# Cancelling etale morphisms

Let $f:Y\to Z$ be an etale 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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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 $g_1(x)=g_2(x)$? –  Laurent Moret-Bailly Nov 8 '12 at 21:50
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. –  Will Sawin Nov 8 '12 at 22:46
Laurent: agreed. –  Roman Fedorov Nov 9 '12 at 4:44
@Will Savin: you are right, I misunderstood the statement. –  Damian Rössler Nov 9 '12 at 18:07