Franklin's Notes


Given two objects $x,y$ in a category $\mathsf C$, a coproduct of $x$ and $y$, denoted $x\amalg y$ or $x+y$, can be defined as a universal arrow $u:\langle x,y\rangle\to \Delta z$ from $\langle x,y\rangle$ to the diagonal functor $\Delta:\mathsf C\to\mathsf C\times\mathsf C$. Often the term "coproduct" is used to refer to the object $z$ itself, rather than the universal arrow $u$. The more familiar definition of a coproduct is as follows: the object $x\amalg y$ and the arrows $i:x\to x\amalg y$ and $j:y\to x\amalg y$ comprise a coproduct of $x$ and $y$ iff for any two morphisms $f:x\to a$ and $g:y\to a$, there exists a unique morphism $h:x\amalg y\to a$ such that the following diagram commutes:

Because coproducts are defined as universal arrows, they are necessarily unique up to isomorphism. Coproducts of more than $2$ objects can be defined as universal arrows from $\langle x_i\rangle_{i\in I}$ to the functor category $\mathsf C^X$, where $X$ denotes a set regarded as a discrete category. (When $|X|=2$, this reduces to the usual binary coproduct.) When the $x_i$ are all equal, this is called a copower.

Here are some examples of categories with coproducts:

Although the category $\mathsf{Poset}$ does have coproducts, the category $\mathsf{LinOrd}$ does not have coproducts. (In fact, the coproduct $1\amalg 1$ does not exist in $\mathsf{LinOrd}$.)

If, in some category $\mathsf C$, the coproduct $w\amalg x$ exists for all objects $x$ where $w$ is a fixed object, then there is a functor which acts on objects by sending $x\to w\amalg x$. We may denote this functor by $w\amalg -$. This functor acts on morphisms as follows: if $f:x\to y$, then $f'$, the image of $f$ under this functor, is defined to be the unique morphism making the following diagram commute:

To see why this is actually functorial, consider the following diagram, for arbitrary morphisms $f,g$:

By the uniqueness of the arrow $w\amalg x\to w\amalg z$ making the diagram commute, and the fact that both $g'f'$ and $(gf)'$ accomplish this, we deduce that it must be the case that $(gf)'=g'f'$, implying that our construction is functorial.



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