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Suppose that a space $X$ deformation retracts to a point $x_0\in X$. I want to prove that for any open neighbourhood $U$ of $x_0$, there exists an open neighbourhood $V$ of $x_0$ with $V\subset U$, such that the inclusion of $V$ in $U$ is based homotopic to the constant map. Any ideas?

Mathmo
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  • Hint: Restrict the homotopy which gives the def. retract to $V$. – Ajay Kumar Nair Aug 25 '20 at 11:29
  • I don't see how that ensures that the restriction is based with base point $x_0$? – Mathmo Aug 25 '20 at 12:19
  • If $F$ is the homotopy, isn't $F(x_0,t) = x_0 \forall t$? – Ajay Kumar Nair Aug 25 '20 at 12:21
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    I think all we can say is that $x\rightarrow x_0$ is homotopic to the identity map on the whole of $X$ (the terminology I'm used to would call the deformation retraction "strong" if we required the homotopy to keep $x_0$ fixed). – Mathmo Aug 25 '20 at 12:26
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    Do you mean that $x_0$ is a strong deformation retract of $X$? Being a deformation retract is weaker (see https://en.wikipedia.org/wiki/Retraction_(topology)) and does not allow to prove what you want. – Paul Frost Aug 25 '20 at 15:29
  • Interesting that you say the strong property is needed. Is there a reference or counterexample showing that the stronger condition is needed? The question came from a textbook exercise, which I believe is using the weaker condition, so I would like to understand this more. – Mathmo Aug 25 '20 at 17:27
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    This will help. https://math.stackexchange.com/q/126342/336894 – Ajay Kumar Nair Aug 26 '20 at 05:18
  • Thanks for the link! – Mathmo Aug 26 '20 at 16:26

2 Answers2

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Assume that $F:X\times I\rightarrow X$ is a contraction of a space $X$ relative to a point $\ast\in X$. i.e. $F$ is a continuous map with $F(x,0)=x$ and $F(x,1)=\ast$ for all $x\in X$ and $F(\ast,t)=\ast$ for all $t\in I$.

Claim: For any neighbourhood $U$ of $\ast$ there is a neighbourhood $V$ of $\ast$ with $U\subseteq V$ such that $U$ is contractible in $V$ relative to $\ast$.

Proof: Since $F(\ast\times I)=\ast\in U$ we have $\ast\times I\subseteq F^{-1}(U)$. Note that $F^{-1}(U)$ is open since $F$ is continuous. Using the compactness of $I$ we appeal to the tube lemma to find an open set $V\subset X$ with $\ast\times I\subseteq V\times I\subseteq F^{-1}(U)$. Then $F(V\times I)\subseteq U$, so $F|_{V\times I}$ is the claimed contraction of $U$ in $V$ relative to $\ast$.

Tyrone
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This is a supplement to Tyrone's answer. That $A$ is a deformation retract of $X$ means that there exists a homotopy $H : X \times I \to X$ such that $H(x,0) =x$, $H(x,1) \in A$ and $H(a,1) =a$ for all $x \in X, a \in A$. If $A = \{x_0\}$, this says nothing else than $X$ is contractible to the point $x_0$ (which is equivalent to $X$ contractible to any point).

Now let $X$ be the comb space as in Is Armstrong saying that the comb space is not contractible?

In my answer to this question you can see that $X$ is not pointed contractible to the point $p$. The proof can easily modified to show that for $U = X \cap (I \times (0,1])$ there does not exist an open neighborhood $V$ of $p$ as required in your question.

Paul Frost
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