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Let $\mathcal C$ be a category with fiber products. We say that a morphism $X \to Y$ in $\mathcal C$ is an effective epimorphism if the sequence of sets

$$ \text{Hom}(Y,S) \to \text{Hom}(X,S) \rightrightarrows \text{Hom}(X \times_Y X, S) $$

is exact for every $S$. The geometric interpretation is that a morphism $X \to S$ descends uniquely to a morphism $Y \to S$ precisely when it is constant along the fibres of $X \to Y$.

Let $Y$ be a reduced noetherian scheme, and let $X$ be the sum of the irreducible components of $Y$, each induced with the reduced subscheme structure. Is it true that $X \to Y$ is an effective epimorphism? (In other words, to give a morphism $Y\to S$ is the same as giving a morphism $Y_i \to S$ for every irreducible component $Y_i$, in such a way that these morphism agree on the intersections.)

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  • $\begingroup$ This appears to be the definition of effective epimorphism, rather than strict epimorphism. Of course there is no difference in this context. $\endgroup$ – Zhen Lin Aug 24 '14 at 20:40
  • $\begingroup$ Dear @Zhen : Thank you $\endgroup$ – Bruno Joyal Aug 24 '14 at 20:42

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