Groups are cocomplete by adjoint functor theorem without explicit use of free products

adjoint-functorsalternative-proofcategory-theorygroup-theory

$$
\newcommand{GRP}{\mathsf{GRP}}
\newcommand{I}{\mathcal{I}}
\newcommand{im}{\mathrm{Im} \;}
$$

This is a follow up for the question Royal Road to Free Groups and Free Products. I was exploiring categorical method suggested by Eric Wofsey in comments, but got some complications.

My goal is to prove that category of groups $\GRP$ is cocomplete by applying general adjoint functor theorem but I have problem with constructing set with solution set condition (SSC).

The Idea is to prove that every small category $\I$ the diagonal functor $\Delta : \GRP \to \GRP^\I$, which maps a group to corresponding constant functor, satisfies SSC and, hence admits a left adjoint. It was allready proved this left adjoint is a colimit functor on $\I$.

Functor $\Delta$ would satisfy SSC if for every diagram $G : \I \to \GRP$ there a (small) indexing set $I$ with a collection of groups $(X_i)_{i \in I}$ each equipped with a natural transform $f_i : G \Rightarrow \Delta(X_i)$, such that for every group $H$ with natural transform $\phi : G \Rightarrow \Delta(H)$ it is possible to find $X_i$ with and a homomorphism $t : X_i \to H$ so there is a factorization $ \phi = t \circ f_i $.

One obvious contender for SSC is a free product $\coprod_{i \in I} G_i$ factorized by a normalizer of the set
$$
A = \left\{ a\big(G_{i,j}(h)(a)\big)^{-1} \Big| i,j \in \I, a \in G_i,h : i \to j \right\}
$$
and $t$ defined by

$$
t \left( \prod^n_{k=1} a^{i_k}_k \right) = \prod^n_{k=1} \phi_{i_k}(a_k^{i_k}),
$$
where $a^{i_k}_k \in G_{i_k}$.
But this is a standart construction of the colimit and is not fun, as it is interesting to apply this theorem without direct use of such colimits as coproduct and coequalizer.

Some of my thoughts so far:

for every $(H,\phi)$, if I denote $N_i = \ker \phi_i$, there are isomorphisms $\frac{G_i}{N_i} \cong \im \phi_i$. So such families of normal group must classify images of $\phi$ up to isomorphism. Moreover, by naturality of $\phi$ it holds that $\phi_i = \phi_j \circ G_{i,j}(h)$ for every $h : i \to j$ and so
$$
(1) \quad (G_{i,j}(h))^{-1}(N_j) = N_i
$$
holds. So it must be possible to take set $I = \{(N_i)_{i \in \I} : \forall i \in \I \; . \; N_i \triangleleft G_i\text{ and (1) holds }\}$ as our indexing set. This set is small as it is a subset of $\prod_{i \in \I} 2^{G_i}$.However, I don't know how to construct $X_N$ an $f_{N,i}$ correctly. It seems that decomposition of form
$$ f_{N,i} : G_i \xrightarrow{\pi_{N_i}} \frac{G_i}{N_i} \xrightarrow{?} X_N $$
But I don't know how to combine all quotients into a single group $X_N$ without use of free products so $t_i$ is indeed a homomorphism. I tried constructing $X_N$ as products, but resulting $t_i$ fails to be a homomorphism for nonabelean groups, and use of free products makes whole construction redundant.

Help me to construct SSC $X_i$ and $f_{N,i}$, with $X_i$ not being some explicit quotients of free product, or explain why this task is impossible.

p.s. It is possible to forget about morphisms $h$ of $\I$ for simplicity. Working out this case will prove existence of free products.

p. p. s. I took liberty to define mathjax shorthands:

\GRP $\to \GRP$

\I $\to \I$

\im $\to \im$

Best Answer

Hint : use the fact that there is a fixed (infinite) bound on $|G(i)|, i\in \mathrm{Ob}(\I)$, and so there's at most a set of $\I$-indexed families of groups on which $G(i)$ surjects, and so at most a set of natural transformations from $G$ to those, and that each natural $G\implies \Delta(H)$ factors through such a natural transformation.

Related Solutions

[Math] Proving that the free group on two generators is the coproduct $\mathbb{Z}*\mathbb{Z}$ in $\textbf{Grp}$

You're working too hard. I'll assume you're convinced that $\mathbb{Z}$ is the free group on one generator, which means you're convinced that

$$\text{Hom}(\mathbb{Z}, G) \cong G$$

(where the RHS should really be the underlying set of $G$ but I'm omitting that I'm applying the underlying set functor). One way to state the universal property of the coproduct is that

$$\text{Hom}(X \sqcup Y, Z) \cong \text{Hom}(X, Z) \times \text{Hom}(Y, Z)$$

from which it follows that

$$\text{Hom}(\mathbb{Z} \sqcup \mathbb{Z}, G) \cong G \times G$$

and this is also the universal property of the free group on two generators, so you're done by the Yoneda lemma. The argument is exactly the same for the free group on $n$ generators, and in fact on a set's worth of generators.

