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Let be $(X,\tau)$ and $(Y,\sigma)$ two topological space and let be $f:X\rightarrow Y$ a continuous and open function: if $X$ is second-countable, then is $Y$ second-countable?

First we observe that if $f$ is a surjective continuous and open funcion beewten two any topological space $X_\tau$ and $Y_\sigma$ then the image $f(\mathcal{B})$ of a basis $\mathcal{B}$ for $\tau$ is a basis for $\sigma$: infact $$ (\forall A\in\sigma\wedge \forall y\in A)\exists B\in\mathcal{B}:f^{-1}(y)\in B\subseteq f^{-1}(A)\in\tau\Rightarrow(\forall A\in\sigma\wedge \forall y\in A)\exists B\in\mathcal{B}:y\in f(B)\subseteq A\wedge f(B)\in\sigma\Rightarrow f(\mathcal{B})\quad\mathscr{is\quad a\quad basis\quad for\quad\sigma}. $$

Then we observe that for any function $\phi$ and for any set $A$ it resut that $|\phi(A)|\le|A|$: someone could demonstrate it using the Choice Axiom.

Well from this two observation we clami that if $X$ is second-conuntable and $f$ is surjective the the image $f(\mathcal{B})$ of a conutable basis $\mathcal{B}$ for $\tau$ is a countable basis for $\sigma$.

But if $f$ is not surjective, what happens?

Following a reference from the 5th chapter of "General Toplogy" by Stephen Willard.

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Could someone help me, please?

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The statement says "the image is second countable", so if $f: X \to Y$ is open and continuous, $f[X]$ must be open in $Y$ and a second countable subspace of $Y$. About $Y \setminus f[X]$ the statement says nothing. So if $f$ is surjective (often assumed implicitly for open maps, even) $Y$ is second countable.

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This is not true in general, let $Y$ be any space which is not second countable and let $f$ be the inclusion $X\hookrightarrow X\sqcup Y$.

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  • $\begingroup$ Okay, so it is true only if $f$ is surjective, right? $\endgroup$ – Antonio Maria Di Mauro Feb 12 '20 at 15:25
  • $\begingroup$ yes, I'm not sure if there's any other reasonable property of $f$ that implies the image is second countable $\endgroup$ – Alessandro Codenotti Feb 12 '20 at 15:26
  • $\begingroup$ I share what you said: however at the 5th chapter of "General Topology" by Stephen Willard it is only written that "The continuous open image of a second countable space is second countable". Who knows, maybe this is an oversight? $\endgroup$ – Antonio Maria Di Mauro Feb 12 '20 at 15:30
  • $\begingroup$ @AntonioMariaDiMauro the image of a second countable space through a continuous $f$ is second countable, because $f$ is always surjective on its image, but you were asking a different question in the body of your post $\endgroup$ – Alessandro Codenotti Feb 12 '20 at 15:53
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    $\begingroup$ If $f$ is not surjective just replace $Y$ with $f(X)$, so we can assume $f$ to be surjective without loss of generality, which is what the author is doing. Note that this is the same that happens when we say that "the image of a compact space is compact" or "the image of a connected space is connected" $\endgroup$ – Alessandro Codenotti Feb 12 '20 at 16:07

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