Is restriction of a projection to a compact subset a quotient map? 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.
 A: 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.
