Intereting Posts

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**In an additive category, why is finite products the same as finite coproducts?**

This is relatively easy to prove when the category is R-mod, but my intuition/creativity fails to see how the method can be extended to arbritrary additive categories

Specifically, a category (in Weibel, “An introduction to homological algebra”) is called additive if the Hom-sets are abelian groups, composition of morphisms distribute over addition, and such that it has a distinguished zero object (that is, an object that is both initial and terminal).

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After giving this definition, Weibel claims, without further explanation, that “this structure is enough to make finite products the same as finite coproducts”.

How is this?

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Note that a product $A \times B$ is the same as a pull-back diagram

```
A x B -> B
| |
v v
A ---> 0
```

with maps $p:A \times B \to A$ and $q: A \times B \to B$.

In particular there is a map $i: A \to A \times B$ such that $pi = 1_{A}$ and $qi = 0$ as well as a map $j: B \to A \times B$ such that $qj = 1_{B}$ and $pj = 0$. Now note that $p(ip + jq) = pip + 0 = p$ and $q(ip + jq) = q$ so that $ip + jq = 1_{A \times B}$.

Let us check that $i: A \to A \times B$ and $j: B \to A \times B$ define a coproduct. Given $f: A \to D$ and $g: B \to D$ we get a map $d: A \times B \to D$ by setting $d = fp + gq$. Since $di = (fp + gq)i = fpi +gqi = f$ and $dj = g$ it remains to prove uniqueness of $d$. But this is clear as any other such map will satisfy $(d-d')1_{A \times B} = (d-d')(ip+jq) = 0$. Summing up, we have proved that the product of two objects is also a coproduct.

Now if we want to say that finite products and coproducts exist and coincide, we need a zero object, since otherwise the empty product (the terminal object) and the empty coproduct (the initial object) would not coincide.

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