# Showing that an inclusion is null homotopic

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:

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.

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Matt: I don't mean this condescendingly, but you really should sit down and draw some pictures before asking all these questions. Why should an arbitrary neighborhood $U$ of $x$ be contractible in itself? It need not even be connected... – t.b. Mar 10 '11 at 12:04
@Theo: I think you are right. I assumed that a neighbourhood had to be connected, why, I don't know. I'm not so comfortable with pictures, I like symbols and "implies" arrows. I wonder how I can get used to algebraic topology : / – Rudy the Reindeer Mar 10 '11 at 12:09
I can understand that to some extent. Let me just mention that even if a neighborhood is connected, it need not be null-homotopic, you can still have "holes" in it (and far worse pathologies). Algebraic topology is a great place where you can sharpen your intuition by drawing pictures and you'll get confidence in your pictures by making the pictures into formal arguments. One reason why Hatcher's book is so great is that it tries to convey this message to the reader. In my mind, algebraic topology without pictures is like food without salt. – t.b. Mar 10 '11 at 12:18
@t.b. It's funny. Reading this I barely recognise my old self from 4 years ago. – Rudy the Reindeer Jul 22 '14 at 15:49
@RudytheReindeer Did you ever prove this one? I'm having trouble with it. – morphic Sep 28 at 1:31

Thanks for the hint. I used it to construct $V$. Now I'm stuck with the last step, i.e. that the inclusion is nullhomotopic. – Rudy the Reindeer Jun 20 '11 at 13:30