# Perfect Image of Separable Metric Spaces

In a previous post on countable network, it was shown that having a countable network is equivalent to being the continuous image of a separable metric space. Since there is an example of a non-metrizable space with countable netowrk, the continuous image of a separable metric space needs not be a separable metric space. However, the perfect image of a separable metrizable space is separable metrizable. First some definitions. A continuous mapping $f:X \rightarrow Y$ is a closed mapping if $f(H)$ is closed in $Y$ for any closed set $H \subset X$. A continuous surjection $f:X \rightarrow Y$ is a perfect mapping if $f$ is closed and $f^{-1}(y)$ is compact for each $y \in Y$.

Let $f:X \rightarrow Y$ be a perfect mapping where $X$ has a countable base $\mathcal{B}$. Assume $\mathcal{B}$ is closed under finite unions. Because $f$ is a closed mapping, $f(X-B)$ is closed and $f(B)$ is open in $Y$ for each $B \in \mathcal{B}$. We show that $\mathcal{B}_f=\lbrace{f(B):B \in \mathcal{B}}\rbrace$ is a base for $Y$. Let $y \in Y$ and $U \subset Y$ be open with $y \in U$. For each $x \in f^{-1}(y)$, choose $B_x \in \mathcal{B}$ such that $f(B_x) \subset U$. Since $f^{-1}(y)$ is compact, we can choose $B_{x(0)},...,B_{x(n)}$ that cover $f^{-1}(y)$. Let $B=B_{x(0)} \cup ... \cup B_{x(n)}$, which is in $\mathcal{B}$. We have $y \in f(B) \subset U$. Thus the topology on $Y$ can be generated by $\mathcal{B}_f$.

Update (11/24/2009):
The proof in the above paragraph is faulty. Thanks to Dave Milovich for pointing this out. Here’s the corrected proof.

Let me first prove a lemma.

Lemma. Let $f: X \rightarrow Y$ be a closed mapping and let $V \subset X$ be open. Then $f_*(V)=\lbrace{y \in Y:f^{-1}(y) \subset V}\rbrace$ is open in $Y$. Furthermore, $f_*(V) \subset f(V)$.

Proof of Lemma. Since $f$ is a closed mapping, $f(X-V)$ is closed. We claim that $f(X-V)=Y-f_*(V)$. It is clear that $f(X-V) \subset Y-f_*(V)$. To show that $Y-f_*(V) \subset f(X-V)$, let $z \in Y-f_*(V)$. Then $f^{-1}(z)$ cannot be a subset of $V$. Choose $x \in f^{-1}(z)-V$. Then we have $z=f(x) \in f(X-V)$. Thus $f(X-V)=Y-f_*(V)$ and $f_*(V)$ is open. It is straitforward to verify that $f_*(V) \subset f(V)$.

Now I prove that the perfect image of a separable metric space is a separable metric space. Let $f:X \rightarrow Y$ be a perfect mapping where $X$ has a countable base $\mathcal{B}$. Assume $\mathcal{B}$ is closed under finite unions. We show that $\mathcal{B}_f=\lbrace{f_*(B):B \in \mathcal{B}}\rbrace$ is a base for $Y$.

Let $y \in Y$ and $U \subset Y$ be open with $y \in U$. For each $x \in f^{-1}(y)$, choose $B_x \in \mathcal{B}$ such that $x \in B_x$ and $f(B_x) \subset U$. Since $f^{-1}(y)$ is compact, we can choose $B_{x(0)},...,B_{x(n)}$ that cover $f^{-1}(y)$. Let $B=B_{x(0)} \cup ... \cup B_{x(n)}$, which is in $\mathcal{B}$. Since $f^{-1}(y) \subset B$, we have $y \in f_*(B)$. We also have $f_*(B) \subset f(B) \subset U$. Thus the topology on $Y$ can be generated by the countable base $\mathcal{B}_f$.