LUSTERNIK-SCHNIRELMANN CATEGORY IN HOMOTOPY THEORY
9
(4.3)
where only
91
and hm are allowed to be e. We then call m the length of x; and we
prove, by induction on m, that every element of
j- 1 (~0)
belongs to F(S).
First let m
=
1. Then if
X
E
r 1 ~G
and
X
=
g1h
11
,
jx
=
(g, h), g
=
h, so
XEs.
Assume now that all elements of
r 1 (~G)
of length
m
belong to F(S),
and let
X
be as in (4.3),
X
E
r 1 (~G).
Set
= (
I
")-1x(hl h"
)-1
= ((
l/)-1h ")
I
h "(
I
hi
-1)
Y
9191 m m 91 1 9z m-1
9m
m ·
It is then plain that
y- 1
is an element of
j-
1
(~0)
of length
m-
1, so that
y- 1
E
F(S). From this it immediately follows that x E F(S), so that, finally, F(S)
=
r 1 (~G)
and
r 1 (~G)
is hence free. 0
We now revert to the
PROOF OF THEOREM 4.2. Let LS-catX
=
2 with structure map p: X
----+
X V X. Since j p ':::'
~:
X
----+
X x X, it follows that there exists a homomorphism
p: G----+ G*G, with G
=
1r1X, such that the diagram
G GxG
ij
G*G
commutes. But then p is one-one since
~
is one-one, and p embeds
G
as a subgroup
of
j- 1 (~0).
But a subgroup of a free group is free. 0
5. Modifications of the definition
In this section we confine ourselves to 2 definitions which, respectively, weaken
and strengthen the notion of LS-category.
5.1. The weak version.
Let q: xn
----+
xn) be the projection of xn onto
the smashed product of n copies of X (thus xn)
=
xn jTn(X)). We then say that
wcatX ~ n if q~ ':::'
0:
X----+ xnl.
Obviously, wcat
X
~
n if cat
X
~
n. Obviously, too, the conclusion of Theorem 4.1
remains valid if we weaken the hypotheses by only requiring that wcat X
~
n. It
thus remains to show, by an example, that wcat X
~
n is a strictly weaker condition
than cat X
~
n. We will, in fact, exhibit a space X such that wcat X = 2, but
cat
X~
3.
Let M be a Poincare 3-sphere, suitably triangulated. For definiteness take
1r1M
=As.
Form
X
by removing the interior of a 3-simplex
s
of
M.
Obviously we
still have
1r1
X
=As,
H
1
X
=
0. We claim that HnX
=
0,
n
~
1; the only cases he
have to consider are n
=
2, 3.
Let z be a 2-cycle of
X,
thus of
M.
Then z
=
8c, where cis a 3-chain of
M.
Now M, triangulated, is a 3-cycle, so that z
=
8(c+kM), for any integer k. Plainly
we may choose k so that s appears in c
+
kM with coefficient 0. Then c
+
kM is a
3-chain of X, so that z is a 2-boundary of X and H
2
X
=
0.
The removal of s ensures that no non-zero 3-chain of X can be a cycle. Thus
H3X
=
0.
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