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

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).

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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The proof can be found here. –  Rasmus Mar 5 '11 at 23:06

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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Thanks! The maps $i,j$ follow from the universal property of pull-backs, right? –  Fredrik Meyer Mar 6 '11 at 3:49
@Fredrik: Yes, but actually it would have been better to directly use the universal property of the product (I don't know why I phrased it this way). –  t.b. Mar 6 '11 at 4:19