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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 $\Sigma$.

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    $\begingroup$ Isn't an isotopy a proper map? $\endgroup$
    – Fernando Muro
    Commented May 27, 2022 at 19:56
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    $\begingroup$ @Random - those maps are not isotopic because one preserves orientation and the other does not. $\endgroup$
    – Sam Nead
    Commented May 28, 2022 at 16:53
  • $\begingroup$ My bad! I forgot the word orientation-preserving: If two orientation-preserving homeomorphisms of a surface are homotopic relative to the boundary, then they are isotopic relative to the boundary. $\endgroup$
    – Random
    Commented May 28, 2022 at 17:28

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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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    $\begingroup$ @Random - those maps are homotopic, but not isotopic. $\endgroup$
    – Sam Nead
    Commented May 28, 2022 at 16:58
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    $\begingroup$ My bad! I forgot the word orientation-preserving: If two orientation-preserving homeomorphisms of a surface are homotopic relative to the boundary, then they are isotopic relative to the boundary. $\endgroup$
    – Random
    Commented May 28, 2022 at 17:28
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    $\begingroup$ I have a question? So, an isotopy (defined as a homotopy through homeomorphisms) between surfaces may not be proper but should fix ends. Am I right? $\endgroup$
    – Random
    Commented May 29, 2022 at 13:20
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    $\begingroup$ @Random - Also, IMHO you should not delete your comments once the thread has grown around them. (Yes, they were wrong, but in a non-trivial way. Learning is happening!) $\endgroup$
    – Sam Nead
    Commented May 29, 2022 at 16:49
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    $\begingroup$ I got this paper recently ams.org/journals/proc/1963-014-04/S0002-9939-1963-0181989-2/… Also, I won't delete comments in future $\endgroup$
    – Random
    Commented May 29, 2022 at 16:50

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