Invariant | Dan Ma's Topology Blog

The Bow-Tie space is the continuous image of a separable metric space and yet is not metrizable (see here). Though taking continuous image can fail to preserve separable metrizability, we show that the perfect image of a separable metric space is a separable metric space. We prove the following theorem, which says that under a perfect map the weight will not increase. The result about separable metric space is a corollary of Theorem 1.

Theorem 1
Let $f:X \longrightarrow Y$ be a perfect map onto the space $Y$ . Then $w(Y) \le w(X)$ , i.e., the weight of $Y$ is no greater than the weight of $X$ .

Proof of Theorem 1

All spaces under consideration are Hausdorff. Let $X$ and $Y$ be spaces. Let $f: \longrightarrow Y$ be a map (or function) from $X$ onto $Y$ . The map $f$ is said to be a closed map if $f(C)$ is closed in $Y$ for any closed subset $C$ of $X$ . The map $f$ is a perfect map if $f$ is continuous, $f$ is a closed map, and $f^{-1}(y)$ is compact for every $y \in Y$ . In words, the last condition is that every point inverse is compact. A point inverse $f^{-1}(y)$ is also referred to as a fiber. Thus, we can say that a perfect map is a continuous closed surjective map with compact fibers. The following lemma is helpful for proving Theorem 1.

Lemma 2
Let $f:X \longrightarrow Y$ be a closed map such that $f(X)=Y$ . Let $V \subset X$ be open. Let $f_*(V)=\{ y \in Y: f^{-1}(y) \subset V \}$ . Then $f_*(V)$ is open in $Y$ and $f_*(V) \subset f(V)$ .

Proof of Lemma 2
We show that $Y \backslash f_*(V)$ is closed in $Y$ . To this end, we show $f(X \backslash V)=Y \backslash f_*(V)$ . Note that $f(X \backslash V)$ is closed in $Y$ since $f$ is a closed map. First, we show $f(X \backslash V) \subset Y \backslash f_*(V)$ . Let $t \in f(X \backslash V)$ . Then $t=f(x)$ for some $x \in X \backslash V$ . Since $x \notin V$ and $x \in f^{-1}(t)$ , we have $f^{-1}(t) \not \subset V$ . This implies that $t \notin f_*(V)$ .

We now show that $Y \backslash f_*(V) \subset f(X \backslash V)$ . Let $z \in Y \backslash f_*(V)$ . Since $z \notin f_*(V)$ , $f^{-1}(z) \not \subset V$ . Choose $x \in f^{-1}(z) \backslash V$ , which implies that $x \in X \backslash V$ . Thus, $z=f(x) \in f(X \backslash V)$ .

To complete the proof of the lemma, we show $f_*(V) \subset f(V)$ . Let $w \in f_*(V)$ . We have $f^{-1}(w) \subset V$ . As a result, $w=f(x)$ for some $x \in V$ . $\square$

Proof of Theorem 1
Let $\mathcal{B}$ be a base for $X$ . We derive a base $\mathcal{B}_1$ for $Y$ such that $\lvert \mathcal{B}_1 \lvert \le \lvert \mathcal{B} \lvert$ , i.e., the cardinality of $\mathcal{B}_1$ is no more than the cardinality of $\mathcal{B}$ . This implies that the minimal cardinality of a base in $Y$ is no more than the minimal cardinality of a base in $X$ , i.e., $w(Y) \le w(X)$ .

We assume that the base $\mathcal{B}$ is closed under finite unions. We show that $\mathcal{B}_1=\{ f_*(B): B \in \mathcal{B} \}$ is a base for $Y$ . Note that $f_*$ is defined in Lemma 2. Let $U \subset Y$ be an open set. Let $y \in U$ . 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, there exists finite $F \subset f^{-1}(y)$ such that $f^{-1}(y) \subset \bigcup_{t \in F} B_t=B$ . Note that $B \in \mathcal{B}$ . Since $f^{-1}(y) \subset B$ , we have $y \in f_*(B)$ . We also have $f_*(B) \subset f(B) \subset U$ . Thus, every open subset $U$ of $Y$ is the union of elements of $\mathcal{B}_1$ . This means that $\mathcal{B}_1$ is a base for $Y$ . Theorem 1 is established. $\square$

Corollary 3
Let $f:X \longrightarrow Y$ be a perfect map onto the space $Y$ . Then if $X$ is a separable metric space, then $Y$ is a separable metric space.

Comment About Lemma 2
A perfect map is not necessarily an open map. If the perfect map $f$ in Theorem 1 is an open map, then Lemma 2 is not needed and $\mathcal{B}_1=\{ f(B): B \in \mathcal{B} \}$ would be a base for $Y$ . However, we cannot assume $f$ is an open map simply because it is a perfect map. To see this, let $X=\mathbb{R}$ be the real line with the usual topology. Collapse the closed interval $[1,2]$ to one point called $p$ . The resulting quotient space is $Y$ where $Y=(-\infty,1) \cup \{ p \} \cup (2,\infty)$ . In $Y$ , the open neighborhoods of points in $(-\infty,1) \cup (2,\infty)$ are the usual Euclidean neighborhoods. The open neighborhoods of the point $p$ are the usual Euclidean open sets containing the interval $[1,2]$ . The resulting quotient map $f$ is an identity map on $(-\infty,1) \cup (2,\infty)$ and it maps points in $[1,2]$ to the point $p$ . It can be verified that $f$ is a perfect map. For the open set $V=(1,2)$ , $f(V)=\{ p \}$ , which is not open in $Y$ . For the open set $V=(0,1.5)$ , $f(V)=(0,1) \cup \{ p \}$ , which is not open in $Y$ . Lemma 2 says that for any open $X \subset$ , $f(V)$ may not be open but has an open subset $f_*(V)$ if $f(V)$ has non-empty interior. The interior sets $f_*(V)$ can work as a base in $Y$ .

Invariant and Inverse Invariant

Let $\mathcal{P}$ be a property of topological spaces. We say that $\mathcal{P}$ is an invariant of the perfect maps or that $\mathcal{P}$ is invariant under the perfect maps if the property $\mathcal{P}$ is preserved by perfect maps, i.e., for each perfect map $f:X \longrightarrow Y$ where $Y=f(X)$ , if the space $X$ has $\mathcal{P}$ , so does $Y$ . On the other hand, $\mathcal{P}$ is an inverse invariant of the perfect maps if this holds: for each perfect map $f:X \longrightarrow Y$ where $Y=f(X)$ , if $Y$ has $\mathcal{P}$ , so does $X$ .

The notions invariant and inverse invariant defined here are for perfect maps. In general, the notions are much broader and can be defined in relation to any class of continuous maps. For example, we know that the continuous image of a separable space is separable. We can say that separability is an invariant of the continuous maps or that separability is invariant under continuous maps. We can now restate Theorem 1 and Corollary 3 as follows.

Theorem 4…..Restatement of Theorem 1
The property “weight $\le \mathcal{K}$ ” is an invariant of the perfect maps.

Corollary 5
The second axiom of countability is invariant under the perfect maps, but is not an invariant of the continuous maps.

The Bow-Tie space is the continuous image of a separable space but cannot have a countable base (see here). In light of Theorem 1, the continuous map that maps a separable metric space to the Bow-Tie space (shown here) cannot be a perfect map. With respect to that map, the upper half plane in the domain (the separable metric space) is closed but its continuous image in the Bow-Tie space is open and not closed.

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Revised March 31, 2023
Revised March 12, 2024

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