$\newcommand{\O}{\operatorname{\mathcal{O}}}$ $\newcommand{\A}{\operatorname{\mathbb{A}}}$ $\renewcommand{\subset}{\subseteq}$

I'm trying to solve Exercise 5.11 from Gathmann's 2019/20 Algebraic Geometry notes...


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 $\phi$?


1 Answer 1


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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .