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In point set topology, we have the following result, which is easily proved.

Theorem. Let $Y$ be Hausdorff space and $f,g:X \to Y$ be continuous functions. If there exists a set $A\subset X$ such that $\bar{A} = X$ and $f|_A = g|_A$, then $f=g$.

I was trying to understand what would be the natural generalization of this fact in the category of schemes. We know that the correct analogous of a Hausdorff space is a separated scheme. So I was thinking in a statement like this:

"Let $Y$ be a separated scheme and $f,g:X \to Y$ be morphisms of schemes. If there exists a set $A\subset X$ such that $\bar{A} = X$ and $f|_A = g|_A$, then $f=g$."

The first problem is that we must be careful with this restriction "$f|_A$", since $f$ is a morphism and I want to consider the morphism of sheaves also. Then I saw that Liu's book on Algebraic Geometry has the following statement:

"Let $Y$ be a separated scheme, $X$ a reduced scheme, and $f,g:X \to Y$ morphisms of schemes. If there exists a dense open subset $U$ such that $f|_U=g|_U$, then $f=g$."

Now this makes sense, since we are dealing with open subsets now. But I still find this result too restrictive. So I came up with this:

"Let $Y$ be a separated scheme, $X$ a reduced scheme, and $f,g:X \to Y$ morphisms of schemes. If there exists a morphism $\varphi:S \to X$ such that $\varphi(S)$ is dense in $X$ and $f\circ \varphi = g\circ \varphi$, then $f=g$."

It's easy to see that Liu's proof of the result concerning only the open set also applies to this context. Finally, let's go to the questions:

  1. Is this really the best generalization? Is there any other results in this direction that are at least slightly different?

  2. I can see where the "reduced" hypothesis enters in the proof, but I found it a little strange. Is it just a technical point or can be "understanded" in some sense? Maybe counterexamples of this fact when this hypothesis isn't valid would help to clarify , but I didn't think of any.

P.S. Sorry for the bad english.

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    Related: [Example of a dense but not schematically dense open subset](https://math.stackexchange.com/questions/2946046/example-of-a-dense-but-not-schematically-dense-open-subset)2018-11-27

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To answer your question (1): Let $X=\mathrm{Spec}(B)$ be such that $B\to O_X(U)$ is not injective for some dense open subset $U$ (this can't happen if $B$ is reduced), let $Y=\mathrm{Spec}\mathbb Z[t]$. Fix an element $b\in B$ non-zero such that $b|_U=0$.

Let $\varphi: \mathbb Z[t] \to B$ be any ring homomorphism and let $\psi : \mathbb Z[t] \to B$ be defined by $\psi(t)=\varphi(t)+b$. Then the corresponding morphisms $f, g : X\to Y$ coincide on $U$ but are not equal.

Standard example of such $B$: $B=k[x,y]/(x^2, xy)$ and $U=X\setminus \{ (0,0)\}=D(y)$.

For your question (1), you can replace the hypothesis $X$ reduced by $S\to X$ schematically dominant. When $X$ is noetherian, this means that the image of $S\to X$ contains the associated points of $X$ (= maximal points of $X$ if the latter is reduced).

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    @John, sorry I didn't give a reference for schematically dominant, but you did the right thing.2012-07-16