# Idea

Copairing is dual to pairing.

# Definition

Let $X$ and $Y$ be objects of some category $C$, and suppose that the coproduct $X \sqcup Y$ exists in $C$.

Let $T$ be some object of $C$, and let $a\colon X \to T$ and $b\colon Y \to T$ be morphisms of $C$. Then, by definition of coproduct, there exists a unique morphism $[a,b]\colon X \sqcup Y \to T$ such that the obvious diagrams commute.

This $[a,b]$ is the copairing of $a$ and $b$.

# Notation

When convenient, it is nice to write it vertically; all of the following are seen:

$\left({a \atop b}\right) ,\quad \left[{a \atop b}\right] ,\quad \left\{{a \atop b}\right\} .$

The vertical notation can be combined with pairing to create a matrix calculus for morphisms from a coproduct to a product. This works best when products and coproducts are the same, as described at matrix calculus.

# Examples

One often sees a function defined by cases as follows:

$f(x) = \left\{\array{ g(x) & \text {if}\; \phi(x) \\ h(x) & \text {if}\; \psi(x) .}\right.$

Such a definition is valid in general iff the domain of $f$ is the (internal) disjoint union of its subsets $\{x : \phi(x)\}$ and $\{x : \psi(x)\}$. In that case, let $X$ be the first subset, let $Y$ be the second, and let $T$ be the target of $f$; let $a\colon X \to T$ and $b\colon Y \to T$ be restrictions of $g$ and $h$. Then we have $X \sqcup Y$ as the domain of $f$, and $f$ itself is the copairing $[a,b]$.

Revised on November 15, 2009 19:24:48 by Toby Bartels (173.60.119.197)