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We say that $X$ is contractible if it is homotopy equivalent to a singleton set.

Let $A \subseteq X$. $A$ is a retract of $X$ if there exists a continuous map $r:X \rightarrow A$ such that $r(a) = a \; \forall a \in A$.

A retract $A$ is a deformation retract of $X$ if there exists a retraction $r$ such that $i \circ r \simeq Id_X \; \text{rel} \; A$.

This is the same as saying that $A$ is a deformation retract if there exists a continuous map $F:X \times I \rightarrow X$ such that $F(x,0) = r(x)$, $F(x,1) = x \; \forall x \in X$ and $F(a,t) = a$ for all $a \in A, t \in I$.

I am trying to prove the following:

Let $X$ be a topological space. $X$ is contractible if and only if it deformation retracts to a point.

I think that I recall this being true, but I can't find the exact statement, and I'm now having my doubts. A counter-example to the above would also be a satisfactory answer.

The backward implication seems fine: If $X$ deformation retracts to a point, then we immediately have that the associated retract $r$ is a homotopy equivalence. We therefore get that $X$ is homotopy equivalent to a singleton set, and is contractible.

For the other implication, if $X$ is contractible, we have that there exist (continuous) maps $f:X \rightarrow \{1\}$ and $g:\{1\} \rightarrow X$ such that $f \circ g \simeq Id_X$ and $g \circ f \simeq Id_{\{1\}}$. This is where I get a bit stuck.

I think that we have that $g(1)$ is fixed by $f \circ g$. We have that $f \circ g:X \rightarrow X$ is continuous, so it's a retract onto $\{g(1)\}$. I think I'm nearly there now, but I can't see how to show that $F(a,t) = a$ for all $t \in I$.

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    $\begingroup$ There's a counterexample in the exercises to chapter 0 in Hatcher. $\endgroup$ – Lord Shark the Unknown Sep 11 '18 at 16:54
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    $\begingroup$ Not exactly a duplicate, but your question is indirectly answered here (this is what the above comment references). $\endgroup$ – Aloizio Macedo Sep 11 '18 at 17:17
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The trick to understanding why this is false is that if $r:X \times I \to X$ is a deformation retract to $pt$ then this implies that for every neighborhood $U$ of $pt$, there exists an open set $V$ containing $pt$ where the inclusion $V \hookrightarrow U$ is nullhomotopic.

This can be seen by the tube lemma. In particular, note that $\{pt\} \times I \subset r^{-1}(U) $ by definition. Applying the tube lemma gives a neighborhood that $V \times I$ is mapped into $U$ by $r$. Showing this is contractible is just restricting the map we already have.

Now, the trick is to come up with a space that is contractible, but not locally connected (or locally contractible) anywhere.

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  • $\begingroup$ To add to Andres's answer, for such an example consider the comb space (en.wikipedia.org/wiki/Comb_space). It is contractible. Moreover it is pointed contractible if given the basepoint $(0,0)$. However, when given the basepoint $(0,1)$ it is not pointed contractible: the neighbourhood $U$ of $(0,1)$ does not exist, since the space is not locally connected. $\endgroup$ – Tyrone Sep 12 '18 at 9:40

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