I'm going over some old exercises in Chapter I of Hartshorne and am stuck on the reverse direction of Exercise I.3.3(b), which should follow straight from definitions. If $\varphi:X\to Y$ is a morphism of (quasiprojective) varieties over an algebraically closed field $k$, then $\varphi$ is an isomorphism if and only if it's a homeomorphism and the induced maps $\varphi_{P}^*:\mathcal O_{Y,\varphi(P)}\to\mathcal{O}_{X,P}$ are isomorphisms for all $P\in X$.
The forward direction is easy. For the converse, we want to show that $\varphi^{-1}$ is a morphism, i.e. if $g:V\to k$ is a regular function on some open $V\subset X$, then $g\circ\varphi^{-1}$ is regular on $\varphi(V)$.
Suppose $P\in V$. Since $\varphi_P^*$ is an isomorphism, there's some $\langle U,f\rangle\in\mathcal{O}_{Y,\varphi(P)}$ such that
$$ \langle V,g\rangle=\varphi_P^*(\langle U,f\rangle)=\langle \varphi^{-1}(U),f\circ\varphi\rangle \tag{1} $$
hence $(f\circ \varphi)\vert_{V\cap \varphi^{-1}(U)}=g\vert_{V\cap \varphi^{-1}(U)}$. Since $\varphi$ is a homeomorphism, composing with $\varphi^{-1}$ restricted to $\varphi(V\cap \varphi^{-1}(U))=\varphi(V)\cap U$ gives
$$f\vert_{\varphi(V)\cap U} = (g\circ\varphi^{-1})\vert_{\varphi(V)\cap U}$$
As $f$ is regular on $\varphi(V)\cap U$, so is $g\circ\varphi^{-1}$.
But we need $g\circ\varphi^{-1}$ to be regular on all of $\varphi(V)$. We obviously want to be able to take $U=\varphi(V)$ in eq. (1), but a priori we don't know that $(\varphi_P^*)^{-1}$ maps pairs $\langle V,g\rangle$ to ones of the form $\langle \varphi(V),g\circ\varphi^{-1}\rangle$ (this would be assuming the result); perhaps $g\circ\varphi^{-1}$ turns out not to be regular on $\varphi(V)$ and we have to pass to a smaller open subset of it.
So my confusion is in not knowing what we can say about $(\varphi_P^*)^{-1}$. What am I missing?
Best Answer
You're not missing anything. Indeed, the definition of regular function is local: a function $f:X \to k$ is regular on $X$ if it is regular at every point $P \in X$. Since your argument proves that $g \circ \varphi^{-1}$ is regular at every point of $\varphi(V)$, you're done!
Notice that this exercise is a special instance of a much more general statement, namely that a morphism of sheaves (of sets) is an isomorphism if and only if it induces an isomorphism of all stalks.