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??

  • $\begingroup$ 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. $\endgroup$ Dec 27, 2021 at 16:21
  • $\begingroup$ Oh right, of course! I've fixed it. $\endgroup$
    – Kenta S
    Dec 27, 2021 at 17:36

1 Answer 1


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.

  • $\begingroup$ 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 $A\to B$. $\endgroup$
    – Kenta S
    Dec 23, 2021 at 1:02
  • 4
    $\begingroup$ @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. $\endgroup$ Dec 23, 2021 at 1:04

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