Question:
Show that if a space $X$ deformation retracts to a point $x ∈ X$, then for each neighborhood $U$ of $x$ in $X$ $\exists$ a neighborhood $V ⊂ U$ of $x$ such that the inclusion map $V \rightarrow U$ is nullhomotopic.
Could anyone lay out the main arguments and connections needed for this? I'm not quite sure what to do.
This question has been asked before, see here.
However, I'm not understanding it, if it is even correct. It also refers to a tube lemma, which I would like to avoid for the time being.
Thanks.
Best Answer
I think using the tube lemma is a pretty direct and simple way to do this. The idea is that the deformation retract $\phi : X\times I \to X$, with $\phi(y,0)=y$, $\phi(y,1)=x$ and $\phi(x,t)=x$ (for all $y \in X$, $t \in I$), almost gives you what you want, which is a homotopy $\psi : V\times I \to U$ with $\psi(y,0)=y$ and $\psi(y,1)= x$ (for all $y \in V$), just by restricting $\phi$ to $V \times I$. The only problem is that $V$ must be chosen small enough that the image of $V \times I$ is contained in $U$. And this problem is solved easily by the tube lemma, since the image of $\{x\} \times I$ lies in $U$.
Is there any reason you want to avoid the tube lemma? The proof is not difficult, by the way.