Showing that an inclusion is null homotopic

Question

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!

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Edit Here is what I've done using Matt E's help:

1. $U$ neighbourhood $\implies$ $\exists$ open set $\tilde{U}$ such that $x \in \tilde{U} \subset U$.

2. $id_x \simeq const.$ $\implies$ $\exists h_t : X \times [0,1] \rightarrow X$ continuous

3. $h_t^{-1}(\tilde{U})$ open, $[0,1]$ compact $\implies$ tube lemma applies $\implies \exists$ open set $O$ such that $\{ x \} \times [0,1] \subset O \times [0,1] \subset h_t^{-1}(\tilde{U}) \times [0,1]$

4. $\implies$ $V := O$ 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.

Answer 1

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.

Answer 2

I don't want to give a complete answer here, and I second Theo Buehler's advice in the comments above. But I will give a hint: the definition of deformation retract involves continuous maps. And in a question in topology involving continuous maps, if you have to produce new open sets from old, this is typically done by invoking the definition of continuity in terms of open sets.

If you are not naturally visual in your thinking, it will take time to learn how to think topologically/geometrically, and to learn how to translate visual intuitions into the formal language of spaces and continuous maps --- but it is well worth the effort and time!

Answer 3

This is how I think about it intuitively, but it might be wrong, please leave a comment if this is wrong.

If $U$ is contractible, let $V$ be $U$, then we are done. But there might be some holes inside $U$ that make $U$ not contractible. However, these holes can be overcome by the deformation retraction of $X$ to $x$ since $X$ is contractible. So after the deformation retraction of $X$ has overcome these holes, take the intersection with $U$ and let it be $V$, then $V$ is contractible.