This is Prop 3.11 of Knutson's Algebraic Spaces:

Let $X\rightarrow Y$ be a map of algebraic spaces. Let $g:Y\rightarrow X$ be a section of f, i.e, a map satisfying $fg=id$. Show that if $f$ is separated, then $g$ is a closed immersion.

Can one give me a proof of this? The proof should be fairly simple (unfortunately, I have no experience with alg. spaces).

  • 2
    $\begingroup$ I don't know, whether that helps in your situation, but the first thing I would try would be to take the proof of this fact for schemes, and see if it survives in the realm of algebraic spaces. Unfortunately, I'm not familiar with algebraic spaces, so I can't tell if the proof for schemes applies to algebraic spaces. But maybe it's a profitable thing to try so. $\endgroup$ – Nils Matthes Jul 25 '13 at 6:30
  • $\begingroup$ Yes, the same proof works. After realizing that the graph might be useful, it is a direct application of the usual closure properties with respect to composition and fiber products (which also hold for algebraic spaces, thanks to descent). It also holds for algebraic stacks etc. $\endgroup$ – Martin Brandenburg Aug 5 '13 at 17:48
  • $\begingroup$ Can you write down a proof for me? As I said i have no exp. with alg. spaces. I know that what I wrote above is direct consequence of descent, but I want to see a detailed proof. $\endgroup$ – Marci Aug 15 '13 at 22:58

In order to apply Knutson's Lemma I 1.21 we need to know that the identity $id_Y$ is a closed immersion (of algebraic spaces), the composition of two closed immersions is a closed immersion and that the base change of a closed immersion is a closed immersion.Let's show that the base change of a closed immersion is a closed immersion.

Let $X \to Y$ be a closed immersion of algebraic spaces and $Z \to Y$ any map. Then there are representable etale coverings $V \to X$ and $U \to Y$ and a closed immersion of schemes $V \to U$. Let $W \to Z$ be a representable etale covering.

First assume that $W \to Z \to Y$ factors through $U$. Now $W \times_U V \to Z\times_Y X$ is a representable etale covering and $W \times_U V \to W$ is a closed immersion of schemes since the base change of a closed immersion of schemes is a closed immersion. So $Z\times_Y X \to Z$ is a closed immersion of algebraic spaces.

For the general case, set $W'=W\times_Y U$. Note that $W'$ is a scheme, of course $W' \to Y$ factors through $U$ and $W' \to W$ is an etale covering. So $W' \times_U V \to W'$ is a closed immersion of schemes and $W' \times_U V=W'\times_Y V \to W\times_Y V$ is an etale covering. Also $W\times_Y V$ is a scheme and $W\times_Y V=W\times_Z(Z\times_Y V) \to Z \times_Y X $ is a representable etale covering. So the maps $W'\times_U V \to W\times_Y V \to W$ and $W'\times_U V \to W' \to W$ form a cartesian diagram of schemes (haven't figured out to how draw diagrams here yet) s.t. both horizontal arrows ( $W'\times_U V \to W\times_Y V$ and $W' \to W$) are etale covering maps and the left vertical arrow ($W'\times_U V \to W'$) a closed immersion. Since closed immersions satisfy effective descent in the etale topology the right vertical arrow ($W\times_Y V \to W$) is a closed immersion. Hence $Z\times_Y X \to Z$ is a closed immersion of algebraic spaces.


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