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Question:

Show that if a space $X$ deformation retracts to a point $x ∈ X$, then for each neighborhood $U$ of $x$ in $X$ $\exists$ a neighborhood $V ⊂ U$ of $x$ such that the inclusion map $V \rightarrow U$ is nullhomotopic.

Could anyone lay out the main arguments and connections needed for this? I'm not quite sure what to do.

This question has been asked before, see here.

However, I'm not understanding it, if it is even correct. It also refers to a tube lemma, which I would like to avoid for the time being.

Thanks.

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2 Answers2

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I think using the tube lemma is a pretty direct and simple way to do this. The idea is that the deformation retract $\phi : X\times I \to X$, with $\phi(y,0)=y$, $\phi(y,1)=x$ and $\phi(x,t)=x$ (for all $y \in X$, $t \in I$), almost gives you what you want, which is a homotopy $\psi : V\times I \to U$ with $\psi(y,0)=y$ and $\psi(y,1)= x$ (for all $y \in V$), just by restricting $\phi$ to $V \times I$. The only problem is that $V$ must be chosen small enough that the image of $V \times I$ is contained in $U$. And this problem is solved easily by the tube lemma, since the image of $\{x\} \times I$ lies in $U$.

Is there any reason you want to avoid the tube lemma? The proof is not difficult, by the way.

  • Haven't covered the "tube lemma" yet, unless it is indirectly a part of other things that you would see in an intro topology course and has been given that name. It also isn't listed in Hatcher's book. Thanks for the response btw. – user42538 Feb 06 '13 at 22:21
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    It doesn't really belong to algebraic topology, so it makes sense that Hatcher doesn't cover it, but you can find it in a point-set topology book like Munkres. –  Feb 06 '13 at 23:09
  • Thanks, I'll go check the library for it. – user42538 Feb 07 '13 at 20:45
  • Hello Adeel, thanks a lot for your great answer! For somehow I am a bit stuck on the issue - why we want the image of $V \times I$ contained in $U$? Thank you! – 1LiterTears Aug 30 '13 at 16:04
  • @jellyfish: because we want a map $V \times I \to U$ (to define the homotopy). –  Aug 30 '13 at 16:29
  • Er still can't visualize it. Do we need $V \times I$ smaller or equal to $U$? That would imply $V$ has to strictly smaller than $U$? – 1LiterTears Aug 30 '13 at 16:33
  • @Jellyfish: you want to prove two maps $V \to U$ are homotopic. For this you need a homotopy $V \times I \to U$. By the definition of a deformation retract you get a map $\varphi : X \times I \to X$ and you consider the restriction onto $V \times I$. If $\varphi(V \times I)$ is contained inside $U$, then by restricting you get the map $\psi : V \times I \to U$ you wanted. If this wasn't true, you wouldn't get a map $V \times I \to U$. –  Aug 30 '13 at 17:24
  • Wouldn't we need $U$ to be compact to apply the Tube Lemma? Can someone please explain why we can apply the Tube Lemma here? – Ankit Sahu Feb 22 '24 at 10:48
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You can see a version without the Tube Lemma here that I posted.

The idea is to realize that the entire deformation retraction, DR, must eventually equal a subset of $U$. But it's more complicated since we don't what the deformation retraction is doing when we aren't looking.

The whole DR may retract to a point in the first 1/100 of a second and then stay there. Or the DR may retract to a point and then expand and then retract back to a point a 100 times, as long as it does so continuously.

So $b$ is the first time the DR goes to a point, and $a$ is some time before when the DR must enter $U$ for the last time. (If the DR leaves $U$ it must come back in eventually so we just shift $a$ forward more.)

We can then reparameterize the DR to the new time interval, which is a homotopy from the inclusion to a constant function.

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