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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 : / –  Matt N. 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
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2 Answers 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!

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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. –  Matt N. Jun 20 '11 at 13:30
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up vote 3 down vote accepted

I think I finally found a proof:

Let $f: X \times [0,1] \rightarrow X$ denote the homotopy from $id_X$ to $c$ where $c(x) = x_0 \forall x \in X$ and let $U$ denote an open neighbourhood of $x_0$. Then $f^{-1}(U) = \tilde{U} \times [0,1]$ for some open set $\tilde{U}$ containing $x_0$.

Now one needs what is called the "tube lemma": If $X,Y$ are topological spaces and $Y$ is compact and $N$ is open in $X \times Y$ such that $\{ x_0\} \times Y \subset N$ then there exists an open set $O$ such that $\{ x_0\} \times Y \subset O \times Y \subset N$.

By this there exists an open set $O$ such that $\{ x_0 \} \times [0,1] \subset O \times [0,1] \subset \tilde{U} \times [0,1]$. Now restrict the domain of $f$ to $O \times [0,1]$ and the range to $U$. Then $f_0 = id_X$ is the inclusion of $O$ into $U$. $f_1 = c$ implies that the inclusion $i: O \hookrightarrow U$ is null-homotopic.

Is there a proof without the tube lemma?

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why f^{-1}(U) = \tilde{U} \times [0,1] i don't understand why its [0,1] why not any other open subset of [0,1]. –  rmznyzgyr 13 hours ago
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