Suppose we have a pointed category $\mathcal{A}$ and a collection $\{A_i\}_{i\in I}$ of objects which have both a product $\prod_iA_i$ and a coproduct $\coprod_iA_i$. The product has projections $\pi_i$ and the coproduct has inclusions $\iota_i$.
According to the Wikipedia article on biproducts, (in the case of just two objects at least) there is a unique morphism $\alpha:\coprod_iA_i\to\prod_iA_i$ satisfying $\pi_j\circ\alpha\circ\iota_i=\sigma_{ij}$ for all $i,j$, where $\sigma_{ij}$ is the identity if $i=j$ and the zero morphism otherwise. The zero morphism $A\to B$ here is the composite $A\to 0\to B$, where $0$ is the zero object.
I have managed to construct two such morphisms, yet am unsure how to show that they are equal, and more generally that there is only one such morphism satisfying the composition above. Does this hold in the general case of an arbitrary collection of objects?
My understanding is that $\{A_i\}_{i\in I}$ has a biproduct iff such a (unique) morphism from coproduct to product is an isomorphism, so we can then identify the product and coproduct. Are other assumptions needed?
Best Answer
No additional assumptions are needed. What is happening in detail is the following.
Given a coproduct with injections $j_i\colon A_i\to\sqcup_i A_i$ and product with projections $\prod_k B_k\to B_k$, every morphism $f\colon\sqcup_i A_i\to\prod_k B_k$ determines a $2$-dimensional array of morphisms whose $(i,k)^\mathrm{th}$ entry is the morphism $f_k^i=\pi_k\circ f\circ j_i\colon A_i\to B_k$. If two such arrays are the same for two morphisms $f$ ang $g$, i.e. if $\pi_k\circ f\circ j_i=\pi_k\circ g\circ j_i$ for all $i$ and $k$, then the universal property of the product implies $f\circ j_i=g\circ j_i$ for all $i$, and then the universal property of the coproduct implies $f=g$.
In a category with a distinguished "zero" morphism between any two objects, for any family of objects $A_j$ you can define a $2$-dimensional array $\delta_k^i=\begin{cases}0\colon A_i\to A_k&i\neq k\\\mathrm{id}_{A_k}\colon A_i\to A_k&i=k\end{cases}$. When the zero morphisms are canonical (e.g. form a two-sided ideal, i.e. $f0=0=0g$ for all $f$ and $g$), then the above array is canonical, hence so is the unique morphism $\sqcup_j A_j\to\prod_j A_j$ corresponding to it.