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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