Lemma 12.3.4. Let $\mathcal{A}$ be a preadditive category. Let $x, y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$. If the product $x \times y$ exists, then so does the coproduct $x \amalg y$. If the coproduct $x \amalg y$ exists, then so does the product $x \times y$. In this case also $x \amalg y \cong x \times y$.

Proof. Suppose that $z = x \times y$ with projections $p : z \to x$ and $q : z \to y$. Denote $i : x \to z$ the morphism corresponding to $(1, 0)$. Denote $j : y \to z$ the morphism corresponding to $(0, 1)$. Thus we have the commutative diagram

$\xymatrix{ x \ar[rr]^1 \ar[rd]^ i & & x \\ & z \ar[ru]^ p \ar[rd]^ q & \\ y \ar[rr]^1 \ar[ru]^ j & & y }$

where the diagonal compositions are zero. It follows that $i \circ p + j \circ q : z \to z$ is the identity since it is a morphism which upon composing with $p$ gives $p$ and upon composing with $q$ gives $q$. Suppose given morphisms $a : x \to w$ and $b : y \to w$. Then we can form the map $a \circ p + b \circ q : z \to w$. In this way we get a bijection $\mathop{\mathrm{Mor}}\nolimits (z, w) = \mathop{\mathrm{Mor}}\nolimits (x, w) \times \mathop{\mathrm{Mor}}\nolimits (y, w)$ which show that $z = x \amalg y$.

We leave it to the reader to construct the morphisms $p, q$ given a coproduct $x \amalg y$ instead of a product. $\square$

There are also:

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