# (Are there) subtleties in the definition of 'biproduct'

I always thought that a biproduct of two objects $A_1,A_2$ in some category $\mathcal{C}$ is an object $P$ with two maps $p_i:P\to A_i$ making it a product and two maps $j_i:A_i\to P$ making it a coproduct. Namely, that it is just an object (with the structure maps) which is simultaneously a product and a coproduct. I was surprised that the nLab entry for biproduct is somewhat more complicated. First, it assumes that $\mathcal{C}$ has zero morphisms (which is equivalent to having a zero object?) Then, it requires the existence of some product $A_1 \times A_2$ and some coproduct $A_1 + A_2$. Finally, it constructs the canonical map $\phi:A_1 + A_2 \to A_1 \times A_2$ which in matrix notation is the $2\times 2$ identity matrix and this map should be an isomorphism. My question is whether this complication is actually necessary.

More concretely, is there an example of $(\mathcal{C},A_i,P,p_i,j_i)$ as above such that $\phi$ (say, from $P$ to $P$) is not an isomorphism?

I think that a standard universal properties argument shows that no such example exists, but then I don't understand why the nLab entry is taking this strange path. Additionaly, is refers to a result that if we have any natural family of isomorphisms $A_1+A_2 \simeq A_1\times A_2$ for all $A_1,A_2$ then all the $\phi$-s are isomorphisms as well, but if I am correct then there is no need for naturality, as this is true 'pointwise'.

• Well, consider an infinite set $X$. Then for cardinality reasons, $X + X \cong X \times X$, yet we do not say that it is a biproduct. – Zhen Lin Oct 11 '14 at 14:25
• This is perhaps a motivation for not defining the concept of biproducts in a category without zero object. Can you turn this into an explicit counterexample of the sort I asked for? I would be happy to accept it :) – KotelKanim Oct 11 '14 at 14:35
• @ZhenLin: But it is a byproduct of the axiom of choice! ;-) – Asaf Karagila Oct 11 '14 at 14:40
• @KotelKanim: "zero morphisms (which is equivalent to having a zero object?)" - No. A silly counterexample is the empty category. But there are more complicated examples. – Martin Brandenburg Oct 11 '14 at 14:41
• @KotelKanim: The definition of a biproduct at the nlab is not really complicated. It just mimics exactly what happens for abelian groups. – Martin Brandenburg Oct 11 '14 at 14:44

Let $X$ be an infinite set and let $x \in X$. Then for cardinality reasons, $(X, x) + (X, x)$ is isomorphic to $(X, x) \times (X, x)$ in $\mathbf{Set}_*$ yet the canonical comparison $(X, x) + (X, x) \to (X, x) \times (X, x)$ is not an isomorphism. So we do not have a biproduct.
Some other comments. The condition that a category have zero morphisms is slightly weaker than the condition that it have a zero object; the latter implies the former and the two are equivalent if the category has either an initial or a terminal object, but for example a nonzero ring $R$ regarded as a one-object linear category has a zero morphism but not a zero object.