11
$\begingroup$

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??

$\endgroup$
2
  • $\begingroup$ 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

15
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.