# Definition of codiagonal in a category

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 $(x,y) \in X \oplus X$?

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 $\Tiny\left( \begin{matrix} 1 \\ 1 \end{matrix} \right)$.

• but what does "$x+y$" mean in a general context? For vector spaces it is clear, but what about other categories? – rmdmc89 Mar 27 '17 at 17:38
• @AguirreK: It doesn't have meaning in general context; it only makes sense in an abelian category (or similar). Which, of the top of my head, is the only context in which it makes sense to write $(x,y)$ for an element of $X \amalg X$ anyways. Everything after the first paragraph is discussing that particular case (since it's the one the OP asked about). – user14972 Mar 28 '17 at 1:07

$(x, y)$ gets sent to $x + y \in X$.

• but what does "$x+y$" mean in a general context? For vector spaces it is clear, but what about other categories? – rmdmc89 Mar 27 '17 at 17:39
• @AguirreK: it doesn't mean anything in a general context. I was answering the OP's specific question about vector spaces. – Qiaochu Yuan Mar 27 '17 at 20:37