Re: subset of contractible

On 11 Lip, 12:21, Sonya84 <sonianard...@xxxxxxxxx> wrote:
On 11 Lug, 12:06, Zbigniew Karno <Zbigniew.Ka...@xxxxxxxxx> wrote:





On 11 Lip, 11:54, Sonya84 <sonianard...@xxxxxxxxx> wrote:

On 11 Lug, 11:41, Sonya84 <sonianard...@xxxxxxxxx> wrote:

Suppose X is a contractible topological space
and S subset of X is a topological subspace of X.

Is S contractible itself?

Since X is contractible we have that X is pathwise-connected, so -
without loss of generality - we can suppose X is homotopically
equivalent to a point p in S.

Sonia

since X is contractible to p, by definition there exists an homotopy
h: X x [0, 1] --> X such that h(*, 0) = id_X while h(*,1) = c_p
(c_p denotes the constant which maps every x in X to p in X).

So h|S: S x [0, 1] --> X is such that h(*, 0) = id_S while h(*,1) =
c_p.

The problem is that it ISN'T true that - in general - for every t
in ]0, 1[ and for every x in S, h(x, t) belongs to S.

So, why S need to be contractible if such is X ?

This means that S is conractible in X,
however S is not contractible itself.

In the case S = S^n, does not exist a homotopy
F : S x [0,1] --> S joining id_S with a constant
map.

Regards, Z. Karno- Nascondi testo tra virgolette -

- Mostra testo tra virgolette -

Consider the tangent bundle TM
to an infinite dimensional Hilbert manifold M
(modelled on an infinite dimensional Hilbert space H).

The structure group of the fiber bundle TM is a subgroup of GL(H).

By a theorem of Kuiper GL(H) is contractible.

I would like to show that TM is trivial, i.e., TM is diffeomoprhic to
M x H.- Ukryj cytowany tekst -

- Poka cytowany tekst -

Have you any reasonable connection between
the contractibility of the structure group
of TM and the triviality of TM ?

Regards, Z. Karno

.

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