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 $f \restriction C = f|(C \to f[C])$ a quotient map?

## Background

In what at first seems like a simple question, any counter-example or proof eludes me. This question was originally asked in Math StackExchange, but has gone unanswered. That question arose in turn in the context of another question.

## Partial results

### $f\restriction C$ is a 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 a 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 a 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 a 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.

### The 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 a 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.