# Homotopy of homeomorphisms implies homeomorphism $X \times [0,1] \to X \times [0,1]$?

Let $h_0$ and $h_1$ be self-homeomorphisms of a topological space $X$. Let us say that $h_0$ is homotopic to $h_1$, and write $h_0 \sim h_1$ if there exists a 1-parameter family $h_t$, $t \in [0,1]$ of self-homeomorphisms of $X$ such that $(x,t) \mapsto h_t(x) :X \times [0,1] \to X$ is continuous. This is just the usual definition of homotopy, except that I am requiring the intermediate maps to also be homeomorphisms. It is natural to want the following thing, which would be true, for example, if we had succeeded in topologizing $\mathrm{Homeo}(X)$ so as to get a topological group in which path-equivalence coincides with homotopy equivalence.

Hoped for statement: $h_0 \sim h_1$ implies $h_0^{-1} \sim h_1^{-1}$.

My first idea to approach this was to consider $H:X \times [0,1] \to X \times [0,1]$ given by $H(x,t) = (h_t(x),t)$. This is a continuous bijection restricting to a homeomorphism $X \times \{t\} \to X \times \{t\}$ for each $t$. If $H$ were a homeomorphism, one could use the inverse mapping $H^{-1} : X \times [0,1] \to X \times [0,1]$ to obtain the desired homotopy. However, I do not see any reason for $H$ to be an open map. So, my question for you is:

Question: Is $H$ a homeomorphism?

Note that the hoped for statement does not actually turn on the question having a positive answer. Indeed, my friend noticed a cute argument which circumvents the issue. Simply consider the map $X \times [0,1] \to X$ defined as the composition $$X \times [0,1] \overset{h_0^{-1} \times \mathrm{id}}{\longrightarrow} X \times [0,1] \overset{(x,t) \mapsto h_t(x)}{\longrightarrow} X \overset{h_1^{-1}}{\longrightarrow} X.$$ This gives a homotopy from $h_1^{-1}$ to $h_0^{-1}$ through homeomorphisms, witnessing $h_0^{-1} \sim h_1^{-1}$.

• Doesn't $(x,t) \mapsto h_t^{-1}(x)$ give you a homotopy $h_0^{-1} \sim h_1^{-1}$? – Clive Newstead Dec 4 '17 at 23:21
• @CliveNewstead: That's basically a restatement of the question. Continuity of your $(x,t) \mapsto h_t^{-1}(x)$ is equivalent to continuity of my $H^{-1}$. Anyway, some people might prefer this way of looking at it, so thanks for the comment. – Mike F Dec 4 '17 at 23:32
• For a nice space, the map should be continuous (for example compact Hausdorff). In general, it is not true, this reference is from the wikipedia page on the homeomorphism group: cs.vu.nl/~dijkstra/research/papers/2005compactopen.pdf – Justin Young Dec 6 '17 at 18:22
• The above only concerns the inverse map on $Homeo(X)$, the homotopy question is still not clear. – Justin Young Dec 6 '17 at 18:29
• @JustinYoung: +1, I agree with all that. For any who don't want to click the link, the reference shows that, for $X$ the Cantor set minus a point, inversion is not a continuous operation on $\mathrm{Homeo}(X)$ in the compact-open topology. However, if $X$ is a space whose path components are singletons, any continuous map $X \times [0,1] \to X$ is constant along the $[0,1]$ direction, so the answer to the Question is trivially positive. – Mike F Dec 6 '17 at 18:38