# How to define maps between products in a complete category using maps on their factors?

Suppose that $C$ is a complete category, and let $\{A_\alpha\}$ and $\{B_\beta\}$ be two indexed sets of objects. If we are given a collection of morphisms $f^\alpha_\beta: A_\alpha \to B_\beta$, is there always an associated morphism $f: \Pi A_\alpha \to \Pi B_\beta$? How would we define this morphism?

Clearly, by the definition of a product, we can obtain unique morphisms $g_\alpha: A_\alpha \to \Pi B_\beta$, but I don't know how to proceed further.

This question is inspired by the definition of the differential on the total complex $\textrm{Tot}^{\Pi}(C)$ of a double complex $\{C_{p,q}\}$ in an abelian category, whereby we may define a map $d: \textrm{Tot}^{\Pi}(C)_n \to \textrm{Tot}^{\Pi}(C)_{n-1}$ using the vertical and horizontal differentials, $d_v$ and $d_h$, on the level of objects.

• It depends what you mean by "associated". You can always precompose your maps $g_\alpha$ with the projection $\pi_\alpha: \Pi A_\alpha \to A_\alpha$. But it won't be canonical in general, I think; you have no way to "add" the maps $f_\beta^*$ whose target is a particular $B_\beta$. This is what works nicely in an abelian category. – Mr. Chip Oct 23 '17 at 0:56
• @Mr.Chip How would it work in an abelian category? What feature do we have there? – Mathmank Oct 23 '17 at 2:08
• (AFAIK) You're asking the wrong question! The objects in the total complex are the coproduct, not the product, of the objects along the diagonal. And a very relevant additional detail is, for each specific source object $A_\alpha$, all but finitely many of the $f^\alpha_\beta$ are zero. (and the same is true for each specific target object $B_\beta$) – user14972 Oct 24 '17 at 3:11

No, this is typically not possible. For instance, consider the case where the first collection $\{A_\alpha\}$ is empty and $\{B_\beta\}$ has just one object $B$. Then you are asking for a canonical map from a terminal object $1$ to any object $B$. There is no such canonical map (and in fact there may not be any map $1\to B$ at all, e.g. in $Set$ if $B=\emptyset$).
One special case where such a map exists is if $\{A_\alpha\}$ is finite and your category is additive. In that case there is a canonical isomorphism from the coproduct $\coprod A_\alpha$ to the product $\prod A_\alpha$. Your morphisms $g_\alpha$ induce a map $\coprod A_\alpha\to \prod B_\beta$, and hence a map $\prod A_\alpha\to\prod B_\beta$ via the isomorphism.
In fact, we can weaken the assumption that $\{A_\alpha\}$ is finite a bit. Suppose merely that for any $\beta$, all but finitely many of the maps $f^\alpha_\beta$ are $0$ (this is true in your motivating example). Then for each $\beta$, we can define a map $\prod A_\alpha\to B_\beta$ by first projecting onto the finite subproduct $\prod_{\alpha\in F}A_\alpha$ where $F=\{\alpha:f^\alpha_\beta\neq 0\}$, and then taking the map $\prod_{\alpha\in F}A_\alpha\to B_\beta$ given by the previous paragraph. These maps $\prod A_\alpha\to B_\beta$ then combine to give a single map $\prod A_\alpha\to\prod B_\beta$.