Proposition 41.6.3. Let S is be a scheme. Let \pi : X \to S be unramified and separated. Let Y be an S-scheme and y \in Y a point. Let f, g : Y \to X be two S-morphisms. Assume
Y is connected
x = f(y) = g(y), and
the induced maps f^\sharp , g^\sharp : \kappa (x) \to \kappa (y) on residue fields are equal.
Then f = g.
Proof. The maps f, g : Y \to X define maps f', g' : Y \to X_ Y = Y \times _ S X which are sections of the structure map X_ Y \to Y. Note that f = g if and only if f' = g'. The structure map X_ Y \to Y is the base change of \pi and hence unramified and separated also (see Morphisms, Lemmas 29.35.5 and Schemes, Lemma 26.21.12). Thus according to Theorem 41.6.2 it suffices to prove that f' and g' pass through the same point of X_ Y. And this is exactly what the hypotheses (2) and (3) guarantee, namely f'(y) = g'(y) \in X_ Y. \square
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