I am reading a book called Toric Varieties by Cox et. al, where I came across this definition of a morphism. Let $U_i \subseteq V_i$ be Zariski open sets of affine varieties for $i=1,2$. A mapping $\Phi:U_1 \rightarrow U_2$ is morphism if $\phi \rightarrow \phi \circ \Phi$ defines a map from $\phi^*:\mathcal{O}_{V_2}(U_2) \rightarrow \mathcal{O}_{V_1}(U_1)$, where $\mathcal{O}_{V_i}(U_i)$ is the set of all regular functions from $U_i \rightarrow \mathbb{C}$.
Now I should mention the definition of regular function also.
Consider $V=\operatorname{Spec}(R)$ be an affine variety. Let $f \in R$ be a polynomial function on $R$, then $V_f=V\setminus V(f)$ is a Zariski open set. Now, we know that these are affine open sets which form basis for Zariski topology in $V$. Now, we will call a function $U \rightarrow \mathbb{C}$ is a regular function, if for every $p \in U$, there exists $f_p \in R$ such that $p \in V_{f_p} \subseteq U$ and $\phi|_{V_{f_p}}\in R_{f_p}$. So essentially, for every $p$, $\phi(p)=a_p/f_p^{n_p}$ with $f_p(p) \neq 0$ and $n_p \geq 0.$
I don't understand how a morphism is a continuous function in the Zariski Topology.
I will be grateful if you can help me on this.
Best Answer
For $V_1$ and $V_2$ affine varieties, to show a morphism $\Phi: V_1\to V_2$ is continuous.
Enough to show that for $D(f)\subset V_2$ (where $D(f)=V_2-V(f)$ for $f$ a polynomial function on $V_2$), $\Phi^{-1}(D(f))$ is open in $V_1$.
Now $\phi^*: \mathcal{O}(V_2)\to \mathcal{O}(V_1)$ takes regular functions on $V_2$ to $V_1.$ Let $g:=\phi^*(f)$ regular function on $V_1$. Check $D(g)=\Phi^{-1}(D(f))$.
Now to see in general: for $V_1$ and $V_2$ (not necessarily affine) varieties, a morphism $\Phi: V_1\to V_2$ is continuous.
So we want to see that for any open subset $U_2\subset V_2$, $\Phi^{-1}(U_2)$ is open in $V_1$.
Cover $U_2$ by basic open sets $D(f_i)$'s. Now $$\Phi^{-1}(U_2)=\bigcup_{i}\Phi^{-1}(D(f_i)) .$$
So $$\Phi^{-1}(U_2)=\bigcup_{i}D(\phi^*(f_i)).$$
Hence $\Phi^{-1}(U_2)$ is open in $V_1$.