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. $\square$
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Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)
There are also: