Being second countable is invariant under perfect mapping Firstly, I will give the definition of perfect mapping:

Let $f$ be a closed mapping from a topological space $X$ to another topological space $Y$. We call it a perfect mapping if for every point $y \in Y$, the subspace $f^{-1}(y)$ of $X$ is compact. 

My question is this: If $X$ has countable base, i.e., $\mathcal W(X) = \omega$, and $f$ is a perfect mapping from $X$ to $Y$, how to show that $Y$ also has countable base?
Thanks ahead:)  
 A: I'll copy here a proof of a more general result from the book Engelking, Ryszard (1989): General Topology. Heldermann Verlag, Berlin. ISBN 3885380064.


Theorem 3.7.19.  If there exists a perfect mapping $f \colon X \to Y$ onto a space $Y$, then $w(Y) < w(X)$.

Here $w(X)$ is the weight of a topological space, i.e. 
$$w(X)= \min \{|\mathcal B|; \text{$\mathcal B$ is a base for $X$}\}+\aleph_0.$$ 
Proof. Let $w(X) = \mathfrak m$. Since the validity of our theorem is obvious for $\mathfrak m < \aleph_0$, we
can assume that $\mathfrak m \ge \aleph_0$. Let $\{U_s\}_{s\in S}$ be a base for $X$ such that $|S| = \mathfrak m$ and let $\mathcal T$ be the
family of all finite subsets of $S$. Since $|\mathcal  T| = \mathfrak m$ it suffices to show that the family $\{W_T\}_{T\in\mathcal{T}}$,
where $W_T = Y \setminus f(X \setminus \bigcup_{s\in T} U_s)$, is a base for $Y$. It follows from the definition that the
sets $W_T$ are open. Let us take a point $y \in Y$ and a neighbourhood $W \subset Y$ of $y$. The
inverse image $f^{-1}(y)$ is a compact subset of $f^{-1}(W)$; thus there exists a $T \in \mathcal T$ such that
$f^{-1}(y)\subset \bigcup_{s\in T} U_s \subset f^{-1}(W)$.
Clearly $y \in W_T$, and since
$$Y\setminus W = f(X\setminus f^{-1}(W)) \subset f(X\setminus \bigcup_{s\in T} U_s)$$
we have $W_T\subseteq W$.

The above is taken verbatim from Engelking's book. But in the place where he writes $w(X)<\aleph_0$, he probably means $|S|<\aleph_0$. (Since $w(X)$ is infinite by definition.)
If a topological space has a finite base, then the topology is finite. The map $f$ is quotient, since every surjective closed map is quotient. This means that for every open set $U\subseteq Y$ the preimage $f^{-1} (U)$ is open. The assignment $U\mapsto f^{-1}(U)$ is an injection from the topology of $Y$ to the topology of $X$. Hence $Y$ has, in this case, only finitely many open sets. (And thus it has finite base.)
