# Is the inverse of an isotopy of embeddings continuous?

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!

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.