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 $h_0^{-1} \sim h_1^{-1}$.