Please check: $f:X\longrightarrow Y$ perfect map. Let $X$ be second countable, prove $Y$ is second countable

Question

Let $B=\{B_n\}_{n\in\mathbb{N}}$ be countable base for $\tau_x$ y $V=\{V_n\mid V_n=\bigcup_{i\in F} B_i\}$, $F\subset\mathbb{N}$ finite and $|V|\leq\aleph_0$. We define $W_n=T-f(X-V_n)$ y $W=\{W_n\}$ $|W|\leq\aleph_0$. Let,s see $W$ is a base for $\tau_Y$: $\forall U\in\tau_Y\;\forall y\in W \,\exists W_x\in U$ such $x=\bigcup_{n\in\mathbb{N}} B_n$. $y\in W\Longrightarrow f^{-1}(y)\in f^{-1}(W)$ and $f^{-1}(y)$ compact $\Longrightarrow$ $f^{-1}(y) \subset\bigcup_{n\in\mathbb{N}} B_n\Longrightarrow f^{-1}(y)\in V_k\in V$. Finally $f^{-1}(y)\cap (X-V_k)=\emptyset\Longrightarrow y\notin f(X-V_k)\Longrightarrow y\in Y-f(X-V_k)=W_k$.

Answer 1

You have the right idea, but your proof is just a mess of symbols and arrows and has no clear structure. Tell the proof like a story, with more explanation.

The same proof (slightly different notation) presented in my style:

Let $\{B_n: n \in \Bbb N\}$ be a countable base for $X$. Let $\mathcal{B}$ be the set of all finite unions of base sets:

$$\mathcal{B}=\{\bigcup_{n \in F} B_n \mid F \subseteq \Bbb N \text{ finite }\}$$

and note that $|\mathcal{B}|\le \aleph_0$ because $\mathbb{N}$ has countably many finite subsets.

Now define $$\mathcal{B}' = \{Y\setminus f[X\setminus B] \mid B \in \mathcal{B}\}$$

And note that this is an at most countable (as $\mathcal{B}$ is) set of open subsets of $Y$ (as $X\setminus B$ is closed and $f$ is a closed map).

We claim that $\mathcal{B}'$ is a base for $Y$ and if we show this, it follows that $Y$ is second countable too. In order to show this, let $y \in Y$ and $O$ open in $Y$ with $y \in O$.

Then $f^{-1}[O]$ is open in $X$ (continuity of $f$) and contains $f^{-1}[\{y\}]$, which is compact (we must assume $f$ is onto so that this set is non-empty too), and so we can pick $n(x) \in \Bbb N$ such that $$x \in B_{n(x)} \subseteq f^{-1}[O]$$ for all $x \in f^{-1}[\{y\}]$. Then the $B_{n(x)}$ cover the compact set $f^{-1}[\{y\}]$ and so there is a finite subset $F$ of $\Bbb N$ such that

$$f^{-1}[\{y\}] \subseteq \bigcup_{n \in F} B_n\subseteq f^{-1}[O]$$

And $B_y:=\bigcup_{n \in F} B_n \in \mathcal{B}$ so that $Y\setminus f[X\setminus B_y] \in \mathcal{B}'$. The proof is finished by noting that

$$y \in Y\setminus f[X\setminus B_y] \subseteq O$$

(suppose $y \notin Y\setminus f[X\setminus B_y]$ then $y \in f[X\setminus B_y]$, so $y=f(x')$ for some $x' \notin B_y$, but then also $x' \in f^{-1}[\{y\}]$ so $x' \in B_y$ contradiction. The inclusion is shown similarly).

QED.