Two continuous functions agree on an open subset of an irreducible space.

Question

I have the varieties $X,Y,Z$ where $X$ is complete. I have the morphism of varieties $f:X\times Y\rightarrow Z$, I have a closed subset $W\subset Y$ such that $f=g\circ\textrm{pr}_Y$ on $X\times(Y\setminus W)$ where $g(y)=f(x_0,y)$ for some $x_0\in X$. Is this sufficient to say that $f=g\circ\textrm{pr}_Y$ on all of $X\times Y$?

Answer 1

Assuming by variety, you mean that the associated topological space is irreducible, then yes, $f = g \circ \text{pr}_Y$ on all of $X \times Y$. (Also, I'm assuming that you implicitly have $W$ is a proper closed subset of $Y$, i.e. $W \neq Y$.)

This follows from the more general fact that if $f, g : X \rightarrow Y$ are two morphisms of varieties, then $Z = \{ x \in X : f(x) = g(x) \}$ is a closed subset of $X$. Indeed, $Z$ is equal to the inverse image of the diagonal (a closed subset) of $Y \times Y$ under the natural map $f \times g : X \rightarrow Y \times Y$, and so is closed since $f \times g$ is continuous. If $Z$ contains a non-empty open set and $X$ is irreducible, then we must have $Z = X$, for the only closed subset of an irreducible topological space that contains a non-empty open subset is the whole space. (Apologies that the notation in this paragraph is not consistent with your notation, but I think it easier to follow the general principle if I change the setup.)