Is the inverse of an isotopy of embeddings continuous?

Question

I was dealing with isotopies and monodromies when the following question arose, which is more subtle than it might seem at first sight.

Let $X$ and $Y$ be topological spaces. Suppose that $\Phi: [0,1] \times Y \to X$ be an isotopy, i.e. a continuous map such that for every $t \in [0,1]$, the map $\Phi_t := \Phi|_{\{t\} \times Y}: Y \to X$ is an embedding of topological spaces.

My question is the following:

Under which conditions the map $$\{(t,x) \in [0,1] \times X \ | \ x \in \mathrm{image} \ \Phi_t\} \longrightarrow Y; \quad (t,x) \longmapsto \Phi_t^{-1}(x)$$ is continuous?

Of course, if the embeddings $\Phi_t$ are all homeomorphisms, then Aren's theorem together with the exponential law solves the problem. Recall that Aren's theorem asserts that if $T$ is a Hausdorff, locally connected and locally compact topological space, then the group of homeomorphisms $T \to T$ is a topological group with the compact-open topology.

I suspect that something similar should be used for the more general question stated above, but I don't know how to solve it...

Does anyone have any clue? Any help will be highly appreciated!

Answer 1

You can consider the map $\hat{\Phi}\colon Y\times I\rightarrow X\times I,\,(y,t)\mapsto(\Phi_t(y),t)$, which is sometimes called the track of the isotopy. It is clearly a continuous injection. Your question is equivalent to asking whether the inverse map $\hat{\Phi}^{-1}\colon\mathrm{im}(\hat{\Phi})\rightarrow Y\times I$ is continuous (nothing interesting happens in the second coordinate), i.e. whether $\hat{\Phi}$ is a topological embedding. This makes it obvious that $Y$ being a compact and $X$ a Hausdorff space is a sufficient condition. More generally, we could assume $\Phi$ is proper (pre-images of compact subsets are compact), $Y$ is Hausdorff and $X$ is locally compact Hausdorff, for that implies that $\hat{\Phi}$ is also proper and then closed.

I don't think there is much hope beyond these cases. Here is a typical counter-example. Let $$F:(0,\infty)\times I\rightarrow(0,\infty),\,(x,t)\mapsto\arctan(x)+t(x−\arctan(x)).$$ This yields an isotopy $\Phi\colon((0,\infty)\sqcup(0,\infty))\times I\rightarrow\mathbb{R}^2$ given on the first component by $(x,t)\mapsto(F(x,t),t)$ and on the second component by $(x,t)\mapsto(\pi-F(x,t),-t)$. This is an isotopy, yet $\hat{\Phi}^{-1}$ is not continuous. Indeed, $(1/2,-t,t)\rightarrow(1/2,0,0)$ as $t\rightarrow0$, yet the preimages are in the first component for $t=0$ and in the second component for $t>0$. You can also modify this to produce a connected example.