the same as in Theorem 7.

We now complete (b) by proving that h '(II nr) C E. Let x e II n r and f e h-'(x) be given. Let x' = x '(f) e b. Then x = h(f) = hlx-'(f) = h1(x'). Therefore x and x' are both represented by a path, to, auch that to(s) (Po) = Po, to(l)= i. Since x e II there is a homotopy, , , with tu(O), tu(1) C B, such that t1(s)(P) C P. Let q(s, t) = tl-t(l - s)(po). Then g(12) C P, g(0, 0) = po, whence the map g: 12 > Q represents an element a E 7r2(Q, P). Also g(s, 1) = to(- s) (po) = Po and g(0, t) = po since tu(1) e B. Let a given element a E 7r1(P) be represented by a map X: (I, 0, 1) -- (P, Po, Po). Since f = x(x') it follows from the original definition of x that f(a) E 7r2(Q, P) is represented by the map ,u: 12 -_ Q, which is given by L(s, t) = to(1 - s)X(t). The element af(a)a-a is represented by the map Ao : 12 -_ Q, which is given by

/Lo(s, t) = g(s, 4t) = tj -(1 -s)(po) if 0 < 4t < 1

=,(s, 2t- 1/2) = to(l-s)X(2t- 1/2) if 1 < 4t 3

= g(s, 4- 4t) = 4t-3(1 - 8)(po) if 3 < 4t < 4.




Let P,' be the homotopy, which is defined by

,zut(s, t) = ~1s(l -S)(po) if 0 ? 4t ? 1 - 'a

= t s (1 ) X t if 1 - ( + <-4t _ 3 + u = ~~ - 2 F 'it/

= 44t3(1 -S)(po) if 3 + u < 4t < 4.

It may be verified that ,IA(I2) C P and q,,(O, 0) = Po, whence /'l e af(a)a-a. But ,/1(s, t) = 41(1 - s)X(t) and since X(t). C P, 4(1 -s)(P) C P it follows that Al(I2) C P. Therefore af(a)a-a = 1 or f(a) = ,#-a withf = a-1. Therefore f e E, whence h-'(H l r) C E. Therefore h-l(f nr) = E.

We now prove (c). Since h(ea) =t'(aa) it follows that h(ea) = I if aa = 1. Therefore Eo C h-1(1). Conversely let h(f) = 1 and let {u mean the same as in the preceding paragraph, with x = h(f) = 1. Since h(f) = 1 we may assume that (u is a deformation of 4o into the constant path, given by {l(t) = i. In particular {1(t) (Po) = PO . We proved that f ea-I, where a is represented by the- map g: -2 Q, which is given by g(s, t) = t1-,(1 - s)(po). Since {u(O), ~~(1) C B, to(s)(po) = po, as before, and since we now have {1(s)(po) = PO, it follows that g(P2) = PO . Therefore aa = 1, whence f e Eo. Therefore h-1(1) = E0 and the proof is complete.

Let [Jo = H/H n f and ro = r/Hf n r. Since H and r are invariant in He it follows that the factor group Hr/H f r, and hence the factor group He/H ln if the natural homomorphism 4: 7ri(P) -*> 7r(Q) is onto, is the direct product H1o X Fo . Therefore, since h(E) = H n r and Th(f) = Ef according to Theorem 9, and in consequence of (7.17), we have the corollary:

COROLLARY. The homomorphism x -* Tx, of He into the group of automorphisms of 7r2(Q, P), determines, in a natural way, a homomorphism of Ho X ro into the factor group of the group of auttomorphisms by the group of inner automorphisms.

Since h(E) = H n r, h-'(1) C E, it follows that h determines an isomorphism of F/E onto Fo.

9. A special case of Hn(Q, P)

Let n > 2 and let P = Sl-7 U * * U * where S'1, * , S are (n-1)- spheres with the single common point, PO, no two meeting anywhere else. Let irj(Q) = 1 and let the natural homomorphism 02: 72(p) -72(Q) be onto, so that 72(Q) = 0 if n > 3. Then, as when n = 2, we take L = 7r,-(P) A = rn(Q, P) in ?7, with a special reference to the final paragraph of ?7. We also take a' = a for each a e L, a e A.

THEOREM 10. There is an isomorphism h: F -- H., of the group F, defined in ?7, onto IH,, such that

Th(f) = Ef for each f E F, where T. means the same as in ?5.




