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 $\Sigma$.



Isotopic homeomorphisms of surface induces same map on the space of ends
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 $\mathcal{E}(\Sigma)$.


