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.

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2 thoughts on “Perfect Image of Separable Metric Spaces

  1. A closed map need not be an open map, even if it’s a perfect surjection between separable metric spaces. Consider the quotient map on the real line that collapses the interval [2,4] to a single point. The map is perfect, yet some open sets, such as (1,3) and (2,4), are mapped to non-open sets.

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