Since r'(Q) = 1 and since P is a polyhedron it follows from Theorem 6(c) that He = IIr. Since 7r,(P) 1, because n - 1 > 1, it follows from Theorem 7 that HI = 1, whence H. = r. Therefore each element x e H. is represented by a path I, such that t(t) (po) po . Let to, ti be two such paths, both of which represent the same element x e He. Let t. be a homotopy such that t8(O)X t,(1) C B and let go: 12 _ Q be given by go(s, t) = t8(t) (po). Since ti(t) (po) = (j) (po) = po (j = 0, 1) it follows that go(t2) = po. Therefore go represents an element a e 7r2(Q). Since 4)2: 72(P) -> r2(Q) is onto there is a deformation g., rel. 12 of go into a map gi 1 I> p. It follows from an extension argument26 of the type used in proving Lemma 6, but with an extra dimension, that the homotopy, t8 can be modified to give a homotopy, t', of t" = to into

i= 0, such that t'(O), X'(1) C B, t'(t (po) E P. Therefore we may restrict the paths, which represent elements of He, to paths, I, such that t(t)(po) = po and any two such paths, which represent the same element of He, are related by a homotopy, ' , such that t'(0), t'(1) C B, tt(t)(po) e P.

We now define a transformation h-1: H. -* F in almost exactly the same way as we defined x: --* F in proving Theorem 9. The fact that h-1(x) depends only on x e H. , and not on a particular choice of t e x, follows from the fact that any two paths o , 4, C x, with to(t) (po) = {l(t) (po) -po , are related by a homotopy, I , such that t,(O), {s(1) C B, t8(t)(po) e P. The rest of the theorem, namely that h-' is an isomorphism onto F and that Th(f) = Ef , follows from arguments which are essentially the same as those used in proving the analogous parts of Theorem 9.

10. Another special case

By way of a final example we consider the case in which P = Sn-1, an (n-i)- sphere, and is contractible in Q, without assuming that iri(Q) = 1. Since Sn-1 bounds a cell in Q it is contractible into po, with the latter held fixed. That is to say, there is a path, p, in A, which joins the constant map, p(O): S n-1 - Po to the identity, p(l) = i, and is such that p(t)(po) = po. Also the space B is contractible in A. For the homotopy X1 B -* A, which is given by Xt(b) = p(l - t)b for each map b e B, deforms the identity, X0: B -* B, into the constant map

: B - p(O)b = p(O). It follows that any closed curve in any arc component of B is contractible in A and hence that Lemma 5 is valid if c is any arc component of B. Let H' be the sub-group of He , whose elements are represented by paths with both end points in e. Let t be such a path and let 77o be a path in e, which joins i to t(0) and tq a path in e, which joins t(1) to i. Let tq , al be any other paths in e, which join i to t(0) and t(1) to i, respectively. Then

/t /_ /-1( t -1

77~l o7-o000000i77

since the closed curves t7o'77 and 711X77 are contractible in A. It follows that

26 See also ?11 below.




704X71 and 7o0X771 represent the same element of iri(A, i). It easily follows from this that H' is isomorphic to iri(A, i), and hence to the Abe27 group K"(Q). An isomorphism Hf -e K,(Q) is defined by making x e H" correspond to the element of Kn(Q), which is represented by the path ptp-', where t e x, t(O) = t(1) = i and p means the same as before.

It follows from (4.1) that x -+ x(O)(x e He) is a homomorphism of He on G and Hf is obviously its kernel. Since P = S'-1 the group G consists of those arc components of B, whose elements are maps (S' 1, po) (S"-1, Po) of degree i 1. Therefore G is of order 2. Let r: (S-1, Po) -- (S"1, po) be a homeomorphism of degree -1 such that r2 = i. Let -q = p1rp. That is to say, 77 is the path given by

=(t) p(l - 2t)r if O _ 2t < 1

p(2t-1) if 1 _ 2t < 2.

Then -q(O) = r, 7(1) = i, whence -q represents an element y e H. and G consists of the elements e and y(O). The element y- 1 is represented by the path28

{p-1(pr)JI r

= (p'1r) (pr2)

-1 = p rp

= H1*

Therefore y-1 = y, or =2 = 1, whence He is a trivial extension29 of H" by the group G.

As proved by Abe, the group Kn(Q), and hence H', is a trivial extension of 7r,(Q) by wi(Q). To simplify the notation we identify the (multiplicative) groups irn (Q) and ri(Q) with the corresponding sub-groups of H". Then any element x e H" has a unique expression of the form x = ao, with a e 7r,(Q) o a e ri(Q). Let g: (I"l, In) -* (Sn', po) be a fixed map such that g I (In-1 n-l ) is a homeomorphism on Sn-i - Po. Then a is represented by a path of the form p-1ip, where t(O)(S'-') = t(1)(Sn-') = U(t)(po) = Po and a, regarded as an element of Tn(Q), is represented by the map h: I' -* Q, which is given by h(tjI... , tn) = (ti)g(t2, ... , tn). The element a- is represented by a path, p-1 p, such that t(t) (Sn-1) = (t), where ,u: (I, 0, 1) -* (Q, Po X Po) is a map repre-

