态射的限制仍是态射：拟射影簇闭子簇情形的证明问询

Alright, let's work through this proof step by step using the tools you've referenced: the open subcase result, affine case proof, and the open cover criterion for morphisms.

First, let's restate the problem clearly:

Let $X,Y$ be quasi-projective varieties, $f:X\to Y$ a morphism. Suppose $X'\subset X$, $Y'\subset Y$ are closed subvarieties with $f(X')\subset Y'$. We need to show $f|_{X'}:X'\to Y'$ is a morphism.

We know two key facts going in:

• The restriction of a morphism to an open subvariety is a morphism, and we already have the proof for the affine case of this problem.
• If ${\mathcal{U}i}{i\in I}$ is an open cover of $X$, ${\mathcal{V}i}{i\in I}$ is an open cover of $Y$, and $f(\mathcal{U}_i)\subset \mathcal{V}i$ for all $i$, then $f$ is a morphism iff each $f|{\mathcal{U}_i}:\mathcal{U}_i\to \mathcal{V}_i$ is a morphism.

Step 1: Build Compatible Open Covers #

Since $X$ and $Y$ are quasi-projective, we can take an affine open cover ${\mathcal{V}i}{i\in I}$ of $Y$ (every quasi-projective variety has an affine open cover). For each $\mathcal{V}_i$, define $\mathcal{U}_i = f^{-1}(\mathcal{V}_i)$: this is an open subvariety of $X$ because morphisms are continuous, so preimages of open sets are open. The collection ${\mathcal{U}_i}$ is an open cover of $X$.

Now restrict these covers to our closed subvarieties:

• For each $i$, let $\mathcal{U}_i' = X' \cap \mathcal{U}_i$. This is an open subvariety of $X'$ (intersecting a closed subvariety with an open subvariety of the ambient space gives an open subvariety of the closed subvariety).
• Similarly, $\mathcal{V}_i' = Y' \cap \mathcal{V}_i$, which is an open subvariety of $Y'$.

We can verify the inclusion holds locally:
$$f|_{X'}(\mathcal{U}_i') = f(X' \cap f^{-1}(\mathcal{V}_i)) = f(X') \cap \mathcal{V}_i \subset Y' \cap \mathcal{V}_i = \mathcal{V}_i'$$
So ${\mathcal{U}_i'}$ is an open cover of $X'$, ${\mathcal{V}i'}$ is an open cover of $Y'$, and $f|{X'}$ maps each $\mathcal{U}_i'$ into $\mathcal{V}_i'$.

Step 2: Apply the Affine Case Result #

Now look at each local restriction $f|_{\mathcal{U}_i'}: \mathcal{U}_i' \to \mathcal{V}_i'$:

• $\mathcal{V}_i$ is affine, and $\mathcal{V}_i' = Y' \cap \mathcal{V}_i$ is a closed subvariety of an affine variety, so $\mathcal{V}_i'$ is affine (closed subvarieties of affine varieties are affine).
• $\mathcal{U}i$ is an open subvariety of $X$, so $f|{\mathcal{U}_i}:\mathcal{U}_i\to \mathcal{V}_i$ is a morphism (this is the open subcase result you mentioned).
• $\mathcal{U}_i' = X' \cap \mathcal{U}_i$ is a closed subvariety of $\mathcal{U}i$, and $f|{\mathcal{U}_i}(\mathcal{U}_i') \subset \mathcal{V}_i'$.

This is exactly the setup for the affine case proof you already know! Applying that result, each $f|_{\mathcal{U}_i'}: \mathcal{U}_i' \to \mathcal{V}_i'$ is a morphism.

Step 3: Use the Open Cover Criterion #

We've now shown:

• ${\mathcal{U}_i'}$ covers $X'$, ${\mathcal{V}_i'}$ covers $Y'$
• For every $i$, $f|_{X'}(\mathcal{U}_i') \subset \mathcal{V}_i'$, and the restriction to each pair is a morphism.

By the open cover criterion for morphisms (the second fact you noted), this implies that $f|_{X'}:X'\to Y'$ is a morphism.

内容的提问来源于stack exchange，提问作者rmdmc89