Lemma 1.15. Let R!, \p be as in 1.14. Let X denote a subcomplex of R!
which is the union of some of the b.(R'). Then i p acts trivially on the
Z(-)-homology groups
integers by adding
Z(-)-homolog y group of X. Here Z(—) is the ring obtained from the
Proof of 1.15: The proof uses double induction. By 1.13j, and induction,
ty acts trivially on the Z(-)-homology of any of the following spaces.
(a) b.(R'), for any b.(R') contained in R!.
(b) bi(Rl) f l Rf. _1, for any b^R') in R! but not in R'._r
(c) X n R!^.
Let {b.(Rf)|i€l} denote the subcomplexes of X which are as in (b)
above. So X = (XflR! -,) U ( U b.(R')). Proceeding inductively over the
3~L i€l X !
index set I, we assume \p acts trivially on the Z (--)-homology groups of
X ^ (XnRl -. ) U C U b.(R'))
3 i€l,
where I denotes the smallest £ integers of the index set I. Note that
(c) above shows this is true for I - 0. Consider the Mayer-Vietoris
sequence for the triple (X , ,X ,b. (R')) , where I.'JU {j} = L+i. This
is a long exact sequence of Z(-)-homology groups
3 . 3
- HqObj(R')) - Hq(X£)
Hq(bj(R')) - Hq(X£+1) +
K B. Cn
q q q
By induction on £ (this is the second induction hypothesis), \p acts
trivially on H (X ) for all q. Then by (a), (b) above, i| ; must act
trivially on A ,B . Let t: C C be a generator of the action by i p on
C . Then 3(t(a)-a) = 0 for any a C . THen, by exactness of the
sequence above, t(a) = a + y, where y is in the image (B +C ). So t(y) =y.
Then computing:
a = tn(a) + ny * n-y = 0 * y = 0.
This shows ^ acts trivially on C above.
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