Showing that an inclusion is null homotopic
I asked this question here a while ago. Meanwhile I've come up with the following and was wondering if you could have a look and tell me if it's correct.
claim: $X$ homotopy equivalent to a point $\ast$ $\implies$ for all neighbourhoods $U$ of $\ast$ there exists a neighbourhood $V$ of $\ast$ such that the inclusion $i: V \hookrightarrow U$ is null homotopic.
Let $h_t :id_X \simeq c$ where $c(x) = \ast$ denote the homotopy.
Let $U$ be a neighbourhood of $\ast$. This means there exists an open set $O$ such that $\ast \in O \subset U$.
claim: $i: O \hookrightarrow U$ is null homotopic.
proof: restrict $h_0$ to $O$: $h_0 |_O$ then $i = i \circ h_0|_O \simeq i \circ c = c$.
Thanks for your help.