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

Question

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.

Could someone help me, please?

Answer 1

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