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The Stacks project

Lemma 4.13.3. Let \mathcal{C} be a category, and let f : X \to Y be a morphism of \mathcal{C}. Then

  1. f is a monomorphism if and only if X is the fibre product X \times _ Y X, and

  2. f is an epimorphism if and only if Y is the pushout Y \amalg _ X Y.

Proof. Let suppose that f is a monomorphism. Let W be an object of \mathcal C and \alpha , \beta \in \mathop{\mathrm{Mor}}\nolimits _\mathcal C(W,X) such that f\circ \alpha = f\circ \beta . Therefore \alpha = \beta as f is monic. In addition, we have the commutative diagram

\xymatrix{ X \ar[r]^{\text{id}_ X} \ar[d]_{\text{id}_ X} & X \ar[d]^{f} \\ X \ar[r]^{f} & Y }

which verify the universal property with \gamma := \alpha = \beta . Thus X is indeed the fibre product X\times _ Y X.

Suppose that X \times _ Y X \cong X . The diagram

\xymatrix{ X \ar[r]^{\text{id}_ X} \ar[d]_{\text{id}_ X} & X \ar[d]^{f} \\ X \ar[r]^{f} & Y }

commutes and if W \in \mathop{\mathrm{Ob}}\nolimits (\mathcal C) and \alpha , \beta : W \to X such that f \circ \alpha = f \circ \beta , we have a unique \gamma verifying

\gamma = \text{id}_ X\circ \gamma = \alpha = \beta

which proves that \alpha = \beta .

The proof is exactly the same for the second point, but with the pushout Y\amalg _ X Y = Y. \square


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