Let $f:Y\to Z$ be an étale morphism, $g_1, g_2:X\to Y$ be morphisms. Assume that for all closed points $x\in X$ we have $g_1(x)=g_2(x)$ and that $f\circ g_1=f\circ g_2$. Can I conclude (under some reasonable conditions on $X$, $Y$, and $Z$) that $g_1=g_2$?
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2$\begingroup$ Take $X=Y=\mathrm{Spec}(\mathbb{C})$, $Z=\mathrm{Spec}(\mathbb{R})$, $g_1=$ identity, $g_2=$ complex conjugation. Perhaps the question is: what do you exactly mean by $g_1(x)=g_2(x)$? $\endgroup$– Laurent Moret-BaillyNov 8, 2012 at 21:50
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$\begingroup$ Damian: If $f$ is the identity then $f \circ g_1 =f \circ g_2$ implies $g_1=g_2$, so I'm not sure what you mean by that. $\endgroup$– Will SawinNov 8, 2012 at 22:46
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$\begingroup$ Laurent: agreed. $\endgroup$– Roman FedorovNov 9, 2012 at 4:44
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$\begingroup$ @Will Savin: you are right, I misunderstood the statement. $\endgroup$– Damian RösslerNov 9, 2012 at 18:07
1 Answer
As pointed out in the comments, by Laurent Moret-Bailly, the example $\mathrm{Spec}(\mathbb{C}) \rightrightarrows \mathrm{Spec}(\mathbb{C}) \to \mathrm{Spec}(\mathbb{R})$ shows that we need to be more careful about what $g_1(x) = g_2(x)$ means. The correct notion is that for each finite type map $p : \mathrm{Spec}(k) \to X$ from a field we have that $g_1 \circ p = g_2 \circ p$. If $X,Y,Z$ are finite-type $k$-schemes with $k = \bar{k}$ then this condition is the same as saying that $g_1$ and $g_2$ agree on the underlying topological space of closed points.
There are also weird schemes $X$ with no closed points. Then consider the two maps $X \rightrightarrows X \sqcup X$ mapping to each of the two components. These agree after applying the natural étale map $X \sqcup X \to X$. Therefore we should assume that $X$ is Jacobson (meaning its closed points are dense in every closed subset).
Let $E \subset Y$ be the locus on which $g_1$ and $g_2$ agree, explicitly this is the pullback along $(g_1, g_2) : X \to Y \times_Z Y$ of the relative diagonal $\Delta_{f} : Y \to Y \times_Z Y$. Because $f$ is étale $\Delta_{f}$ is open (if $f$ is also separated then $\Delta_f$ is also closed). Therefore $E \subset X$ is an open immersion (also a closed immersion if $f$ is separated). For any point $x \in X$ we assumed that $g_1(x) = g_2(x)$ which really means that $\mathrm{Spec}(\kappa(x)) \to X \rightrightarrows Y$ are the same map. Therefore $\mathrm{Spec}(\kappa(x)) \to X$ factors through $E$ which implies that $E = X$ because it is an open subscheme containing every closed point and the closed points are dense in $E^C$ so $E^C$ must be empty.