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?


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.

  • $\begingroup$ Do you really mean $f(C \to f[C])$, as opposed to, say $f : C \to f[C]$? What do $\mathcal T_Z|Z'$ and $\mathcal T_X|g(Z')$ mean? $\endgroup$
    – LSpice
    Jan 21 at 21:29
  • $\begingroup$ A restriction symbol was missing; $f|(C \to f[C])$. $\mathcal{T}_Z|Z'$ is the subspace topology of $Z'$ on $Z$. $\endgroup$
    – kaba
    Jan 21 at 23:41

1 Answer 1


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.

  • $\begingroup$ Nice, thanks! This trick of "bijective but not homeomorphic" seems like a generally useful way to produce counter-examples for something being a quotient map. $\endgroup$
    – kaba
    Jan 21 at 20:12
  • 1
    $\begingroup$ @kaba Indeed, whenever you have a surjective function $f : X \to Y$ that is not a quotient map, you get a bijective non-homeomorphism $X/\sim \to Y$. So in a sense, bijective non-homeomorphisms account for all examples. $\endgroup$ Jan 21 at 21:59

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