- #1

Bacle

- 662

- 1

I have not found a clear explanation for this:

Are any two contractible subspaces A,B<X homotopy-equivalent to each other?

Clearly, homotopy equivalence is an equivalence relation. BUT: what if

A<X is H-E (Homotopy-Equiv.) to a point p in X, while B is H-E to a point q

p=/q ? (like in the case where X is disconnected, and A,B are in different

components of X ) . If we had p=q, then the answer is trivially yes; we compose

a homotopy of A with p , with the reverse homotopy of B with p.

Also: when we say that the identity map id:X-->X from any topological space

to itself is not nullhomotopic, i.e., homotopic to a point. Is this always equivalent

to saying that X is not contractible? I have a proof saying yes, but one part

of the proof is kind of shaky.

Hope this is not too dumb.