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.