I want to understand a proof that establishes the fact that every map between abstract algebraic varieties (ie, a ringed space on k-algebras which is locally isomorphic to a Zariski closed on the affine space) as ringed spaces is a regular map. The definitions I am using are:
- A morphism of ringed spaces is a pair $(f,\phi):(X,\mathscr{O}_X)\to(Y,\mathscr{O}_Y)$ where $f:X\to Y$ is continuous and $\phi:\mathscr{O}_Y\to f_*\mathscr{O}_X$ is a sheaf morphism.
- A regular map between the varieties $(X,\mathscr{O}_X)$ and $(Y,\mathscr{O}_Y)$ is a continuous function such that its pullback defines a sheaf map $f^*: \mathscr{O}_Y\to f_*\mathscr{O}_X$, ie, for all open $V\subset Y$ and all regular functions $u\in\mathscr{O}_Y(V)$ the pullback $f ^*(u)=u\circ f\in f_*(\mathscr{O}_X(V)):=\mathscr{O}_X(f^{-1}(V))$ is regular on $X $.
The theorem then consists in proving that every morphism of ringed spaces between varieties is regular and that always $\phi=f^*$.
The first simplification made in the proof is to assume that both $Y$ and $X$ are affine varieties. I would be grateful if you could explain to me in detail why this can be assumed and how to extend it to the general case where $X,Y$ are arbitrary varieties. Then what is proved is that the preimage of a maximal ideal in $\mathscr{O}_X$ by $\phi$ is also maximal and finally uses this to conclude that $\phi=f^*$. Please, I would be grateful if you could explain me in detail why the aforementioned assumption can be made. Thanks.
They are trying to prove something about morphisms of sheaves (namely $\phi=f^\ast$), and sheaves are local in nature - if two maps of sheaves agree locally then the maps are equal as a global section $s$ is glued uniquely from local sections (and since varieties locally are affine varieties, we can assume the varieties are affine)
In more detail: let $\{V_\alpha\}_\alpha$ be a open cover of $Y$ by affine varieties, and $\{U_{\alpha, i}\}_i$ an open cover of $f^{-1}(V_\alpha)$ by affines. For each $\alpha, i$ suppose we have proven that the restriction of $(f, \phi)$ to $U_{\alpha,i} \to V_\alpha$ satisfies $\phi^*=f$, let $g_{\alpha_i}$ be this common map (composed with a restriction)
$$g_{\alpha, i}:\mathcal{O}_Y\vert_{V_\alpha} \to (f_*\mathcal{O}_X)\vert_{V_\alpha} \to \mathcal{O}_X\vert_{U_{\alpha,i}}$$ Let $V$ in $Y$ be open, and let $s \in \mathcal{O}_Y(V)$. Then $\phi(s)\in \mathcal{O}_X(f^{-1}V)$ is the unique section that when restricted to $U_{\alpha, i}$ equals $g_{\alpha,i}(s)$. The same can be said of $f(s)$. So since local sections glue uniquely for the sheaf $f_*\mathcal{O}_X$, we must have $\phi(s)=f(s)$.