2
$\begingroup$

Let $(X, \mathcal{T}_X)$ and $(Y, \mathcal{T}_Y)$ be topological spaces, $Z = X \times Y$, $\mathcal{T}_Z$ be the product topology on $Z$, $f : Z \to X$ be defined by $f(x, y) = x$, and $C \subset Z$ be compact. Is $f \restriction C$ a quotient map?

Background

The motivation for this question comes from this question. A positive answer for this question provides a positive answer for that question, and a negative answer illuminates it.

Partial results

$f\restriction C$ is quotient when $X$ is Hausdorff

Suppose $X$ is Hausdorff. Then $f\restriction C$ is a continuous map from a compact space to a Hausdorff space, hence a closed map, hence a quotient map.

$f\restriction V$ is quotient when $Y$ is compact and $V \subset Z$ is closed

Suppose $Y$ is compact. Then $f$ is closed. Let $V \subset Z$ be closed. Then $f\restriction V$ is a closed map, hence a quotient map.

$f\restriction U$ is quotient when $U \in \mathcal{T}_Z$

Suppose $U \in \mathcal{T}_Z$. It can be shown that $f$ is open. Then $f\restriction U$ is an open map, hence a quotient map.

$f\restriction C$ is quotient when $C$ is closed

Let $\pi_X : Z \to X$ be defined by $\pi_X(x, y) = x$ and $\pi_Y : Z \to Y$ be defined by $\pi_Y(x, y) = y$. Let $C_X = \pi_X(C)$ and $C_Y = \pi_Y(C)$. By continuity, $C_X$ and $C_Y$ are compact. Therefore $D = C_X \times C_Y$ is compact. By a previous section, $\pi_X \restriction D$ is closed. Since $C$ is closed in $D$, $\pi_X \restriction C$ is closed. Therefore $f\restriction C$ is a quotient map.

Previous strategy fails when $C$ is not closed

The previous proof does not generalize to the case when $C$ is not closed. Let $X = Y = \{0, 1\}$ and $\mathcal{T}_X = \mathcal{T}_Y = \{\emptyset, \{0\}, \{0, 1\}\}$. Then $\{(0, 0), (1, 0), (0, 1)\}$ is compact, but not closed in $X \times Y$.

Compact slices are not sufficient to be quotient

Let $X = Y = \mathbb{R}$, $Z' = \{(0, 1)\} \cup \{(1/n, 0) : n \in \mathbb{N}^{> 0}\}$, and $g = f \restriction Z'$. Then $(\{x\} \times Y) \cap Z'$ is compact for each $x \in X$ as a singular subset. Let $V = \{0\}$, and $U = g^{-1}(V) = \{(0, 1)\}$. Then $U \in \mathcal{T}_Z|Z'$, and $V \not\in \mathcal{T}_X|g(Z')$. Therefore $g$ is not a quotient map.

Edit: Since there were no answers, the question is now posted also at MathOverflow.

$\endgroup$

1 Answer 1

1
$\begingroup$

This question was answered at MathOverflow by user NameNo. I'll rephrase that answer here.

The answer is no, and here is a counter-example. Let $X = Y$ be finite, but non-homeomorphic. Let $C = \{(x, x) : x \in X\}$. As a finite set, $C$ is compact. Let $f_X : Z \to X$ be defined by $f_X(x, y) = x$ and $f_Y : Z \to Y$ be defined by $f_Y(x, y) = y$. Now $f_X \restriction C$ and $f_Y \restriction C$ are both bijections. However, since a bijective quotient map is a homeomorphism, $f_X \restriction C$ and $f_Y \restriction C$ cannot both be quotient maps. Without loss of generality (by swapping $X$ and $Y$), we may assume that $f_X$ is not a quotient map.

To be concrete, let

$$ \begin{aligned} X & = Y = \{0, 1\}, \\ \mathcal{T}_X & = \{\emptyset, \{0\}, \{0, 1\}\}, \\ \mathcal{T}_Y & = \{\emptyset, \{0\}, \{1\}, \{0, 1\}\}, \\ C & = \{(0, 0), (1, 1)\}, \\ V & = \{1\} \not\in \mathcal{T}_X. \end{aligned} $$ Then

$$(f_X \restriction C)^{-1}(V) = \{(1, 1)\} = (\{0, 1\} \times \{1\}) \cap C \in \mathcal{T}_Z|C.$$

Therefore $f_X \restriction C$ is not a quotient map.

