Lemma 5.4.2. Let $f : X \to Y$ be continuous map of topological spaces. The following are equivalent:

1. $f$ is separated,

2. $\Delta (X) \subset X \times _ Y X$ is a closed subset,

3. 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 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 #8648 by on

I don't know if this may be a triviality, but I think it's interesting to note that the analogue of the result for schemes, "a morphism of schemes is separated iff its fibers are separated schemes," is not true. Even though separated morphisms of schemes have separated fibers, the converse doesn't hold: one considers $X=$ the affine line with doubled origin over a field $k$ and $f:X\to\mathbb{A}_k^1$, the morphism squashing the double origin to the single origin of $\mathbb{A}_k^1$. Then, one verifies that $f^{-1}(t)$ is a singleton for every point $t\in\mathbb{A}^1_k$ distinct from the origin (in the case $t=\eta$ is the generic point, $f^{-1}(\eta)$ is the generic point of $X$) and $f^{-1}(0)$ are the two origins of $X$. (Finite schemes are always affine, whence separated.)

Comment #8650 by on

(Sorry: a finite scheme is not necessarily affine. However, schemes with 1 or 2 points always are.)

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