I am reading the Hartshorne's algebraic geometry now, and by his definition for the morphism of varieties:
A continuous map $\varphi:X\to Y$, such that for every open set $V\subseteq Y$ and every regular function $f:Y\to \mathbb{K}$, $f\circ \varphi:\varphi^{-1}(V)\to \mathbb{K}$ is also a regular function.
However, the open set $\varphi^{-1}(V)$ may be empty, then how can we consider the function $f\circ \varphi:\emptyset\to \mathbb{K}$ as a regular function?
If such a function on empty set could be a regular function, why Hartshorne emphasis the open set $\varphi_U^{-1}(V)$ is nonempty when $\langle U,\varphi_U\rangle$ is a dominant rational map? And he also emphasis that $f\circ\varphi_U$ is a regular function on $\varphi_U^{-1}(V)$ because $\varphi_U^{-1}(V)$ is nonempty ($\varphi_U(U)$ is dense in Y).
Best Answer
As Captain Lama's answer, empty function $\emptyset\to \mathbb{K}$ is a regular function. However, empty function $\emptyset\to \mathbb{K}$ should not be in the field $K(X)$ of a variety $X$, since if $\langle\emptyset,\emptyset\to \mathbb{K}\rangle$ in the field $K(X)$, then the equivalence relation "$\sim$" on $K(X)$ would not be well-defined:
For two elements $\langle U,\varphi_U\rangle,\langle V,\phi_V\rangle$ in $K(X)$ where $\varphi_U\neq\phi_V$ on $U\cap V$, then we have $$\langle U,\varphi_U\rangle\sim \langle\emptyset,\emptyset\to \mathbb{K}\rangle \text{ and } \langle\emptyset,\emptyset\to \mathbb{K}\rangle\sim \langle V,\phi_V\rangle, $$ so we could conclude $\langle U,\varphi_U\rangle\sim \langle V,\phi_V\rangle$ which contradicts to $\varphi_U\neq\phi_V$ on $U\cap V$.