nLab quotient projections out of compact Hausdorff spaces are closed precisely if the codomain is Hausdorff

Statement

Proposition

(quotient projections out of compact Hausdorff spaces are closed precisely if the codomain is Hausdorff)

Let

$$\pi \;\colon\; (X, \tau_X) \longrightarrow (Y, \tau_Y)$$

be a continuous function between topological spaces such that

1. $(X, \tau)$ is a compact Hausdorff topological space;

2. $\pi$ is a surjection and $\tau_Y$ is the corresponding quotient topology.

Then the following are equivalent

1. $(Y, \tau_Y)$ is itself a Hausdorff topological space;

2. $\pi$ is a closed map.

Proof

The implicaton $\left( (Y, \tau_Y)\, \text{Hausdorff} \right) \Rightarrow \left( \pi \, \text{closed} \right)$ follows since maps from compact spaces to Hausdorff spaces are closed. We need to show the converse.

Hence assume that $\pi$ is a closed map. We need to show that for every pair of distinct points $y_1 \neq y_2 \in Y$ there exist open neighbourhoods $U_{y_1}, U_{y_2} \in \tau_Y$ which are disjoint, $U_{y_1} \cap U_{y_2} = \emptyset$.

First notice that the singleton subsets $\{x\}, \{y\} \in Y$ are closed. This is because they are images of singleton subsets in $X$, by surjectivity of $f$, and because singletons in a Hausdorff space are closed, and because images under $f$ of closed subsets are closed, by the assumption that $f$ is a closed map.

It follows that the pre-images

$$C_1 \coloneqq \pi^{-1}(\{y_1\}) \phantom{AA} C_2 \coloneqq \pi^{-1}(\{y_2\}) \,.$$

are closed subsets of $X$.

Now since compact Hausdorff spaces are normal it follows that we may find disjoint open subset $U_1, U_2 \in \tau_X$ such that

$$C_1 \subset U_1 \phantom{AAA} C_2 \subset U_2 \,.$$

Moreover, by this lemma we may find these $U_i$ such that they are both saturated subsets. Therefore finally this lemma says that the images $\pi(U_i)$ are open in $(Y,\tau_Y)$. These are now clearly disjoint open neighbourhoods of $y_1$ and $y_2$.

Applications

Example

Consider the function

$$\array{ [0,2\pi]/\sim &\longrightarrow& S^1 \subset \mathbb{R}^2 \\ t &\mapsto& (cos(t), sin(t)) }$$

• from the quotient topological space of the closed interval by the equivalence relation which identifies the two endpoints

  $$(x \sim y) \;\Leftrightarrow\; \left( \left( x = y\right) \,\text{or}\, \left( \left( x \in \{0,2\pi\}\right) \,\text{and}\, \left( y\in \{0, 2\pi\} \right) \right) \right)$$

• to the unit circle $S^1 = S_0(1) \subset \mathbb{R}^2$ regarded as a topological subspace of the 2-dimensional Euclidean space equipped with its

metric topology .

This is clearly a continuous function and a bijection on the underlying sets. Moreover, since continuous images of compact spaces are compact and since the closed interval $[0,1]$ is compact we also obtain another proof that the circle is compact.

Hence since continuous bijections from compact spaces to Hausdorff spaces are homeomorphisms the above map is in fact a homeomorphism

$$[0,2\pi]/\sim \;\simeq\; S^1 \,.$$