TITLE: Definition of codiagonal in a category
QUESTION [7 upvotes]: I'm confused by the definition of a codiagonal in a category with coproducts.
The definition on nLab is as follows. Let $\mathcal{C}$ be a category with coproducts and let $X \in \mathcal{C}$. Then the codiagonal $\nabla$ is the canonical map
$$ \nabla : X \coprod X \stackrel{(Id,Id)}{\to} X. $$
Now how is this map actually defined? For instance, consider $\mathcal{C} = \mathbf{ \mbox{FinVec}}$, and let $X$ be a vector space. What can possibly be the definition of $(Id,Id) : X \oplus X \to X$ ? Where do you send a pair $(x,y) \in X \oplus X$?

REPLY [8 votes]: By definition of coproduct, for any pair of maps $X \to Z$ and $Y \to Z$, there is a unique map $X \coprod Y \to Z$ such that the two maps above are the composites $X \to X \coprod Y \to Z$ and $Y \to X \coprod Y \to Z$.
The two insertion maps $X \to X \oplus X$ are given by $x \mapsto (x,0)$ and $x \mapsto (0,x)$. Therefore, we can calculate that the map $(1,1) : X \oplus X \to X$ satisfies
$$ (x,0) \mapsto x $$
$$ (0,x) \mapsto x $$
and therefore
$$ (x,y) = (x,0) + (0,y) \mapsto x + y $$
In an abelian category, we can give some nice meaning to the notation: if we imagine a direct sums as forming an object of column vectors, then the map $(1,1)$ acts via matrix multiplication. Similarly, we would write the diagonal $X \to X \oplus X$ as $\Tiny\left( \begin{matrix} 1 \\ 1 \end{matrix} \right)$.