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

1. $Y$ is connected

2. $x = f(y) = g(y)$, and

3. 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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