Vtyjkyuk

$$
newcommandGRPmathsfGRP
newcommandImathcalI
newcommandimmathrmIm ;
$$

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
$$
A = left i,j in I, a in G_i,h : i to j right
$$
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 $fracG_iN_i cong im phi_i$. 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_itext 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
$$ f_N,i : G_i xrightarrowpi_N_i fracG_iN_i xrightarrow? X_N $$
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 $to im$

also related to How to use the adjoint functor theorem construct the coproduct in Grp?

Hint : use the fact that there is a fixed (infinite) bound on $|G(i)|, iin mathrmOb(I)$, 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 $Gimplies Delta(H)$ factors through such a natural transformation.