More abstractly, the forgetful functor $\text{Grp} \to \text{Set}$ has a left adjoint, the free group functor. As a left adjoint, it preserves colimits, and in particular coproducts. But every set $X$ is the coproduct of $X$ copies of the $1$-element set, so it follows that the free group $F_X$ is the coproduct of $X$ copies of $\mathbb{Z}$.

Royal Road to Free Groups and Free Products

In general, if $\mathcal{A}$ is a variety of algebras and $U : \mathcal{A} \to \mathbf{Sets}$ is the forgetful functor, then we can construct a left adjoint $F : \mathbf{Sets} \to \mathcal{A}$ as follows:

First, take the set of formal expressions built up from atoms corresponding to elements of $X$, using formal counterparts of the operations in $\mathcal{A}$. (If you're using ZFC as your foundations, you will likely need a set-theoretic argument to show this can be represented by an actual set rather than a proper class.)
Then, take the quotient by the smallest equivalence relation which contains all instances of relations in $\mathcal{A}$ with formal expressions substituted in, and which also makes all the formal operations be respectful of the equivalence relation. (It is straightforward to show the class of relations on the set of formal expressions from (1) satisfying these requirements is closed under arbitrary intersections; so then, for example, you could construct the desired equivalence relation as the intersection of all such relations.)

Then, in many cases, it's possible to represent each equivalence class by a member of a restricted class of expressions which is easier to work with; and if we're lucky, we might even be able to choose a canonical such expression in each equivalence class.

In the case $\mathcal{A} = \mathbf{Groups}$, for example, it is easy to show that any formal expression from (1), modulo the equivalence relation in (2), is equivalent to a (possibly empty) product of elements $x \in X$ or $y^{-1}$ for $y \in X$ which together form a reduced word. You would prove this by "structural induction" on the general formal expression (or, if you prefer, by strong induction on the "total size" of the general formal expression). The harder part is to show that any two distinct reduced words end up in different equivalence classes.

To make this process more concrete in the group case, I think it might help to show how this would be encoded in Coq, where for example (1) is very similar to the way you would define an "abstract syntax tree" type in OCaml or Haskell:

(* formal expressions from (1) *)
Inductive free_group (X:Type) : Type :=
| fg_atom (x:X)
| fg_id
| fg_inv (x : free_group X)
| fg_mul (x y : free_group X).

Arguments fg_atom [X] x.
Arguments fg_id [X].
Arguments fg_inv [X] x.
Arguments fg_mul [X] x y.

(* Equivalence relation described in (2) *)
Inductive fg_equiv (X:Type) : free_group X -> free_group X -> Prop :=
| fg_left_id : forall x : free_group X,
  fg_equiv X (fg_mul fg_id x) x
| fg_right_id : forall x : free_group X,
  fg_equiv X (fg_mul x fg_id) x
| fg_left_inv : forall x : free_group X,
  fg_equiv X (fg_mul (fg_inv x) x) fg_id
| fg_right_inv : forall x : free_group X,
  fg_equiv X (fg_mul x (fg_inv x)) fg_id
| fg_mul_assoc : forall x y z : free_group X,
  fg_equiv X (fg_mul (fg_mul x y) z) (fg_mul x (fg_mul y z))
| fg_equiv_refl : forall x : free_group X,
  fg_equiv X x x
| fg_equiv_sym : forall x y : free_group X,
  fg_equiv X x y -> fg_equiv X y x
| fg_equiv_trans : forall x y z : free_group X,
  fg_equiv X x y -> fg_equiv X y z -> fg_equiv X x z
| fg_equiv_resp_id : fg_equiv X fg_id fg_id
| fg_equiv_resp_inv : forall x x' : free_group X,
  fg_equiv X x x' -> fg_equiv X (fg_inv x) (fg_inv x')
| fg_equiv_resp_mul : forall x x' y y' : free_group X,
  fg_equiv X x x' -> fg_equiv X y y' ->
  fg_equiv X (fg_mul x y) (fg_mul x' y').

(Incidentally, it's also very easy to extend this to describe the group with any given presentation, or in general the object of any variety of algebras with a given presentation: just add the relations from the presentation to the requirements on the equivalence relation in (2).)

Best Answer

Related Solutions

[Math] Proving that the free group on two generators is the coproduct $\mathbb{Z}*\mathbb{Z}$ in $\textbf{Grp}$

Royal Road to Free Groups and Free Products

Related Question