Continuous, surjective, open map takes a basis to a basis. Prove a continuous, surjective, open map $f:X \rightarrow Y$ takes a basis to a basis.
Attempt Let $\beta=\{B_\alpha\}_{\alpha \in \lambda}$ be a basis for $X$. Let $V$ be an open set in $Y$.Let $f(x) \in V$. Then $f^{-1}(V)$ is open in $X$. Then there is a $B_\alpha \in \beta$ with $x \in B_\alpha \subset f^{-1}(V)$. So $f(x) \in f(B_\alpha) \subset V$. So $\{f(B_\alpha)\}_{\alpha \in \lambda}$ is a basis for $Y$.
Is this proof correct? If not can I know what is wrong?
 A: It’s correct as far as it goes, though it would be a little better, however, to start with an arbitrary $y\in V$ and observe that because $f$ is surjective, there is an $x\in X$ such that $f(x)=y$; the rest of the proof can then proceed as before. That way you make explicit use of the fact that $f$ is surjective, something that really is necessary.
You do still need to show that the image of $\mathscr{B}$ is not a base for a strictly finer topology on $Y$. In other words, you have to show that $\mathscr{V}\subseteq\{f[B]:B\in\mathscr{B}\}$, then $\bigcup\mathscr{V}$ is open in $Y$. This is of course a trivial consequence of the fact that $f$ is open, so that each of the sets $f[B]\in\mathscr{V}$ is open in $Y$, and therefore their union is as well.
There is an alternative approach that avoids looking at individual points altogether. Let $V$ be any open set in $Y$, and let $U=f^{-1}[V]$; $f$ is continuous, so $U$ is open in $X$, and therefore there is a family $\mathscr{U}\subseteq\mathscr{B}$ such that $U=\bigcup\mathscr{U}$. Let $B\in\mathscr{U}$; $f$ is open, so $f[B]$ is open in $Y$, and of course
$$f[U]=f\left[\bigcup\mathscr{U}\right]=\bigcup\left\{f[B]:B\in\mathscr{U}\right\}\subseteq V\,.$$
Finally, $f$ is surjective, so in fact
$$f[U]=f\left[f^{-1}[V]\right]=V\,.$$
Thus, every open set in $Y$ is a union of members of $\{f[B]:B\in\mathscr{B}\}$.
Finally, we argue as above that if $\mathscr{V}\subseteq\{f[B]:B\in\mathscr{B}\}$, then $\bigcup\mathscr{V}$ is open in $Y$, so $\{f[B]:B\in\mathscr{B}\}$ is a base for $Y$.
