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 $t \in I$.

  • 2
    $\begingroup$ There's a counterexample in the exercises to chapter 0 in Hatcher. $\endgroup$ – Angina Seng Sep 11 '18 at 16:54
  • 1
    $\begingroup$ 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

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.

| cite | improve this answer | |
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.