1
$\begingroup$

In the algebraic geometry of schemes, we know that monomorphisms which are universally closed (= every pullback is a closed map) and of finite type are closed immersions. See Gortz, Wedhorn "Algebraic Geometry I", Corollary 12.92.

Is something similar true with complex analytic spaces (as defined in Grauret, Remmert "Coherent Analytic Sheaves"). That is, is it true that every universally closed monomorphism (+ perhaps some extra conditions) is a closed immersion?

$\endgroup$
4
  • 1
    $\begingroup$ Dear Onion Dip: The right notion is properness. In CAS there is the structure theorem for quasi-finite morphisms: any $f:X \rightarrow Y$ with fiber over $y$ having $x$ as an isolated point restricts to a finite morphism (i.e., classified by a coherent sheaf of algebras in an evident manner) between opens around $x$ and $y$. Using this, if $f$ has discrete finite fibers and is proper then easily it is a finite morphism. Hence, a proper monomorphism is a finite monomorphism, so consideration of infinitesimal fibers gives the closed immersion property as an exercise with coherent sheaves. $\endgroup$
    – user74230
    Commented Nov 26, 2014 at 0:35
  • $\begingroup$ By the way, it is not true that universally closed morphisms of finite type are closed immersions; one has to assume separatedness (in the analytic setting separatedness is also required in the definition of properness). For example, if $Y$ is the affine line over a non-empty base $S$ and $X$ is a gluing of two copies of $Y$ to itself along $Y-0(S)$ via the identity map then the evident $S$-map $f:X \rightarrow Y$ is finite type and non-separated but it is closed by inspection and hence universally closed since the formation of $f$ commutes with base change over $S$ and $Y$ is $S$-separated. $\endgroup$
    – user74230
    Commented Nov 26, 2014 at 2:40
  • $\begingroup$ @user74230: yes, but a monomorphism is automatically separated. $\endgroup$ Commented Nov 26, 2014 at 9:39
  • $\begingroup$ @LaurentMoret-Bailly: Indeed! $\endgroup$
    – user74230
    Commented Nov 26, 2014 at 18:51

0

You must log in to answer this question.