# Is the image of a second-countable topological space $X$ a second-countable space?

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.

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.
This is not true in general, let $$Y$$ be any space which is not second countable and let $$f$$ be the inclusion $$X\hookrightarrow X\sqcup Y$$.
• Okay, so it is true only if $f$ is surjective, right? – Antonio Maria Di Mauro Feb 12 '20 at 15:25
• yes, I'm not sure if there's any other reasonable property of $f$ that implies the image is second countable – Alessandro Codenotti Feb 12 '20 at 15:26
• @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 – Alessandro Codenotti Feb 12 '20 at 15:53
• 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" – Alessandro Codenotti Feb 12 '20 at 16:07