(Assume that the affine varieties are all in an affine space over the algebraically closed field $k$).
In Gathmann's Algebraic Geometry, is proved that the morphisms of two affine varieties $X\to Y$ are in (a functorial) bijection with the homomorphisms of the coordinate rings $A(Y)\to A(X)$. Plus, one can easily adapt the proof to the case that $X$ is a prevariety and $Y$ an affine variety. But what can be said if $X$ is affine and $Y$ a prevariety?
I'm looking for an example where there is no such bijection. Consider $\mathbb P^1$, that is made gluing two copies of $\Bbb A^1$. First I tried to calculate the ring of regular functions on $\Bbb P^1$. Take a function regular on $\Bbb A^1$, that is an $f(t)\in k[t]$. Since the gluing isomorphism is defined $$\Bbb A^1-\{0\}\to \Bbb A^1-\{0\}: a\mapsto a^{-1},$$ on the other copy of $\mathbb A^1$ we would need a regular function that extends $f(\frac 1t)$, and this is possible only if $f(t)$ is constant. So I'd say that the regular functions on $\Bbb P^1$ are the elements of $k$.
At this point, we see that there are infinite ways to embed the affine line in $\Bbb P^1$, but the homomorphism $k\to k[t]$ is unique.
Is my example correct? In general, even if it isn't a bijection, does a map from morphisms (of prevarieties) to homomorphisms (of the algebras of regular functions) still exist? Thank you
Best Answer
Yes, your example is correct.
Given a map of locally ringed spaces $f:X\to Y$ there's always a map from the global sections of $\mathcal{O}_Y$ to the global sections of $\mathcal{O}_X$. In your case, that means a map from the regular functions on $Y$ to the regular functions on $X$. So there's a function $\operatorname{Mor}(X,Y)\to\operatorname{Mor}(\mathcal{O}_Y(Y),\mathcal{O}_X(X))$, and the result you mention in your post says that this is a bijection if $X$ and $Y$ are affine. In fact, by this result at the stacks project it is a bijection if $Y$ is affine. In general, if $Y$ isn't affine, your example with $\Bbb A^1\to\Bbb P^1$ shows that you cannot expect any sort of result.