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Recall that there is a classifying space functor $B(-)\colon Groups\to sSet_*$ with values in based simplicial sets. I would like to know if there is a ``finite'' version of this result.

Namely, let $Groups^{fp}$ and $sSet^{ft}_*$ denote the categories of finitely presented groups and simplicial sets with finitely many simplices in each degree, respectively. I think it is easy to see that for any $G\in Groups^{fp}$ there is $S\in sSet^{ft}_*$ with the required homotopy type $B(G)\sim S$.

The naive question would be:

${\bf Q\ 1}$ Is there a functor $B^{f}(-)\colon Groups^{fp}\to sSet^{ft}_*$ lifting $B(-)$?

Of course there is no natural solution to the problem and for my needs it will be enough to have a positive result in the following form. Assume $G\colon D\to Groups^{fp}$ is a finite diagram of groups (in my case of free groups). Is it possible to lift $B\circ G$ to $sSet^{ft}_*$? Equivalently:

${\bf Q\ 2}$ Is there a diagram $S\colon D\to sSet^{ft}_*$ of aspherical spaces such that $\pi_1(S_x,pt)\simeq G_x$ depends functorially in $x\in D$?

Put it differently, one has an infty-coherent diagram of spaces of the form $\vee S^1$, does it admits a rectification by simplicial sets with finitely many simplices in each degree?

${\bf UPD}$ As was mentioned by user IJL it is not true in general that a finitely presented group admits a classifying space with finitely many simplices in each degree. Further, the comment by R. van Dobben de Bruyn make it transparent that it is better to restrict the question to small category. So let me restrict ${\bf Q 2}$ to the case of free finitely generated groups and leave it as the main question of the thread.

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    $\begingroup$ Quick thought: I think this is probably impossible, because of automorphism groups. Finitely generated groups such as $F_n$ or $\mathbf Z^n$ have large automorphism groups, but finite type simplicial models for $BG$ typically seem to have fewer automorphisms. I do not know how to turn this into a proof, though. $\endgroup$ Commented Nov 18 at 10:17
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    $\begingroup$ This cannot be done even for constant diagram. You need to realize every automorphism of your (free) group by a simplicial map of degreewise finite simplicial sets, but up to homotopy every such map is determined by its action on 1-simplices; so there are only finitely many possibilities. $\endgroup$ Commented Nov 18 at 10:53
  • $\begingroup$ @DenisT, thank you, I was too optimistic and updated the message. Please forget about the naive part of the question. $\endgroup$ Commented Nov 18 at 11:03
  • $\begingroup$ When you say "small category," do you mean finite? Otherwise, this is generally impossible, for roughly the reason that Denis T said above. $\endgroup$ Commented Nov 19 at 14:52
  • $\begingroup$ @K.Strong, thanks for pointing this, of course I really mean a finite diagram. Corrected. $\endgroup$ Commented Nov 21 at 7:10

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Not every finitely presented group has a classifying space with finitely many cells in each dimension. A group is said to be of type $F_n$ if it has a classifying space with finite $n$-skeleton, of type $F_\infty$ if it is of type $F_n$ for all $n$, and of type $F$ if it has a classifying space with finitely many cells. The first example of a finitely presented group not of type $F_3$ was due to Stallings, and it can be described as the kernel of a map $(F_2)^3\rightarrow \mathbb{Z}$, where $F_2$ denotes the free group on two generators, and the homomorphism takes each standard generator of each $F_2$ to $1\in \mathbb{Z}$.

You could instead ask your question for groups of type $F$ or for groups of type $F_\infty$.

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  • $\begingroup$ Sure, as I mentioned I'm mainly interested in the case of finitely generated free groups. I will edit the message to be clear. $\endgroup$ Commented Nov 18 at 10:47
  • $\begingroup$ Even for finite categories $D$, arbitrary finitely presented groups can arise as retracts of colimits of diagrams of free groups, so the difficulty that I suggest won't go away. $\endgroup$ Commented Nov 18 at 12:18
  • $\begingroup$ Dear @IJL, could you clarify it a bit? As far I understand your objection, a coherent lifting of $B(G_x),x\in D$ to spaces of finite type, somehow provides a finite type space model for classifying space of a retract of $colim_x G_x$. I don't understand the argument, because the classifying space functor doesn't commute with (homotopy) colimits. For example it doesn't imply that $B(colim G_x)$ admits a finite type model. $\endgroup$ Commented Nov 18 at 13:05
  • $\begingroup$ No, I meant to say that the fundamental group of the hocolim of the classifying spaces of the finitely generated free groups might have a fundamental group whose classifying space cannot be finite. This is probably not relevant to your Q2, except that the colimit of the functor $S$ in your question will not be aspherical in general. $\endgroup$ Commented Nov 18 at 13:41

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