On Lemma 4.1 of Hartshorne's algebraic geometry text I'm in the process of teaching myself algebraic geometry from Hartshorne. Lemma 4.1 says that if we let $X$ and $Y$ be two varieties, and let $\phi$ and $\psi$ be two morphisms from $X$ to $Y$, and suppose there is a nonempty open subset $U \subseteq X$ such that $\phi|_U = \psi|_U$, then $\phi = \psi$. 
However, I'm curious about what happens when we change the conditions on $\phi$ and $\psi$. What if instead of having morphisms, we have two regular maps, or just two continuous maps (with respect to the Zariski topology). 
I'm pretty sure that a regular map defined on an open subset has a well-defined extension, but in the case of just continuity, I think that this fails..

The argument is just topological.  Consider the product map $\phi \times \psi : X \rightarrow Y \times Y$.  Then the locus $Z \subset X$ where $\phi$ and $\psi$ are equal is just the preimage of the diagonal of $Y \times Y$, which is closed.  Thus, if $\phi$ and $\psi$ are both continuous, then $Z$ is closed.  Since $Z$ contains the (non-empty) open set $U$ and $X$ is (by hypothesis) connected, $Z = X$.
