Identity map on coproduct as 'linear combination'

Question

Let $\mathcal{C}$ be a category, and suppose for all objects $X, Y$ of $\mathcal{C}$, Mor$_{\mathcal{C}}(X,Y)$ is equipped with the structure of an Abelian group, such that the composition of morphisms is bilinear.

Now it can easily be shown that Mor$_{\mathcal{C}}(X,X) =:$ End$_{\mathcal{C}}(X)$ is a ring (under composition as multiplication). (In the absence of any knowledge about the Abelian group structure, distributivity still arises from bilinearity in this case.)

Now let $(S, i, j)$ be a coproduct (or sum) of objects $X$ and $Y$. It can then be shown (ask me for the proof), that there exist unique morphisms $p : S \rightarrow X$ and $q : S \rightarrow Y$ such that $p \circ i = \text{id}_X, p \circ j = 0_{YX}, q \circ i = 0_{XY}$ and $q \circ j = \text{id}_Y$.

To demonstrate: $i \circ p + j \circ q = \text{id}_S$

The first thing to observe is that by bilinearity, $$(i \circ p + j \circ q) \circ i = i \circ \text{id}_X + j \circ 0 = i$$ and $$(i \circ p + j \circ q) \circ j = i \circ 0 + j \circ \text{id}_Y = j $$

What we need however, is that $(i \circ p + j \circ q) \circ f = f$ for all morphisms $A \xrightarrow{f} S$. If somehow $i$ and $j$ would generate any such morphism, we would be done (as $(i \circ p + j \circ q) \circ (i+j) = i+ j $), but I wouldn't know what 'generate' would mean in this context. The last step would be to show that for all morphisms $S \xrightarrow{g} B$, $g \circ (i \circ p + j \circ q) = g$, but I suppose this will be similar to the first problem.

I'm afraid there is so much to unpack in the suggested other proof (what are preadditive categories, wedges, mediators, what is monic, jointly monic, etc.?) that I don't see how this is relevant to this (very elementary) exercise in my first introduction to category theory. You need to be a lot further along in the subject to understand all that, and my question is still very much worthwile for those who don't know all that jargon.

Answer 1

You can prove it using the universal property of the coproduct (there's a unique function which makes the diagram commute, so if two functions make the coproduct diagram commute, they're equal).

It may help to draw the coproduct diagram for the following situation:

1. You have the inclusions $i$ and $j$ into the coproduct $X+Y$.

2. You have a special morphism $f\equiv (i\circ p)+(j\circ q)$ from $X+Y$ to $X+Y$ (which you can draw as a second copy of $X+Y$).

3. By the bilinearity of function addition and the special property of $p$ and $q$, you know that $f\circ i = i$ and $f\circ j = j$. Therefore you have arrows from $X$ and $Y$ into the second copy of $X+Y$.

4. By the universality of the coproduct, there is a unique morphism $X+Y \rightarrow X+Y$ which makes this diagram commute. Specifically, there must be a unique morphism $s$ for which $s\circ i = (f\circ i)$ and $s\circ j = (f\circ j)$. Evidently, $f$ itself has this property. But so does the identity function $id:X+Y\rightarrow X+Y$. Therefore, both functions are the same.