Is there $(X,\mathcal{T})$ contractible, $\mathcal{T}'\subseteq\mathcal{T}$ such that $(X,\mathcal{T}')$ isn't contractible?

Question

Let $(X,\mathcal{T})$ be a contractible space and let $\mathcal{T}'\subseteq\mathcal{T}$ be a coarser topology on $X$ than $\mathcal{T}$. Is it always true that $(X,\mathcal{T}')$ is also contractible?

I highly doubt that this is the case, because intuitively, if it were true, then it should work with the same homotopies for $\mathcal{T}'$ as for $\mathcal{T}$ because of the generality of the statement (I mean, given the generality, we cannot adapt the homotopy equivalences for $(X,\mathcal{T})$ in a way that makes sense), but there is absolutely no reason for $F:X\times I\to X$ to still be continuous if we equip $X$ with $\mathcal{T}'$ instead of $\mathcal{T}$.

However, I fail to come up with a counterexample; the eight topologies I know over $\mathbb{R}$ don't provide one. Can you give an elementary example where this fails?

Answer 1

Stefan beat me to the (same) answer, but I think it's worth having a different explanation about the contractibility, using the homotopy theory of finite spaces.

---

Playing around with finite spaces, I think I've found an example. Unfortunately, I don't think it occurs "in practice".

Let $X = \{a, b, c, d\}$ be a four point set. Let $\mathcal{T}$ be the topology generated by the basis $\{a\}$, $\{b\}$, $\{a, b, c\}$, and $\{b, d\}$, and let $\mathcal{T}'$ be the topology generated by the basis $\{a\}$, $\{b\}$, $\{a, b, c\}$, and $\{a, b, d\}$. It is easy to verify that $\mathcal{T}' \subsetneq \mathcal{T}$.

I claim that $(X, \mathcal{T})$ is contractible. There is a classification of the homotopy type of finite spaces by their cores. See chapter 2 of Peter May's finite spaces book for an explanation of this and the terminology I am going to employ. Anyway, for $(X, \mathcal{T})$, the point $d$ is a downbeat point, witnessed by $b$. So there is a deformation retract of $(X, \mathcal{T})$ onto $(\{a,b,c\}, \mathcal{T}|_{\{a,b,c\}})$. But this latter space is the (non-Hausdorff) cone on $S^0 = \{a,b\}$. So it is contractible. Another way to see this is that $c$ is a maximum element of the associated poset.

On the other hand, $(X, \mathcal{T}')$ is the (non-Hausdorff) suspension on $S^0 = \{a,b\}$, which is weakly homotopy equivalent to $S^1$. So it is not contractible. Alternatively, note that $(X, \mathcal{T}')$ is minimal finite, so is its own core. Since its core is not a point, $(X, \mathcal{T}')$ is not contractible.

Answer 2

I think I have found an example of two topologies on a space $X$ such that $X$ with the finer topology is contractible, but with the coarser topology it is not. This example was inspired by the pseudocircle $P$, which is the set $X = \{a,b,c,d\}$ with the topology $$ \mathcal T = \{ X, \{a\},\{b\},\{a,b\}, \{a,b,c\}, \{a,b,d\} \} $$ This space can be obtained from the circle $S^1$ by identifying the open sets $\{(x,y)\mid x<0\}$ and $\{(x,y)\mid x>0\}$ to points $a$ and $b$, respectively. Since $P$ is covered by the open sets $\{a,b,c\}$ and $\{a,b,d\}$, choosing $a$ and $b$ as base-points allows us to apply the Seifert van Kampen theorem for groupoids, from which we see that $P$ has fundamental group isomorphic to $\mathbb Z$ (see for example Topology and Groupoids by Ronald Brown).

Starting from the pseudocircle, I tried to make the topology slightly finer, and thus obtain a contractible space. The following modification of $\cal T$ seemed promising $$ \mathcal T' = \mathcal T \cup \{\{b,d\}\} $$ The intuition was to cut $P$ between $a$ and $d$, so that it becomes what we could call a "pseudointerval". You can see it as the outer vertices of the following image

with the yellow sets forming a basis for $\cal T'$. The picture also illustrates how a retracting homotopy $H$ to $c$ looks like: Every edge from an outer point $p$ to the point in the center represents the subspace $\{p\} \times I$ of $X \times I$, and the color shows the value that $H$ assumes. As you can see, $H$ is fixed on the point $c$, while every other point moves along a path towards $c$, with $d$ passing through $b$ on its way. If I didn't miss anything, this should be deformation retraction of our space to $c$.