Definition 5.4.1. A continuous map f : X \to Y of topological spaces is called separated if and only if the diagonal \Delta : X \to X \times _ Y X is a closed map.
5.4 Separated maps
Just the definition and some simple lemmas.
Lemma 5.4.2. Let f : X \to Y be continuous map of topological spaces. The following are equivalent:
f is separated,
\Delta (X) \subset X \times _ Y X is a closed subset,
given distinct points x, x' \in X mapping to the same point of Y, there exist disjoint open neighbourhoods of x and x'.
Proof. If f is separated, by Definition 5.4.1, \Delta is a closed map. The fact that X is closed in X gives us that \Delta (X) is closed in X\times _ Y X . Thus (1) implies (2).
Assune \Delta (X) \subset X \times _ Y X is a closed subset and denote U the complementary open. This means we have an open set W \subset X \times X such that W \cap (X \times _ Y X) = U. However, by definition of the product topology, if (x, x') \in W \cap (X \times _ Y X), we have V and V' open sets of X such that x \in V, x' \in V' and V \times V' \subset W. If we had V \cap V' \neq \emptyset , we would have z \in V \cap V'. However, (z, z) \in X \times _ Y X, so (z,z) \in (V \times V') \cap (X\times _ Y X) \subset U, which is absurd. Therefore V \cap V' = \emptyset , and we have two disjoint open neighborhoods for x and x'. It proves that (2) implies (3).
Finally, we suppose that given distinct points x, x' \in X mapping to the same point of Y, there exist disjoint open neighbourhoods of x and x'. Let F be a closed set of X. We will show that \Delta (F) is a closed subset of X \times _ Y X. Let (x, x') \in X \times _ Y X be a point not contained in \Delta (F). Then either x \not= x' or x \not\in F. In the first case, we choose disjoint open neighbourhoods V, V' \subset X of x, x' and we see that (V \times V') \cap X \times _ Y X is an open neighbourhood of (x, x') not meeting \Delta (F). In the second case, we see that ((X \setminus F) \times X) \cap X \times _ Y X is an open neighbourhood of (x, x') not meeting \Delta (F). We have shown that (3) implies (1). \square
Lemma 5.4.3. Let f : X \to Y be continuous map of topological spaces. If X is Hausdorff, then f is separated.
Proof. Clear from Lemmas 5.4.2 and 5.3.1 as \Delta (X) closed in X \times X implies \Delta (X) closed in X \times _ Y X. \square
Lemma 5.4.4. Let f : X \to Y and Y' \to Y be continuous maps of topological spaces. If f is separated, then f' : Y' \times _ Y X \to Y' is separated.
Proof. Follows from characterization (2) of Lemma 5.4.2. Namely, with X' = Y' \times _ Y X the diagonal \Delta (X') in the fibre product X' \times _{Y'} X' is the inverse image of \Delta (X) in X \times _ Y X. \square
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