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I would like to prove the equivalence of the following two definitions (this question is mostly proof verification)

Let $(f:X\to Y, f^\sharp: \mathscr{O}_Y\to f_*\mathscr{O}_X)$ be a morphism of schemes. Then the following are equivalent

  1. $f:X\to Y$ induces a homeomorphism between $X$ and a closed subset of $Y$ and furthermore $f^\sharp: \mathscr{O}_Y\to f_*\mathscr{O}_X$ is surjective (epimorphism).
  2. $(f,f^\sharp)$ is an affine morphism, and given any open affine subset $\mathrm{Spec} B\subset Y$ with $f^{-1}(\mathrm{Spec B}):=\mathrm{Spec A}\subset X$ affine open, the canonical ring homomorphism $f^\sharp_B:B\to A$ is surjective.

A morphism satisfying one (and therefore both) properties is called a closed immersion.

I want to make sure I got it right. Here is a sketch of proof:

($2\to 1$) $(f,f^\sharp)$ with def. 2 is a finite morphism, hence $f$ is closed. Given any surjection of ring $B\to A$, there is bijection $\mathrm{Spec}A\simeq \mathrm{Spec}B/I$ for some ideal. This means $f$ is also one-to-one. Hence the induced map $X\to f(X)$ is a continuous, closed, bijection and hence a homeomorphism. Moreover if $f_B^\sharp$ is surjective, then the induced map on stalks is the colimit of a bunch of surjections and hence is itself a surjection.

($1\to 2$) I think this is essentially a consequence of Exercise 3.11 (a,b) of Hartshorne. Part (a) says that closed immersions (def. 1) are stable under base change. Part (b) says a closed subscheme (def. 1) of an affine scheme is affine. Combining the two we find that $(f,f^\sharp)$ is an affine morphism. To show that $f^\sharp_B: B\to A$ is surjective, we use the fact that a homomorphism of $B$-modules $M\to N$ is a surjection iff for all prime ideals $\mathfrak{p}\subset B$, the induced maps $M_\mathfrak{p}\to N_{\mathfrak{p}}$ are surjective. This together with the fact that $f^\sharp_y$ are surjective for all $y\in \mathrm{Spec}B$ should finish the job.

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  • $\begingroup$ Looks good to me. $\endgroup$ – paf Aug 24 '16 at 23:19
  • $\begingroup$ I am agree with paf! $\endgroup$ – Armando j18eos Aug 28 '16 at 12:45

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