Lemma 5.3.1. Let X be a topological space. The following are equivalent:
X is Hausdorff,
the diagonal \Delta (X) \subset X \times X is closed.
5.3 Hausdorff spaces
The category of topological spaces has finite products.
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.slogan 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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