Confusion about proving that closed subprevarieties are actually prevarieties…

algebraic-geometrysheaf-theory

$\newcommand{\O}{\operatorname{\mathcal{O}}}$
$\newcommand{\A}{\operatorname{\mathbb{A}}}$
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I'm trying to solve Exercise 5.11 from Gathmann's 2019/20 Algebraic Geometry notes…

https://www.mathematik.uni-kl.de/~gathmann/de/alggeom.php


I think I have understood the statement of the exercise – I think it says the following:

Exercise 5.11. Let $X$ be a prevariety and let $U\subset X$ be an open affine subset.

Let $\phi:(U,\O_X|_U)\to (Z,\O_Z)$ be an isomorphism where $Z\subset \A^n$ is Zariski-closed.

Let $Y\subset X$ be a closed subset.

Then $W:=\phi(U\cap Y)$ is a Zariski closed subset of $\A^n.$

Show that $\phi:U\cap Y \to W$ is an isomorphism of ringed spaces.


So basically we need to show that, for any open $V\subset W,$ the pullback map
$$\phi_U:\O_W(V)\to \O_Y(\phi^{-1}V)$$
is an isomorphism (right?).

But how can we do this without knowing anything about $\phi$?

Best Answer

We do know that $\phi$ induces an isomorphism on all stalks $\mathcal{O}_{X,p}\simeq\mathcal{O}_{Z,\phi(p)}$ for all $p\in U$. By the sheaf properties (and the fact that $\phi$ is already a globally defined function), we can "glue" the stalks to see that $\phi$ induces an isomorphism on the entirety of $V\subseteq W$. This all follows from the important fact that a morphism of sheaves is an isomorphism if and only if it is an isomorphism at all stalks.

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