Canonical Map in a Category with Finite Products, Coproducts and a Zero Object. Let $C$ be a category as advertised in the title, $\left \{ A_{i} \right \}_{1\leqslant i\leqslant n}$ a finite collection of objects in $C$. Why is there then is a canonical map $f:\coprod A_{i}\rightarrow \prod A_{i}$ s.t the matrix of arrows $\pi_{A_{i}}\circ f \circ i _{A_{j}}$ is $\delta _{ij}$, that is, the identity? I have seen this for some familiar  categories like $Ab$, but I have not been able to prove it in the general case. Thanks.
 A: Let me explain something more general. Let $\{X_i\}$ and $\{Y_j\}$ be two collections of objects. In order to specify a map $$f:\coprod X_i\to \prod Y_j,$$ it suffices to say how we map $\coprod X_i$ to each factor, $Y_j$. Thus we need to define a set of maps $$f_{\bullet j}\coprod X_i\to Y_j.$$ But to specify a map out of a coproduct, it suffices to give a map for each component. In particular if we have some random coproduct $\coprod Z_k\to W$, if we want to specify a map, $G:\coprod Z_k\to W$ then we need to specify a collection of maps, $$G_k:Z_k\to W .$$ Therefore, to define a map $$f_{\bullet j}\coprod X_i\to Y_j,$$ it suffices to define a collection of maps, $$f_{ij}:X_i\to Y_j.$$ Now for your question, set $X_i=Y_i$, and $f_{ii}=id$, and $f_{ij}=0$ if $i\neq j$.
A: Edit: This answer is basically the same as Baby Dragon's. I didn't see his/her answer when I started working on my pictures!
Let $A=\{A_i:i\in I\}$ and $B=\{B_j:j\in J\}$ be two collections of $\mathscr C$-objects such that $\coprod A$ exists and $\prod B$ exists. For each $i\in I$ and $j\in J$ let $\ell_{ij}:A_i\to B_j$. For fixed $i \in I$, the universal property of $\prod B$ ensures a unique $\ell_{i\bullet}$ making the diagram

commute for every $j\in J$. Then the universal property of $\prod A$ ensures a unique $\ell_{\bullet\bullet}$ making the diagram

commute. Putting these diagrams together, we see that $\ell_{\bullet\bullet}$ is the unique $\mathscr C$-morphism making the diagram

commute for every $i\in I$ and $j\in J$. Now, assuming that $\mathscr C$ has zero-morphisms, your result is obtained by letting $I=J=\{1,\dotsc,n\}$, $A=B$, and $\ell_{ij}=\delta_{ij}$.
