Restriction of a morphism is a morphism

Question

Let $X,Y$ be quasi-projective varieties and $f:X\to Y$ a morphism. If $X',Y'$ are closed subvarieties of $X,Y$ respectively such that $f(X')\subset Y'$, prove that $f|_{X'}:X'\to Y'$ is a morphism.

I know how to prove this when $X',Y'$ are open subvarieties and I also when $X,Y$ are affine.

I also know the following result: if $\{\mathcal{U}_i\}_{i\in I}$ and $\{\mathcal{V}_i\}_{i\in I}$ are open covers of $X$ and $Y$ respectively with $f(\mathcal{U}_i)\subset \mathcal{V}_i$, then $f:X\to Y$ is a morphism if and only if $f:\mathcal{U}_i\to \mathcal{V}_i$ are morphisms $\forall i$.

Now this is what I'm trying to do: if I can take the above covers $\{\mathcal{U}_i\}_{i\in I}$ and $\{\mathcal{V}_i\}_{i\in I}$ to be affine, then $\mathcal{U}_i\cap X'$ is an affine subvariety of $\mathcal{U}_i$, so $f:\mathcal{U}_i\cap X'\to\mathcal{V}_i\cap Y'$ is a morphism $\forall i$ and consequently $f:X'\to Y'$ is a morphism.

My difficulty is proving that such covers exist, but I'm not even sure that they do.

If there is a better strategy for the whole thing, I'd also be happy to know about it. Thank you!

Answer 1

Step 1. Let $Y$ be affine. Consider the affine open cover of $X$.

Step 2. Consider the affine open cover of $Y$ and its preimage on $X$.

Answer 2

(i) If $X$ is an open or closed subvariety of $Y$, the inclusion $i: X \to Y$ is a morphism.

(ii) The composition of morphisms is a morphism

So let $i: X' \to X$ the inclusion of the subvariety $X'$ in $X$. This is by (i) a morphism. So by (ii) $f \circ i: X' \to Y$ is a morphism. Because of $f(X') \subset Y'$ the image of $f \circ i$ is in $Y'$. Therefore $\left.f\right|_{X'}$ is a morphism.