TITLE: Isotopic homeomorphisms of surface induces same map on the space of ends
QUESTION [2 upvotes]: Let $\Sigma$ be a non-compact orientable connected two-manifold without boundary. Let $f,g\colon \Sigma\to \Sigma$ be two homeomorphisms. Suppose there is a homotopy $H\colon \Sigma\times [0,1]\to \Sigma$ from $f$ to $g$ such that $H(-,t)\colon \Sigma\to \Sigma$ is a homeomorphism for each $t\in [0,1]$.
Now, $f,g$ induce maps $\mathcal E(f),\mathcal E(g)\colon \mathcal E(\Sigma)\to \mathcal E(\Sigma)$, where $\mathcal E(\Sigma)$ denotes the space of ends of $\Sigma$.

Is it true that $\mathcal E(f)=\mathcal E(g)$?
If the above question has a false answer, then do I need to consider
some restriction, like $\pi_1(\Sigma)$ is (in)finitely-generated?

Of course, if $H$  itself is a proper map, then $\mathcal E(f)=\mathcal E(g)$, for any $\Sigma$.

REPLY [3 votes]: Yes, the induced maps $\mathcal{E}(f)$ and $\mathcal{E}(g)$ are equal.  This is because two isotopic transversely oriented separating circles in $\Sigma$ determine the same subset of $\mathcal{E}(\Sigma)$ and because such subsets give a basis for the topology of $\mathcal{E}(\Sigma)$.
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