27 M. ABE, Japanese Journal of Math., 16 (1939), 169-76. 28 N.B. n = (tf) n,(t)-1 = 'l1,(f) l = t-lf,(tn)f = (tf)(nf), where , are any paths in

A and f e B. N.B. also that r-1 = r. 29 A group H is called an extension of a group X by a group Y if there is a homomorphism,

f of H onto Y whose kernel is isomorphic to X. It is a trivial extension if there is a representative of each co-set of this kernel, such that the aggregate of these representatives is a sub-group of H. That is to say it is a trivial extension if, and only if, there is an isomorphism g: Y -. H (into) such thatfg = 1.




senting a- e irl(Q). As before, let y e H. be represented by the path t = p- rp and let x = ao- e H' be represented by p-itp, where t(0) = t(1) = p(O). Then yxy-', or yxy since y2 = 1, is represented by the path

I Vrp)((pip)}r(p'rp) (p'rtp)r(PVrp)

p7'(tr) (pr) (p-r)p

P'(tr)P-

Therefore it is obvious from the above description of a and a. that yay'1 = a-1 -1 yo-y = a or

(10.1) yao-y-1 = a-lo.

It is well known that the path p determines a homomorphism, X: 17r_,-(Sn') -

irn(Q, Sn'), such that ax = 1, which is defined as follows. If 13 e 7r~(Sn') is represented by the map f: (In-l, In-l) (Sn-', Po) then X(13) is represented by the map k: In -* Q, where k(t, , , tr) = p(tl)f(t2, * , t). It follows from a standard argument that30

7n (Q, Sn-) a- 1(l) X XBy (S n-1).

It is also well known3' that a-1(1) = 4/{i7rn(Q)I, 41-1(1) = 44{ir.(Sn-1)}, where VI: 7rn(Q) -_ '7r.(Q, Sn-1) and 4: 7nr(Sn-) -*, xr(Q) are the homomorphisms in which corresponding elements are represented by the same map. Since Sn-l is contractible in Q we have {I rn(Sn-1) } -1, whence J/ is an isomorphism. Also X is an isomorphism, since X(O) = 1 implies,8 = AX(B) = 1.

The group H. operates on 7rn(Q, Sn-1) as follows. Let acoy' (e = 0 or 1) be an arbitrary element of H. , where a e rTn(Q), a- e irl(P). Let i, (13)X () (m = 0, i4 1, *- - ) be an arbitrary element of irn(Q, Sn-), where 13 e rnr(Q) and y is the

generator of 7n-1(Sn-1) which is represented by the above map g: (In-l, In-l) (Sn-l, po). The element i1 (1) e 7rn(Q, P) is represented by a map f: (In, I) ) (Q, po). The elements a, a- C He are represented by paths p-'lp and p-'t'p, such that t(O)(Sn-l) = {(1)(Sn-l) = t(t)(po) = po and {'(t)(S -) = (t), where ,u: (I, 0, 1) -- (Q, po , po) represents a-. Also p(t) (po) = po, whence (p-l'p) (t) (po) = po. Hence it is not difficult to verify that

Ta{i,1(13)} = T(I3), T{(,y)} = aX (y),

T,4(13)} I= T(o, ) TI{X(y)} = (,y)

where 13U1 means the same as a-1,1 in ?1, the group irn(Q) now being multiplicative.

30 This result is valid for an arbitrary, contractible P C Q even if n = 2 and 72(Q, P) is non-commutative. For in this case it follows from (7.2) that a-1(1) is in the centre of zr2(Q, P).

31 See for example W. HuREWIcz and N. STEENROD, Proc. Nat. Acad. of Sciences, 27 (1941) 60-64 or J. H. C. WHITEHEAD, Proc. Lond. Math. Soc., 48 (1944), pp. 281 et seq.




Also

TV{+(d3)} = /(j), To {X(7y)} = x(Y1

since the effect of the deformation (p-'rp)(1 - t): S4' -* Q is to replace the cell swept out by Sa'- in the deformation p(t): S"' -- Q by the cell swept out in the deformation p(t)r: Sn' -- Q. Thus we have

Taa{ifr(,3)X(wym) } = (XB-);() (10.3) Ta{ift'(3)X(ym)=

TV q1(#)X(1 )1 } = (o)Xy- )

