I'm trying to prove this statement
Let $ f,g:X\rightarrow Y$ be two $ S $-scheme morphisms that agree on $ U $, a dense open subset of $ X $. If $ X $ is reduced and $ Y$ separated, then $ f = g $.
I've gone so far as showing that the locus of agreement of $ f $ and $ g $ must be all of $ X $. The locus of agreement is a closed subscheme of $ X $ defined via this universal property
If $h:Z\rightarrow X$ is another $ S $-morphism such that $ f\circ h=g\circ h $, then $ h $ factors uniquely through the locus of agreement.
How can I show that since the locus of agreement is all of $ X$, then $ f=g$? It sounds intuitive, but I'm unable to show it via the universal property above.
Thank you
Best Answer
I was able to answer this soon after posting. The fact that $ X $ is the locus of agreement gives a commutative diagram. By composing each path in the diagram by an appropriate projection I see that the morphisms are indeed equal.