Lemma 28.47.2. Let $S$ be a scheme. Let $X$ and $Y$ be schemes over $S$. Assume $X$ has finitely many irreducible components with generic points $x_1, \ldots , x_ n$. Let $s_ i \in S$ be the image of $x_ i$. Consider the map

\[ \left\{ \begin{matrix} S\text{-rational maps}
\\ \text{from }X\text{ to }Y
\end{matrix} \right\} \longrightarrow \left\{ \begin{matrix} (y_1, \varphi _1, \ldots , y_ n, \varphi _ n)\text{ where } y_ i \in Y\text{ lies over }s_ i\text{ and}
\\ \varphi _ i : \mathcal{O}_{Y, y_ i} \to \mathcal{O}_{X, x_ i} \text{ is a local }\mathcal{O}_{S, s_ i}\text{-algebra map}
\end{matrix} \right\} \]

which sends $f : U \to Y$ to the $2n$-tuple with $y_ i = f(x_ i)$ and $\varphi _ i = f^\sharp _{x_ i}$. Then

If $Y \to S$ is locally of finite type, then the map is injective.

If $Y \to S$ is locally of finite presentation, then the map is bijective.

If $Y \to S$ is locally of finite type and $X$ reduced, then the map is bijective.

**Proof.**
Observe that any dense open of $X$ contains the points $x_ i$ so the construction makes sense. To prove (1) or (2) we may replace $X$ by any dense open. Thus if $Z_1, \ldots , Z_ n$ are the irreducible components of $X$, then we may replace $X$ by $X \setminus \bigcup _{i \not= j} Z_ i \cap Z_ j$. After doing this $X$ is the disjoint union of its irreducible components (viewed as open and closed subschemes). Then both the right hand side and the left hand side of the arrow are products over the irreducible components and we reduce to the case where $X$ is irreducible.

Assume $X$ is irreducible with generic point $x$ lying over $s \in S$. Part (1) follows from part (1) of Lemma 28.40.4. Parts (2) and (3) follow from part (2) of the same lemma.
$\square$

## Comments (2)

Comment #2994 by Minsik Han on

Comment #3117 by Johan on