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

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 $\mathop{Mor}\nolimits (z, w) = \mathop{Mor}\nolimits (x, w) \times \mathop{Mor}\nolimits (y, w)$ 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. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)

There are also: