## 5.3 Hausdorff spaces

The category of topological spaces has finite products.

Lemma 5.3.1. Let $X$ be a topological space. The following are equivalent:

1. $X$ is Hausdorff,

2. the diagonal $\Delta (X) \subset X \times X$ is closed.

Proof. We suppose that $X$ is Hausdorff. Let $(x, y) \not\in \Delta (X)$, i.e., $x \neq y$. There are $U$ and $V$ disjoint open sets of $X$ such that $x \in U$ and $y \in V$. This implies that $U \times V \subset (X \times X) \setminus \Delta (X)$. This shows that $(X \times X) \setminus \Delta (X)$ is an open set of $X \times X$ which is equivalent to say that the diagonal $\Delta (X) \subset X \times X$ is closed in $X \times X$. The converse is similar: The complement $(X \times X) \setminus \Delta (X)$ consist precisely of $(x, y) \in X\times X$ with $x \neq y$. Thus, if $\Delta (X) \subset X \times X$ is closed, then, by the definition of the product topology, for every such $(x, y)$, there are opens $U, V \subset X$ with $(x, y) \in U \times V$ and $(U \times V)\cap \Delta (X) = \emptyset$. In other words, with $x \in U$ and $y \in V$ such that $U \cap V = \emptyset$. $\square$

Lemma 5.3.2. Let $f : X \to Y$ be a continuous map of topological spaces. If $Y$ is Hausdorff, then the graph of $f$ is closed in $X \times Y$.

Proof. The graph is the inverse image of the diagonal under the map $X \times Y \to Y \times Y$. Thus the lemma follows from Lemma 5.3.1. $\square$

Lemma 5.3.3. Let $f : X \to Y$ be a continuous map of topological spaces. Let $s : Y \to X$ be a continuous map such that $f \circ s = \text{id}_ Y$. If $X$ is Hausdorff, then $s(Y)$ is closed.

Proof. This follows from Lemma 5.3.1 as $s(Y) = \{ x \in X \mid x = s(f(x))\}$. $\square$

Lemma 5.3.4. Let $X \to Z$ and $Y \to Z$ be continuous maps of topological spaces. If $Z$ is Hausdorff, then $X \times _ Z Y$ is closed in $X \times Y$.

Proof. This follows from Lemma 5.3.1 as $X \times _ Z Y$ is the inverse image of $\Delta (Z)$ under $X \times Y \to Z \times Z$. $\square$

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