## 4.13 Monomorphisms and Epimorphisms

Definition 4.13.1. Let $\mathcal{C}$ be a category and let $f : X \to Y$ be a morphism of $\mathcal{C}$.

1. We say that $f$ is a monomorphism if for every object $W$ and every pair of morphisms $a, b : W \to X$ such that $f \circ a = f \circ b$ we have $a = b$.

2. We say that $f$ is an epimorphism if for every object $W$ and every pair of morphisms $a, b : Y \to W$ such that $a \circ f = b \circ f$ we have $a = b$.

Example 4.13.2. In the category of sets the monomorphisms correspond to injective maps and the epimorphisms correspond to surjective maps.

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

Comment #8860 by Francisco Gallardo on

Proof of Lemma 08LR, last part of the proof. It says ''commutes and if $W\in \operatorname{Ob}(\mathcal{C})$ and $\alpha,\beta\colon X\to Y$ such that..." It should say $\alpha,\beta\colon W\to X$. Also, maybe I'm not seeing it, but saying that $X\cong X \times_Y X$ isn't enough? Should it be $(X,\operatorname{id}_X,\operatorname{id})\cong (X\times_Y X,\text{pr_1},\text{pr_2})$? Thanks in advance.

Comment #9229 by on

Thanks for the typo which is fixed here. Yes, strictly speaking we need to tell the reader the maps.

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).