11. Proof of Lemma 1

The proof depends on the following straightforward generalization of the extension argument used in proving Lemma 6. Let K be a polyhedron and O: K - Q a given map. Let L C K be a sub-polyhedron and At a homotopy

of AtO such that AO L = Al i L. Let Xt = At I L and let the homotopy Xt be deformable, rel. Xo, Xi, into the constant homotopy, Xt , given by X = Xo = Xi. That is to say, let there be a 2-parameter family of maps, XAt L Q, such that XoG = Xt X t = X/ = Xo, Xsl = X8o = Xo . Then there is a homotopy, j4 :K- Q, of A = AO intoA = A , which is rel. L. For letA K: X I- Q be given by ,u(x,t) = ,t(x), for each xe K. Let M = (K X 0) U (L X I) U (K X 1) and let 4s M Q(O _ s < 1) be the deformation of )o = AJ .11, which is given by

+8(X, j) = (Xj) f X eK, j= O, 1

48(yt) =X8t(y) if y eL, t e I

The homotopy 48 can be extended to a deformation VI,8 K X I Q and jA,

given by ,A'x) = J1(x, t) is the desired homotopy. Now let I, i' be two paths such that (0) = i'()e B, t(1) = i'(1) = i _

Let a e rn(Q, P) and o Ea be given and let g0 = fo I In. Let gt = ( - t)go, g'(t) = -t) go and let ft , ft be extensions32 of gt, g' to I'. Let At be the deformation of tio = fi into ,ul = f, which is given by

At = Af-2t if O < 2t _ 1 = f2 t-1 if I _ 2t < 2.

Then Xt = Pt is given by

Xt = 91-2t = t(2t)go if O < 2t ? 1

- g2t-i = '(2- 2t)go if I < 2t ? 2.

Since (j) = ('(j) (j = 0, 1) it follows that Xo = Xi , and since t _ {' it follows that Xt is deformable, rel. Xo, Xi, into the constant homotopy X = Xo . There-

32 ff , fmay be different extensions of gt in case =t




fore it follows from the preliminary result, with K = In, L = In, that f' is homotopic, rel. In, tofi. Therefore!fi and f, represent the same element ta E irn(Q, P). This disposes of (b) and (c).

We now prove (a). Let foo I fio C a, let gjo _ fjo I In (j = 0 1) and let fit

be an extension to In of gjt = {(1 - t)gjo. We have to prove that foi and fil represent the same element ta E xr,(Q, P).

Since foo I fio C a there is a homotopy fso (In, InP ao) - > (Q, PI po). Let g8o = fso I In and let g9t = {(1 - t)go . Then g9l is a homotopy of go, into gil = fi I Pn. Let fl(fl= foi) be an extension of g8l throughout I n. Since g9o(In) C P, g8o(ao) = Po and since t(O) E B, gs, = (O)so , it follows that g81(In) C P, g81(ao) = po. Therefore fil and fol = fol represent the same element of 7rn(Q, P). The maps fil and fil are obtained from foo by extensions of the two deformations

goo - o10 -->911 and goo-> gol --91.

i,'

FIG. 5

Each of these is deformable into the other, rel. goo, Igi, by means of the 2-parameter family g9t = t(1 - t)g80 . More explicitly, the first is deformed into the second by the homotopy XA , which is given by

X8(t) = g2(1-s)t,2st if 0 < 2t < 1

- gl-2s(1-t),l-2(1-s)(1-1) if 1 ? 2t ? 2.

It follows from the preliminary result that fil is deformable into fil, rel. In. Therefore fi and fJl , and hence fi and fo, represent the same element {a E irn(Q, P). This completes the proof.

12. Proof of Theorem 2

We have to prove that x(a, + a2) = xal + xa2 for any elements x E 71(A, B, e), a, I a2 C ir,(Q, P). Let En-l be the part of In in which either ti = 0 or at least one of t2X tn is either 0 or 1. Let E n be the half of In in which 0 < 2tn ? 1 and En the half in which 1 < 2tn < 2. Let a, and a2 be represented by maps fi and f2 , such that fj(En') = Po(j = 1, 2) fi(E2) = f2(E ) = Po. Then




al + a2 is represented by the map ho , which is given by ho = fj in E' (j = 1, 2). Letgo = fj I (In U EA') (j, k = 1, 2; k 7 j). That is to say, go isfj , restricted to the union of In and the half of I' in which fj is constant. Let t e x be a path such that {(1) = i, let g't = (1- t)go and let fjt be an extension of get to I'. Then fjt is also an extension to I' of the deformation gjt = (l - t)gjo, where gjo = fj . Therefore fji e xaj . On the other hand, since

fit(E l n En) = f2t(El n E n) = -(j -t)(po),

a deformation he: In Q is defined by taking he = fit in E n The deformation he is an extension to In of gt = {(j - t)go, where go = ho In. Therefore hi E X(ai + 2). But hi = fji in En andfj1 e xa. Alsofj1(En1) =(O)fj(En-1) = t(0) (po) = po. Therefore hi e xal + xa2 , whence x(ai + 2) = xal + xa2.

MAGDALEN COLLEGE OXFORD



on operators in relative homotopy groups

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