FINITENESS OF SUBGROUPS OF SELF-HOMOTOPY EQUIVALENCES 11 Proof=?: Let v1 E Vo be a basis element. Then g(vj)- f(vj) = 0 and dHi(vj) + Hid(v1) = Hd(v1) = H(v1) = 0 by Lemma 4.3. Thus Now let Wk E vl be a basis element. Then g(wk) = Ha(wk) = H(wk + r(wk) + Zk) by Lemma 3.2, where Zk E (Vo) = f(wk) + H(id(wk) + di(wk)) since H(zk) = 0 by Lemma 4.3 = f(wk) + dHi(wk) + Hid(wk)· ~: If Vj E Vo is a basis element, Ha(v1) = H(vj + Vj) = f(vj) + Hdi(v1) = f(vj) + dHi(vj) + Hid(vj) = g(vj)· Note that g( Vj) = f( Vj) + dHi( Vj) so that H( Vj) = 0. If Wk E vl is a basis element, then Thus H ends at g. with Zk E (Vo) = f(wk) + H(di(wk) + id(wk)) since H(zk) = 0 = f(wk) + dHi(wk) + Hid(wk) = g(wk)· 0 4.5 Corollary With the hypothesis of the previous proposition, if H is a homotopy from f to g, then 0 We now show by example that the converse of Proposition 3.3(ii) is false. We give examples of maps whose obstruction to being homotopic to the identity { does not vanish, but which are homotopic to {. 4.6 Example Let M = A(v2n-l,v2n,W4n-l) with subscripts denoting degrees, d(v2n-d = 0 = d(v2n) and d(w4n-l) = v~n· Note that M is the minimal model of S 2n-l X S2n. Then M is 2-stage with Vo generated by V2n-l, V2n and vl
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