When is bijective map between closed point of varieties a morphism?

Question

Let $f:X\rightarrow Y$ be a bijective map between complex varieties, when will it be a morphism?

I meet this question when working over Fourier–Mukai transforms in algebraic geometry and some papers. In Corollary 5.23 of the book above, it induces a morphism from a bijection $f$ between smooth projective complex varieties as follows.

Let $P$ be a $X$-flat sheaf on $X\times Y$ such that $P\rvert_{\{x\}\times Y}\cong k(f(x))$ is the skyscraper sheaf for each closed point $x\in X$. Then $f$ induces a morphism $X\rightarrow Y$.

How does it work? Also, I meet in some papers in complex algebraic geometry that people define a morphism by claiming on closed points. Is it something about GAGA?

Answer 1

By Nakayama's lemma, the support of a coherent sheaf $\mathscr F$ on a Noetherian scheme $X$ is the set of $x \in X$ such that $\mathscr F \otimes_{\mathcal O_X} \kappa(x) \neq 0$ (as opposed to $\mathscr F \otimes_{\mathcal O_X} \mathcal O_{X,x} \neq 0$ as in the definition of support). Thus, we see that $\mathcal P$ is supported on a closed subset $\Gamma \subseteq X \times Y$ whose restriction to $\{x\} \times Y$ is the point $(x,f(x))$ for any $x \in X$. This means that the fibres of the first projection $\pi \colon \Gamma \to X$ above closed points are closed points, hence $\pi$ is quasi-finite [Tag 01TI]. Since $\Gamma$ is closed and $Y$ is proper, we also get that $\pi$ is proper, so $\pi$ is finite [Tag 02LS]. Since the fibres have length $1$, we conclude that $\pi$ is an isomorphism, i.e. $\Gamma$ is the graph of a morphism $g \colon X \to Y$. We necessarily have $g(x) = f(x)$ for all $x \in X$, meaning $g$ is the morphism we were looking for.

If you want, you can also check that $\mathcal P$ is a line bundle on $\Gamma$: it corresponds to a coherent sheaf on $\Gamma$ whose fibres all have length $1$, hence $\mathcal P$ is locally free of rank $1$ [Tag 0FWH]. This is probably what Huybrechts means by "choose local sections": pass to an open $U \subseteq X$ with preimage $V \subseteq \Gamma$ on which $\mathcal P|_V$ is free of rank $1$, choose an isomorphism $\mathcal P|_V \cong \mathcal O_V$, and use $\mathcal O_U \stackrel\sim\leftarrow \mathcal O_V \twoheadleftarrow \mathcal O_{U \times Y} \hookleftarrow \mathcal O_Y$ to define a morphism $g_U \colon U \to Y$ that agrees set-theoretically with $f|_U$ since $\mathcal O_V|_{\{x\} \times Y} \cong \mathcal P|_{\{x\}\times Y} \cong \kappa(f(x))$ for $x \in U$. I bypassed this by using the support, which is maybe a bit cleaner.

Answer 2

It's important to distinguish two things. One is what it takes to uniquely specify a morphism, and the other is what it takes to construct a morphism.

When we say a morphism is uniquely specified by the induced map on closed points, we're saying there do not exist morphisms $f_1,f_2$ which match each closed point to the same point but are unequal. This holds for morphisms between varieties (or even reduced schemes of finite type) over an infinite field.

We're not claiming that we can find the algebraic morphism in a useful way given the map on closed points, or give a nice criterion for a map on closed points to come from a map of varieties. I don't think such a criterion exists, beyond the tautological one coming from the definition, outside the case of varieties of dimension $0$ ("a map from the point to the set of closed points of $X$ comes from a map of varieties if and only if it sends the point to a point with residue field the base field")

Instead, in cases like yours, explaining what the map does on closed points is just a convenient way to describe it. To actually find the polynomial equations describing this map, you'd want to look towards the sheaf $P$ and the ideal of functions on $X \times Y$ vanishing on it.