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I've read here and there an I am not sure about this

Wikipedia does not explicity states second countability is a topological invariant, nor the majority of sources I 've searched through do, but I found here,

http://mathonline.wikidot.com/second-countability-under-homeomorphisms-on-topological-spac

if I am not mistaken, that it is.

Is this correct?

some_math_guy
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3 Answers3

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Of course. A homeomorphism maps open sets to open sets.

  • why then is it not listed as a topological invariant along with connectedness, compactness, etc? – some_math_guy Feb 15 '20 at 18:58
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    Whatever list of topological invariants you are referring to will necessarily be incomplete, because there are A LOT of topological invariants. – Lee Mosher Feb 15 '20 at 19:07
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If $f: X \to Y$ is a homeomorphism and $\mathcal{B}$ is a base for $X$, then $f[\mathcal{B}] = \{f[B]: B \in \mathcal{B}\}$ is a base for $Y$ (as $f$ is open and continuous) and vice versa, if $\mathcal{B}'$ is a base for $Y$, $\{f^{-1}[B]: B \in \mathcal{B}'\}$ is a base for $X$ for similar reasons.

So if one of $X$ or $Y$ has a countable base, so has the other. So it's clearly one of the very many topological invariants that we know.

Henno Brandsma
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  • My interest on this is that I was trying to find a simple reason for which a topological manifold is required to have second-countability(at least by most authors), because if it wasn't required, then the functions that constitute the charts, which are required to be homeomorphisms, would be mapping a topological manifold that is not second-countable with $\mathbb R^n$, which is second-countable, and that couldn't be, since second-countability is a topological invariant. Please tell me if I am wrong – some_math_guy Feb 15 '20 at 23:37
  • I am refering to the question you answered here. https://math.stackexchange.com/questions/3547427/why-is-a-topological-manifold-defined-with-a-countable-basis/3547618?noredirect=1#comment7295399_3547618 – some_math_guy Feb 15 '20 at 23:39
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    @juancarlosvegaoliver No, a space can be "locally second countable" (as the charts show) but not second countable as a whole, see the long line example. Being globally second countable, you can prove a theorem that in fact the manifold can be embedded as a subspace in some $\Bbb R^N$, and this corresponds to many people's intuition more. – Henno Brandsma Feb 16 '20 at 07:06
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If $X$ and $Y$ are homeomorphic spaces and $X$ is second countable, then $Y$ is second countable.

This is because: If $X$ is second countable and $f:X\rightarrow Y$ is a continuous open and surjective map, then $Y$ is second countable.

Here is a proof:

Assume that $\mathbb{B}$ is a countable basis on $X$. Clearly, $\{$ $f(B)$ $:$ $B\in \mathbb{B}$ $\}$ is countable. Therefore we must verify that this new set, call it $f(\mathbb{B})$, is a basis on $Y$. As $f$ is an open map, $f(B)$ is open for each $B\in \mathbb{B}$. All is left to show is that every open subset of $Y$ is the union of some collection of members of $f(\mathbb{B})$. To this end, let $V$ be an open subset of $Y$. If $v\in V$ then there exists some $x\in f^{-1}(V)$ such that $f(x)=v$. Since $f$ is continuous, $f^{-1}(V)$ is open in $X$. Therefore there exists some $B\in \mathbb{B}$ for which, $x\in B\subseteq f^{-1}(V)$. Therefore, $f(x)=v\in f(B)\subseteq V$.

M Khaled Bin-Lateef
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