**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 $Y\amalg _ X Y = Y$.
$\square$

## Comments (2)

Comment #8860 by Francisco Gallardo on

Comment #9229 by Stacks project on