# In a preadditive category, products are coproducts

$\renewcommand{\phi}{\varphi}$ $\require{AMScd}$ My question is different from this one.

I want to see how in a preadditive category $\mathcal{C}$ a product is a coproduct, following this proof. I'll write the proof up to the point where I get stuck.

Let $A=A_1,B=A_2$ be objects and suppose their product $P$ does exist, with projections $\pi_i:P\rightarrow A_i$. By considering a wedge

$$\begin{CD}A @<{id_A}<< A @>{0_{AB}}>> B\end{CD}$$

and the obvious analogous one with $A$ and $B$ exchanged, where $0_{A,B}$ is the zero morphism, we see that the product property implies the existence of two unique morphisms $k_i\in\mathcal{C}[A_i,P]$ s.t.

\begin{align*} \pi_j\circ k_i=\begin{cases} \mathrm{id}_i &\text{if }i=j\\ 0_{A_i, A_j} &\text{if }i\neq j \end{cases}. \end{align*} Using these equations and the definition of zero morphism, it's easy to see that the morphism $m\equiv k_1\circ \pi_1+ k_2\circ \pi_2$ satisfies $\pi_i\circ m=\pi_i$.

This is where I get stuck: In the linked proof, they now state that this implies $m=\mathrm{id}_P.$ But I don't see why this should be true.

Specifically, if $\phi\in\mathcal{C}[S,P]$ for some arbitrary object $S$, it is not apparent why $m\circ\phi=\phi$. All that we know is that there are some morphisms $x_i\in\mathcal{C}[S,A_i]$ that factor through $P$ via the universal property, but these are not necessarily unique.

And we learn even less about $m$ by looking at morphisms in $\mathcal{C}[P,S]$.

What am I missing here?

• I don't really understand why you say that the morphisms $x_i$ are not necessarily unique. Universal properties always have a uniqueness requirement! – Arnaud D. Apr 20 '17 at 12:54
• But aren't the projections $\pi_i$ only unique up to unique isomorphism? That for me would translate to "kinda unique", which is not really the same as unique? I guess I'm being stubborn – user435308 Apr 20 '17 at 13:09
• Well, the limit is unique up to unique isomorphism; but if you considered a fixed object $P$ with fixed projections $\pi_i$, the universal property tells you that an arrow $\phi:S\to P$ is uniquely determined by $\pi_1\phi$ and $\pi_2\phi$. – Arnaud D. Apr 20 '17 at 13:16

## 1 Answer

The equality $m=id_P$ follows from the identities $\pi_i\circ m=\pi_i=\pi_i\circ id_P$, for $i=1,2$. The reason is that by the universal property of the product (and especially the uniqueness part), the projections $\pi_i$ must be jointly monic.

• Ok, so this jointly monic part means that whenever $\pi_i\circ f =\pi_i \circ g$ for $i=1,2$, then we get wedges that are actually the same, but have mediators $f$ and $g$, respectively. But mediators are unique, so $f=g$. And thanks for noticing that I needed to see the word fixed :) – user435308 Apr 20 '17 at 13:30
• I hope it's ok to add this question as a comment: We can now for some morphisms $a_i\in\mathcal{C}[A_i,W]$ (call $a_1=a, a_2=b)$ construct the arrow $l\equiv a\circ\pi_1+b\circ\pi_2$. Then $l\circ k_A=a+b\circ 0_{A,B}$, and similarly for $k_B$. The problem is then that $b\circ 0_{A,B}\neq 0_{A,W}$, so I don't see how this $l$ is the mediator for the coproduct (which is what the proof on the Stacks Project site implies) – user435308 Apr 20 '17 at 14:10
• Nah, composition is bilinear, so anything composed with a zero will give zero. Then we get a bijective correspondence between $\mathcal{C}[P,W]$ and $\mathcal{C}[A,W]\times \mathcal{C}[B,W]$ by $(a,b) \mapsto a\circ\pi_1+b\circ\pi_2$ and $l \mapsto (l\circ k_A, l\circ k_B)$. – user435308 Apr 20 '17 at 14:27
• I was about to tell you that :) – Arnaud D. Apr 20 '17 at 14:28