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$\newcommand{\Spec}{\operatorname{Spec}}$I'm working on the following problem from Vakil's notes:

Show that (scheme) morphisms $X\to\Spec A$ are in natural bijection with ring morphisms $A\to\Gamma(X,\mathscr O_X)$.

Of course, any map $X\to\Spec A$ induces a map of global sections $A\to\Gamma(X,\mathscr O_X)$. I am working out the other direction. It is certainly true if $X$ is affine, because affine scheme morphisms $\Spec B\to\Spec A$ are in bijection with ring maps $A\to B$.

In the general case, we cover $X$ with affine open sets $\Spec B_i$. Then for each $i$ we get a map

$$A\to\Gamma(X,\mathcal O_X)\overset{\text{res}}{\to}\Gamma(\Spec B_i,\mathcal O_X)=B_i$$

which in turn gives a map $\pi_i:\Spec B_i\to\Spec A$ (in particular, if we let $\pi_i^{\#}$ be the map $A\to B_i$, then the map is given by $\pi_i(p)=(\pi_i^{\#})^{-1}(p)$).

How can I show that these glue correctly to give a morphism of schemes $\pi:X\to\Spec A$?

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    If I recall correctly, back when I took Hartsh-- I mean, algebraic geometry, our lecturer proved it by showing that any morphism $X \to \operatorname{Spec}A$ factors uniquely through $\operatorname{Spec}\Gamma(X, \mathscr O_X)$, and then you can apply the affine result. – 2017-01-11

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To show the $\pi_i$ glue, you just have to show that if $\operatorname{Spec} C$ is an affine open subset of $X$ that is contained in both $\operatorname{Spec} B_i$ and $\operatorname{Spec} B_j$, then $\pi_i$ and $\pi_j$ restrict to the same map on $\operatorname{Spec} C$ (since you can cover $\operatorname{Spec}B_i\cap\operatorname{Spec} B_j$ with such sets $\operatorname{Spec} C$). To restrict $\pi_i$ to $\operatorname{Spec C}$, you just compose the dual map $A\to B_i$ with the map $B_i\to C$. So you just have to show the compositions $$A\to \Gamma(X,\mathcal{O}_X)\to B_i\to C$$ and $$A\to\Gamma(X,\mathcal{O}_X)\to B_j\to C$$ are equal. But this is obvious, since they are both just the map $A\to\Gamma(X,\mathcal{O}_X)$ composed with the restriction map $$\Gamma(X,\mathcal{O}_X)\to\Gamma(\operatorname{Spec}C,\mathcal{O}_X)=C.$$

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    Wonderful as usual. I'm starting to notice a pattern developing here.. – 2017-01-11