# Morphisms to products and morphisms from coproducts

Let $$\mathbf{Grp}$$ be the category of groups, so the product of $$G$$ and $$H$$ is $$G\times H$$, with projections $$p_1\colon G\times H\to G$$ and $$p_2\colon G\times H\to H$$. These morphisms are part of the definition of a product, but somehow there are also natural morphisms $$G\to G\times H$$ and $$H\to G\times H$$, given by $$g\mapsto (g,1_H)$$ and $$h\mapsto (1_G,h)$$. Is there a nice category-theoretical explanation for these morphisms? Or are they just coincidences?

The dual of this phenomenon occurs as well. The coproduct of $$G$$ and $$H$$ is the free product $$G*H$$, with inclusions $$G\to G*H$$ and $$H\to G*H$$. However, there is also a projection to $$G$$, given by "ignoring" elements of $$H$$: $$G*H\to G*H/[G,H]\cong G\times H\xrightarrow{p_1} G.$$

Here, $$[-,-]$$ is the commutator.

This phenomenon does not always occur. For instance, in the category $$\textbf{Set}$$ of sets, the product is the usual cartesian product $$S\times T$$, but there is no canonical inclusion $$S\to S\times T$$. Similarly, the coproduct is the disjoint union $$S\sqcup T$$, but there is no canonical map $$S\sqcup T\to S$$.

The situation is different in the category $$\textbf{Set}_*$$ of pointed sets, since now there is a canonical map $$S\to S\times T:s\mapsto (s,*_T)$$.

What is happening here??

• By the way: you've got the kernel of the map $G*H\to G\times H$ incorrect; it's not the commutator (the quotient would be $G^{\rm ab}\times H^{\rm ab}$). The kernel of the map $G*H\to G\times H$ is called the "cartesian", and is the subgroup $[G,H]$, generated by all elements of the form $[g,h]$ with $g\in G$ and $h\in H$ (or more formally, $[\iota_G(g),\iota_H(h)]$, where $\iota_G$ and $\iota_H$ are the canonical embeddings into the free product. Dec 27, 2021 at 16:21
• Oh right, of course! I've fixed it. Dec 27, 2021 at 17:36

Both maps are the result of $$\mathsf{Grp}$$ having a "zero object" (the trivial group). A zero object is an object $$\mathbf{Z}$$ which is both initial and final in the category; that is, for every object $$C$$, there is a unique morphism $$i_C\colon \mathbf{Z}\to C$$, and a unique morphism $$t_C\colon C\to\mathbf{Z}$$.
This means that for any two objects $$C_1$$ and $$C_2$$, there is a canonical "zero map", $$z_{C_1,C_2}\colon C_1\to C_2$$, obtained via the composition $$i_{C_2}\circ t_{C_1}\colon C_1\to \mathbf{Z}\to C_2$$.
1. In any category with a zero object, if $$A$$ and $$B$$ have a product $$P$$, then there are canonical morphisms $$\iota_A\colon A\to P$$ and $$\iota_B\colon B\to P$$, obtained by considering the maps $$\mathrm{id}_A\colon A\to A$$ and $$z_{A,B}\colon A\to B$$, to get a map $$A\to P$$ via the universal property; and symmetrically for $$B$$. Since $$\pi_A\circ \iota_A = \mathrm{id}_A$$, it follows that $$\iota_A$$ has a left inverse and therefore is a monomorphism (and also you can conclude that the projection map $$\pi_A$$ is an epimorphism); symmetrically for $$\iota_B$$. The same holds for a product over an arbitrary family, not just a pair of objects.
2. Dually, in any category with a zero object, if $$A$$ and $$B$$ have a coproduct $$C$$, with canonical morphisms $$\iota_A,\iota_B$$, you get maps $$p_A\colon C\to A$$ and $$p_B\colon C\to B$$ obtained by considering the maps $$\mathrm{id}_A\colon A\to A$$ and $$z_{B,A}\colon B\to A$$ to obtain the map $$p_A$$; since $$p_A\circ \iota_A = \mathrm{id}_A$$, we conclude that $$p_A$$ has a right inverse and thus is an epimorphism (and $$\iota_A$$ is a monomorphism). Symmetrically for $$B$$, and for an arbitrary family that has a coproduct in the category.
Since the trivial group gives you a zero object for $$\mathsf{Grp}$$, you can observe that phenomenon. In $$\mathsf{Set}$$, the empty set is the initial object, and singletons are terminal objects, but the lack of a zero object means that you do not observe the same phenomenon. You have the same problem in $$\mathsf{Semigroup}$$ and in $$\mathsf{Ring}^1$$ (rings with unity with unital morphisms), but in $$\mathsf{Monoid}$$ and in $$\mathsf{AbGrp}$$, the zero object gives you the same phenomenon. In the category $$\mathsf{Set}_*$$ of pointed sets, singletons are (isomorphic) zero objects, so again you get the same phenomenon thanks to the corresponding zero morphisms.
• Thanks, this is a very clean explanation! I guess the main point is that when there is a zero object, one has a "canonical" morphism $A\to B$. Dec 23, 2021 at 1:02
• @KentaS: exactly. You actually only need the existence of a canonical morphism for $m_{AB}\colon A\to B$ for any pair of objects, which satisfies $m_{BC}\circ m_{AB}=m_{AC}$, to get these results for products and coproducts. The zero object neatly yields these morphisms. Dec 23, 2021 at 1:04