I'm trying to do exercise 5 on page 18 in Hatcher:
Show that if a space $X$ deformation retracts to a point $x \in X$, then for each neighborhood $U$ of $x$ in $X$ there exists a neighborhood $V \subset U$ of $x$ such that the inclusion map $V \rightarrow U$ is nullhomotopic.
My question:
Does the existence part of the neighbourhood $V$ follow from taking $V=U$, i.e. there is at least one neighbourhood in $U$($U$ itself)? Or does it have to be a proper neighbourhood? If it has to be a proper neighbourhood: is the idea to retract $U$ "a bit", just enough to get a new neighbourhood $V \subset U$?
Thanks for any hints, I appreciate your help!
Edit
Here is what I've done using Matt E's help:
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$U$ neighbourhood $\implies$ $\exists$ open set $\tilde{U}$ such that $x \in \tilde{U} \subset U$.
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$id_x \simeq const.$ $\implies$ $\exists h_t : X \times [0,1] \rightarrow X$ continuous
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$h_t^{-1}(\tilde{U})$ open, $[0,1]$ compact $\implies$ tube lemma applies
$\implies \exists$ open set $O$ such that $\{ x \} \times [0,1] \subset O \times [0,1] \subset h_t^{-1}(\tilde{U}) \times [0,1]$ -
$\implies$ $V := O$ is a neighbourhood of $x$ s.t. $x \in V \subset U$
How do I show that $i : V \rightarrow U$ is nullhomotopic? Thanks for your help.
Best Answer
I'm writing up an answer for this because it is still the first question you find looking into this exercise from Hatcher.
Let $h_t$ be the deformation retract from $X$ to $x$ with $h_0=id_X$, $h_1(x')=x$ for any $x'\in X$ and $h_t(x)=x$ for any $t\in I=[0,1]$. Then call $H$ the map from $X\times I \rightarrow X$ where $H(x,t)=h_t(x)$.
Consider $H^{-1}(U)$ in $X\times I$. This set contains $\{x\}\times I$ because $h_t(x)=x\in U$ for any $t$. By the Tube Lemma there is an open set $V \in X$ such that $\{x\}\times I \subseteq V\times I\subseteq H^{-1}(U)$. We can see that $x\in V\subseteq U$ because $h_0$ is the identity map. This means that $V$ is an open neighborhood of $x$. Additionally $h_t$ restricted to $V$ has an image that is contained in $U$ because $V\times I \subseteq H^{-1}(U)$. Therefore $V$ is a neighborhood of $x$ where the inclusion map of $V\rightarrow U$ is nullhomotopic.