Lemma 4.13.3. Let $\mathcal{C}$ be a category, and let $f : X \to Y$ be a morphism of $\mathcal{C}$. Then
$f$ is a monomorphism if and only if $X$ is the fibre product $X \times _ Y X$, and
$f$ 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
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
commutes and if $W\in \mathop{\mathrm{Ob}}\nolimits (\mathcal C)$ and $\alpha ,\beta : X\to Y$ such that $ f\circ \alpha = f\circ \beta $, we have a unique $\gamma $ verifying
which proves that $\alpha = \beta $.
The proof is exactly the same for the second point, but with the pushout $Y\amalg _ X Y = Y$. $\square$
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)
There are also: