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Let $X\subseteq A^m$ and $Y\subseteq A^n$ be affine varieties with affine algebras $K[X]$ and $K[Y]$, respectively. Why is every $K$-algebra homomorphism $\theta\colon K[Y]\to K[X]$ obtained as the comorphism of a morphism $\phi\colon X\to Y$?

We know that $K[Y]=K[T_1,\ldots,T_n]/\mathscr{I}(Y)$, so if we let $t_i$ denote the equivalence class of $T_i$ then $\theta$ is uniquely determined by its values on the $t_i$. We can define a morphism $\phi\colon X\to A^n$ as $\phi(x)=(\psi_1(x),\ldots,\psi_n(x))$, where $\psi_i=\theta(t_i)$. If we know that $\phi(X)\subseteq Y$, then it is clear that $\phi^*=\theta$ as they agree on the $t_i$, where $\phi^*$ is the induced comorphism of $\phi$. However, how do we know that $\phi(X)\subseteq Y$?

Gregoire Rad
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To show $\phi(X)\subseteq Y$, we have to show that for any $x\in X$ and any $g\in\mathscr{I}(Y)$, $g(\phi(x))=0$. But since $g\in\mathscr{I}(Y)$, we know that the equivalence class of $g$ in $K[Y]$ is $0$, and in particular its image under $\theta$ remains $0$. That means that when we replace each variable $t_i$ of $g$ with $\psi_i$ (i.e., apply $\theta$ to $g$), we get $0$. That is, $g(\psi_1,\dots,\psi_n)=0$. Now $$g(\phi(x))=g(\psi_1(x),\dots,\psi_n(x))$$ is just $g(\psi_1,\dots,\psi_n)$ evaluated at $x\in X$, so it is $0$, as desired.

Eric Wofsey
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