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Showing that an inclusion is null homotopic

I asked this question here a while ago. Meanwhile I've come up with the following and was wondering if you could have a look and tell me if it's correct.

claim: $X$ homotopy equivalent to a point $\ast$ $\implies$ for all neighbourhoods $U$ of $\ast$ there exists a neighbourhood $V$ of $\ast$ such that the inclusion $i: V \hookrightarrow U$ is null homotopic.


Let $h_t :id_X \simeq c$ where $c(x) = \ast$ denote the homotopy.

Let $U$ be a neighbourhood of $\ast$. This means there exists an open set $O$ such that $\ast \in O \subset U$.

claim: $i: O \hookrightarrow U$ is null homotopic.

proof: restrict $h_0$ to $O$: $h_0 |_O$ then $i = i \circ h_0|_O \simeq i \circ c = c$.

Thanks for your help.


marked as duplicate by Willie Wong Aug 29 '11 at 12:51

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ What is preventing $O$ from having multiple connected components? $\endgroup$ – wckronholm Aug 26 '11 at 16:01
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    $\begingroup$ The problem with your proof is that the homotopy $h_t$ might move $O$ outside of $O$, and so you cannot make the claim that $i\circ h_0|O \simeq i\circ c$ as maps $O \to U$. $\endgroup$ – wckronholm Aug 26 '11 at 16:04
  • $\begingroup$ ooh, I see, thank you! $\endgroup$ – Rudy the Reindeer Aug 26 '11 at 18:13
  • $\begingroup$ @Matt: I am closing this as exact duplicate. For future reference, if you want to bring attention to old problems, you can edit the old question text with any new ideas you have had during the intervening months. This would bump it again to the front page. $\endgroup$ – Willie Wong Aug 29 '11 at 12:50

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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