**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 neighbourood 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 neighbourood of $(x, x')$ not meeting $\Delta (F)$. We have shown that (3) implies (1).
$\square$

## Comments (3)

Comment #8647 by Elías Guisado on

Comment #8648 by Elías Guisado on

Comment #8650 by Elías Guisado on