Joint Epi and Joint Mono

The universal property of (binary) products states that a pair of maps $f : X \to A$ and $g : X \to B$ uniquely determines the map

$$(f, g) : X \to A \times B.$$

It works in reverse too. Given a map $s : X \to A \times B$, we can form two maps

$$u = \pi_1 \circ s : X \to A \quad \text{and} \quad v = \pi_2 \circ s : X \to B$$

such that $(u, v) = s$.

This gives rise to an isomorphism of the corresponding homs, specifically

$$\hom(X, A \times B) \cong \hom(X,A) \times \hom(X,B).$$

This gives us a convenient way of establishing equality of morphisms via probing them by projections. If $\pi_1 \circ a = \pi_1 \circ b$ and $\pi_2 \circ a = \pi_2 \circ b,$ then $a = b$. This is the joint monic property of products.

The coproduct situation is entirely analogous. Given a pair of maps $h : A \to X$ and $k : B \to X$ we have the unique copairing

$$[h, k] : A \amalg B \to X.$$

So from $s : A \amalg B \to X$ we can form

$$q = s \circ i_1 : A \to X \quad \text{and} \quad p = s \circ i_2 : B \to X$$

such that $[q, p] = s$.

There is also the corresponding isomorphism of homs:

$$\hom(A \amalg B,X) \cong \hom(A,X) \times \hom(B,X).$$

Hence, if $c \circ i_1 = d \circ i_1$ and $c \circ i_2 = d \circ i_2$, then $c = d$. This is the joint epic property of coproducts.