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According to the answers to this question, every group can be realized as the localization of a partially ordered set at some subcategory of weak equivalences. However, I've been unable to construct anything but free groups, because I don't understand how to utilize the 2-simplices (or composition of morphisms in my poset) to get a desired quotient.

My suspicion is that localizing a poset $\mathcal{P}$ at every morphism produces the fundamental group of the geometric realization of (the nerve of) $\mathcal{P}$. If this is true, then the answer to this question constructs a topological space with fundamental group any finitely presented group, and then we can just craft a poset whose geometric realization is said topological space.

My problem is that I have no idea what the construction is saying (form $X$ by attaching a copy of $D^2$ to $S$ using $f_\beta$? Excuse me?) on top of not really understanding geometric realization well; the question is too complicated to brute force with computer search, and I have been drawing diagrams of posets to try to figure out how to make a cyclic group of order 2 for days with no end in sight.

So, for a given $n \in \mathbb{N}$, what is a poset whose localization is $C_n$?

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  • $\begingroup$ I think it might be helpful for you to split your question up. The bit about constructing 2-complexes with a given fundamental group is quite elementary (i.e., I can answer it $\ddot{\smile}$), so maybe you could usefully separate that bit out. For the cyclic group of order 2 all you need is real projective 2 space. $\endgroup$
    – Rob Arthan
    Commented Sep 20 at 23:47
  • $\begingroup$ @RobArthan My frustration seems to come from how difficult it is to translate from the CW-complex structure to a poset structure. In particular, the data associated to a CW-complex is given in rather hand-wavy ways, for example in your link it says "Real projective spaces have a simple CW complex structure, as $Pn(R)$ can be obtained from $Pn−1(R)$ by attaching an n-cell with the quotient projection $Sn−1 → Pn−1(R)$ as the attaching map. " and I have absolutely no idea what that means in terms of a poset that realizes to it, since I have to express that quotient via mere arrows. $\endgroup$
    – L. E.
    Commented Sep 21 at 0:03
  • $\begingroup$ It also seems like all I need to understand what's going on is a diagram of some poset that localizes to a finite group and everything will become clear even if there is no explanation attached to that diagram. $\endgroup$
    – L. E.
    Commented Sep 21 at 0:08
  • $\begingroup$ I probably misunderstood what you were having difficulty with. What is $f_{\beta}$ in your context? $\endgroup$
    – Rob Arthan
    Commented Sep 21 at 0:10
  • $\begingroup$ @RobArthan The quote is from the answer to "Example of finite fundamental group", $f_\beta$ being one of the relations in the presentation of the group. Ostensibly, the generators of the group translate into the 1-cells and the relations translate into the 2-cells we quotient by. My issue is that, given relations as simple as $g^n = 1$ on a single generator, I cannot translate that to poset data. $\endgroup$
    – L. E.
    Commented Sep 21 at 0:12

1 Answer 1

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[I am not an expert; I am learning about these things in the process of writing this answer, so please take it with a grain of salt.]

I assume that you're looking for a finite poset that can be drawn on the page, rather than some generic recipe like "take the poset of nondegenerate simplices of a subdivision of the nerve of $\mathbb{Z}_n$" (see e.g. nLab or Kerodon).

It seems unlikely (to me, not an expert) that there is a finite poset whose geometric realisation is $B\mathbb{Z}_n$ for $n \geq 2$, since that space is infinite-dimensional (see this MO question).

On the other hand, if we're just looking for some space whose fundamental group is $\mathbb{Z}_2$, then we can look at the real projective plane $\mathbb{R}P^2$. The first step is to find a triangulation of $\mathbb{R}P^2$ as a simplicial complex, which we can find in these slides. For intuition, observe that removing vertex $4$ (or $5$ or $6$, by symmetry) leaves us with a triangulation of the Möbius strip (the one that appears e.g. here); the remaining faces attach a disk along the boundary of the strip.

We can then take the poset of faces of this simplicial complex ordered by inclusion (see Poset topology §1.1). This can be drawn on a finite sheet of paper, but it is not particularly enlightening IMO. It is likely that a simpler poset with the same geometric realisation exists. Localising this at every morphism should then yield a groupoid equivalent to $B\mathbb{Z}_2$. Don't ask me how to prove this.

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  • $\begingroup$ Thank you for leaving this answer, I figured this was the answer but couldn't justify it well enough. I agree that the resulting poset is not particularly enlightening. $\endgroup$
    – L. E.
    Commented Oct 7 at 0:33

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