Let $\varphi:X\to Y$ be a morphism of varieties and let $X$ be irreducible. If $U\subseteq X$ is a non-empty open subset, is it true that $\varphi$ is the only extension to $X$ of $\varphi|_U$?
I know that this is true for affine varieties. I was wondering if it stays true in this broader setting?
(I am using Gathmann's notes).
This question came up while I was trying to prove that every morphism $\mathbb{P}^n \to \mathbb{P}^{m}$ is basically given by $m+1$ homogeneous polynomials.
Best Answer
It depends what you mean by varieties. Here is the correct result:
Theorem (Reduced to separated extension theorem). Suppose $f,g:X\to Y$ are two morphisms where $X$ is reduced and $Y$ is separated. If there exists a dense open $U\subset X$ where $f|_U = g|_U$, then $f=g$.
If a variety for you includes the conditions reduced and separated, then yes! If not, then no - for instance, the two inclusions of $\Bbb A^1$ in to the line with two origins fail this because the line with two origins is non-separated.