Let $f:X\rightarrow Y$ be a morphism of schemes (or algebraic spaces), and $s:Y\rightarrow X$ is a section to $f$, i.e. $fs=1_{Y}$.
Question: Is $s$ an (closed) immersion?
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5$\begingroup$ If $f$ is separated, then the answer is 'yes': it is a standard fact that if a composition of two morphisms is a closed immersion and the second one is separated, then the first one is also a closed immersion. See Hartshorne's book Exercise II.4.8; if you don't want to do exercises, I'm sure you can also locate this is in EGA without troubles. $\endgroup$– Kestutis CesnaviciusSep 14, 2013 at 2:34
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2$\begingroup$ Also, if $f$ is not separated, then $s$ need not be a closed immersion (take $X$ to be two affine lines glued away from the origin, and $f$ the projection to a single affine line). $\endgroup$– Kevin VentulloSep 14, 2013 at 5:49
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4$\begingroup$ In general $s$ is a locally closed immersion. This belongs to the basics of AG and can be found everywhere, therefore I have voted to migrate the question to math.SE. $\endgroup$– Martin BrandenburgSep 14, 2013 at 10:50
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1$\begingroup$ This is false for algebraic spaces: the diagonal of an algebraic space is not necessarily an immersion. $\endgroup$– Marc HoyoisApr 21, 2017 at 18:00
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1 Answer
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Prop. If $f\colon X\longrightarrow Y$ is a separated morphism of schemes, then every section $\sigma\colon Y\longrightarrow X$ is a closed immersion.
Proof. Let $\sigma\in X(Y)$ be a section of the $Y$-scheme $X$. Let's take the fibered product $\left(X\times_X Y,p_X,q_X\right)$, where we are considering $Y$ as a $X$-scheme endowing it with the structural morphism $\sigma\colon Y\longrightarrow X$. The projections $p_X$ and $q_X$ induce a closed immersion of $Y$-schemes $X\times_X Y\longrightarrow X\times_Y Y$, since $\ast$ $f\colon X\longrightarrow Y$ is separated. In a similar way, the identity map of $Y$ and the section $\sigma$ induce a morphism $Y\longrightarrow X\times_X X$ of $X$-schemes. In other words we have the following commutative diagram.
At this point the solution of the exercise follows by the fact that a composition of closed immersions is a closed immersion and by the fact that the morphisms $q^{-1}_X$ and $p^{}_Y$ are isomorphisms, then, in particular, closed immersions.
$\ast$ Let $Y$ be a separated $Z$-scheme. Then for any $Y$-schemes $X_1,X_2$, the canonical morphism $X_1\times_Y X_2\longrightarrow X_1\times_Z X_2$ is a closed immersion.
The claim above is the Proposition 3.9 (f) at pag.101 of Algebraic Geometry and Aritmetic Curves.
answered Apr 21, 2017 at 17:29