Munkres topology 30.12 part (A) Let $f:X \rightarrow Y$ be a continuous open map. Show if $X$ is first countable, then $f(X)$ is first countable.
Let $\{B_n  \ | \ n \in \mathbb{N}\}$ be a countable neighborhood basis for an $x \in X$. Let $V$ be an open set containing $f(x) \in f(X)$. Since $f$ is continuous, $f^{-1}(V)$ is open in $X$ and there is a basis element $B_n$ containing $x$ with $x \in B_n \subset f^{-1}(V)$. Then since $f$ is open, $f(x) \in f(B_n)\subset f(f^{-1}(V))\subset V$. So $\{f(B_n) \ | \ n \in \mathbb{N}\}$ satifies the condition for being a neighborhood basis for $f(x) \in f(X)$ and since $f(x)$ was arbitrary, $f(X)$ is first countable.
I was reading this proof from this source Second Countability Proof, and noticed the proof to be considerably more difficult to understand than my attempt, which makes me wary of my own effort. What is the fault in the proof of my attempt? Upon looking at problem from source it seem that in this case $f(f^{-1}(V))=V$ should be in the third line, but why?
 A: Your argument is along the right lines, but you’ve slid over a couple of important details. First, it’s $f[X]$ that you want to show is first countable, so you have to work in that subspace of $Y$, not in $Y$ itself: if $f$ isn’t surjective, $Y$ may not be first countable. Thus, properly speaking you should start with an arbitrary $p\in f[X]$ and show that it has a countable nbhd base in $f[X]$, not in $Y$.
Since $p\in f[X]$, there is an $x\in X$ such that $p=f(x)$, and since $X$ is first countable, there is a countable nbhd base $\{B_n:n\in\Bbb N\}$ at $x$. For each $n\in\Bbb N$ let $V_n=f[B_n]$, and let $\mathscr{V}=\{V_n:n\in\Bbb N\}$; it’s true that $\mathscr{V}$ is a local base at $p$ in $f[X]$, but it takes just a little more work to prove it than you actually did.
Let $U$ be an open nbhd of $p$ in $f[X]$; we need to show that $V_n\subseteq U$ for some $n\in\Bbb N$. Since $U$ is open in $f[X]$, there is an open $W$ in $Y$ such that $U=W\cap f[X]$. $X$ is an open subset of $X$, and $f$ is an open map, so $f[X]$ is open in $Y$, and therefore $U$ is actually open in $Y$. Thus, $f^{-1}[U]$ is an open nbhd of $x$ in $X$, and there is an $n\in\Bbb N$ such that $B_n\subseteq f^{-1}[U]$. Then $V_n=f[B_n]\subseteq f\big[f^{-1}[U]\big]=U$, as desired, $\mathscr{V}$ is a local base at $p$, and $f[X]$ is first countable.