Here are some of my own observations.

Locally closed subsets fail too

Since the result does hold when $C$ is open or closed, one may wonder whether the result also holds when $C$ is locally closed (i.e. intersection of open and closed subset). In this case too, the answer is no.

Using the same spaces as above, let $C = \{(1, 0), (0, 1)\}$ and $V = \{1\} \not\in \mathcal{T}_X$. Then $C$ is locally closed and $(f_X \restriction C)^{-1}(V) = \{(1, 1)\} \in \mathcal{T}_Z|C$.

Therefore $f_X \restriction C$ is not a quotient map.

Generalized closed subsets fail too

A subset $C \subset Z$ is generalized closed, if whenever $C \subset U$ for some $U \in \mathcal{T}_Z$, then $\overline{C}(Z) \subset U$. Here is a counter-example for when $C$ is generalized closed. Let

$$ \begin{aligned} X & = Y = \{0, 1\}, \\ \mathcal{T}_X & = \{\emptyset, \{0, 1\}\}, \\ \mathcal{T}_Y & = \{\emptyset, \{0\}, \{0, 1\}\}, \\ C & = \{(1, 0), (0, 1)\}, \\ V & = \{1\} \not\in \mathcal{T}_X. \end{aligned} $$

Then $C$ is generalized closed, but

$$(f_X \restriction C)^{-1}(V) = \{(1, 1)\} \in \mathcal{T}_Z|C.$$

Therefore $f_X \restriction C$ is not a quotient map.

But k-closed/k-open subsets work!

The $k$-topology of a topology $\mathcal{T}_X$ is

$$ K(\mathcal{T}_X) = \{A \subset X : \forall \text{ compact } K \subset X : A \cap K \in \mathcal{T}_X|K\} $$

Call $(X, K(\mathcal{T}_X))$ the k-space of $X$. Some facts:

  • K-topology indeed is a topology.
  • A subset $A \subset X$ is k-open, if $A \in K(\mathcal{T}_X)$.
  • A subset $A \subset X$ is k-closed, if $A \cap K$ is closed in $K$ for each compact $K \subset X$. This is equivalent to $X \setminus A \in K(\mathcal{T}_X)$.
  • K-topology refines the original topology: $\mathcal{T}_X \subset K(\mathcal{T}_X)$.
  • The compact subsets in $(X, K(\mathcal{T}_X))$ coincide with the compact subsets in $(X, \mathcal{T}_X)$.
  • The subspace topology in any compact subset of $X$ is the same in both topologies: $K(\mathcal{T}_X)|C = \mathcal{T}_X|C$ for each compact $C \subset X$.
  • A compactly generated space is a space in which the k-topology coincides with the topology: $K(\mathcal{T}_X) = \mathcal{T}_X$.
  • $(X, K(\mathcal{T}_X))$ is compactly generated; i.e. $K(K(\mathcal{T}_X)) = K(\mathcal{T}_X)$.
  • Every locally compact space is compactly generated.
  • Hence so is every compact space.
  • Product of compactly generated spaces may not be compactly generated.
  • Product of a compactly generated space and a locally compact space is compactly generated.

Claim

Suppose $C$ is compact and k-closed. Then $f \restriction C$ is a closed map (and hence a quotient map).

Proof

Let $f_Y : Z \to Y$ be defined by $f_Y(x, y) = y$. Then $f_Y$ is continuous. Let $Y' = f_Y[C]$. Then $Y'$ is compact. Hence, without loss of generality we may assume that $Y$ is compact.

Replace the topologies of $X$ and $Y$ by their k-topologies. Then $X$ and $Y$ are both compactly generated, and $Y$ is compact. Hence $X \times Y$ is also compactly generated. Hence k-closed simply means closed, and we already know that then $f \restriction C$ is a closed map (and hence a quotient map). But since $C$ is compact and $f[C]$ is compact, the subspace topologies in $C$ and $f[C]$ are the same as in the original topologies. Hence $f \restriction C$ is a closed map even on the original topologies.

Claim

Suppose $C$ is compact and k-open. Then $f \restriction C$ is an open map (and hence a quotient map).

Proof

Similarly to the proof for the k-closed case.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.