How do I prove that all smooth manifolds being homotopic implies contractible?

125 Views Asked by At

Let $X$ and $Y$ be smooth manifolds. If all smooth maps from $Y$ to $X$ are homotopic, then show that the identity map on $X$ is homotopic to some constant map(i.e that $X$ is contractible).

I have already proved the converse of this. The strategy there was to show that any arbitrary smooth map from $Y$ to $X$ are homotopic to the constant map, hence they are all homotopic to each other as homotopy is an equivalence relation. But, what strategy is to be used to prove the converse? Any hints/suggestions would be appreciated?

1

There are 1 best solutions below

0
On

This is not true, take $Y$ to be a point, all maps $Y\rightarrow X$ are homotopic if $X$ is (path) connected, but $X$ is not necessarily contractible.