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Possible Duplicate:
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.

proof:

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.

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    What is preventing $O$ from having multiple connected components?2011-08-26
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    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 $O \to U$.2011-08-26
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    ooh, I see, thank you!2011-08-26
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    @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.2011-08-29

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