# Is the 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 $$f \restriction C = f|(C \to f[C])$$ 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.

• 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? Jan 21 at 21:29
• A restriction symbol was missing; $f|(C \to f[C])$. $\mathcal{T}_Z|Z'$ is the subspace topology of $Z'$ on $Z$.
– kaba
Jan 21 at 23:41

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.

• 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.
– kaba
Jan 21 at 20:12
• @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. Jan 21 at 21:59