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

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 $\to \im$

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

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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.

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  • $\begingroup$ Thanks. I am currently working with this idea. But I still don't understand why $|G(i)|$ must be bounded, when $\I$ is infinite. $\endgroup$ – Nik Pronko Sep 10 '18 at 19:22
  • $\begingroup$ the bound can be infinite ! $\endgroup$ – Max Sep 10 '18 at 19:59
  • $\begingroup$ Yes, I understand that the bound is some cardinal $\kappa$. I think I got in now: the union $A = \bigcup_{i \in \I} G_i$ is a set so I can take $\kappa = |A|$ and then $|G_i| \le \kappa$ for all $i \in \I$. $\endgroup$ – Nik Pronko Sep 10 '18 at 20:07

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