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
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In what at first seems like a simple question, any counter-example or proof eludes me. This question was originally asked in [Math StackExchange](https://math.stackexchange.com/q/4350112/51161), but has gone unanswered. That question arose in turn in the context of [another question](https://math.stackexchange.com/questions/4348707/is-projection-of-locally-connected-compact-subset-locally-connected).


Partial results
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### $f\restriction C$ 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](https://math.stackexchange.com/questions/143988/how-to-prove-the-closed-map-lemma), [hence a quotient map](https://math.stackexchange.com/questions/3981375/a-surjective-continuous-and-closed-map-is-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](https://math.stackexchange.com/questions/22697/projection-map-being-a-closed-map). 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](https://math.stackexchange.com/questions/247542/projection-is-an-open-map). 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.



Is the restriction of a projection to a compact subset a quotient map?
A counterexample with finite topological spaces:


recall that on a set $X$, an "Alexandroff discrete" topology ($\mathrm T\_0$ where all intersections of open sets are open)
is the same thing as a partial order; the minimum open set containing $x$ is the principal order filter $Fx$ in the order.
The dual order gives the topology where open and closed sets are interchanged.


Take a finite $X$, and as $Y$ its dual as above; take as $C$ the principal diagonal of points $(x,x)$.
The minimum open set which contains $(x,x)$ is the direct product of the principal order filter $Fx$ and the principal ideal $Ix$.
The only $y$ with both inequalities $x\leq y$, $x\geq y$ is $y=x$, so the diagonal $C$ is discrete.


Essentially, one has the identity map on a finite set $X$, with the discrete topology in the domain and an arbitrary $\mathrm T\_0$ topology
in the codomain. No identifications but no homeomorphism either (except for the antichain poset order on $X$). No quotient.


Edit. Take any two distinct quasi compact topologies on a set $X$,
and the projections from the diagonal as above. At least one
of the two projections is not a quotient.


