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I'm having trouble with the following elementary(?) thing, but its confusing me alot.

For a category $\mathcal{C}$, we can define the classifying space in terms of nerve.

Similarly, for a topological group (or monoid) G, we can think of it as a category with one point, say $*$.

So, $B_{n}G$ is the set of all $* \xrightarrow{g_0} * \xrightarrow{g_1} \cdots \xrightarrow{g_n} *$ (where $g_i \in G$) or as a tuple $(g_0, g_1, \cdots ,g_n)$, i.e. $B_{n}G = G^{n+1}$.

To get its classifying space $BG$, we can take its geometric realization.

The question that I'm bothered with is, how can we think (or describe) the loops in $BG$, that is, how can we describe the elements of $\Omega BG$, explicitly?

I think we could think of it as tuples $(g_0, ... g_n)$ such that $g_0g_1 \cdots g_n = e$, but I'm not sure...

May
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    $\Omega BG\simeq G$... https://math.stackexchange.com/questions/442805/the-loop-space-of-the-classifying-space-is-the-group-omegabg-cong-g – Bob Dobbs Aug 11 '23 at 07:21
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    Yes, I'm familiar with that result, and also the group completion theorem. I would like to know explicitly how they look like... – May Aug 11 '23 at 07:49
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    So, from the homotopy equivalance,, dont we think that each loop corresponds to a group element? – Bob Dobbs Aug 11 '23 at 07:56
  • You're right, but I'm not quite satisfied (sorry if I'm too hand wavy in this regard) as it's not a bijective correspondence. For instance, in https://math.stackexchange.com/questions/1313289/canonical-map-of-a-monoid-to-its-classifying-space, we find a loop $l_m$ for $m \in M$, a (topological) monoid. I'm sure we can find more such loops... – May Aug 11 '23 at 08:51
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    Do you care about precisely representing the loops or do you care about precisely representing the loop homotopy classes? – FShrike Aug 11 '23 at 09:15
  • I think your indexing is wrong: $B_nG$ consists of tuples of length $n$, not length $n+1$. – John Palmieri Aug 11 '23 at 14:52
  • @wanderer Why ask for a bijective correspondence? In algebraic topology and category theory, this is almost always the wrong notion of sameness. A homotopy equivalence does tell exactly what the loops "look like". At least, up to homotopy. But thats what we care about, right? – IsAdisplayName Aug 12 '23 at 00:11
  • Sorry for the late comment, I require bijective correspondence because I'm trying to deconstruct various proofs and getting some ideas... – May Aug 15 '23 at 09:19

1 Answers1

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The $1$-simplices in $BG$ have the form $* \xrightarrow{g} *$, so each one is a loop already. The set of all loops is therefore the set of all concatenations of these: the set of all paths. The homotopies will be determined by the fact that the path $* \xrightarrow{g} * \xrightarrow{h} *$ is homotopic to $* \xrightarrow{gh} *$ because there is a $2$-simplex with faces $g$, $h$, and $gh$. As a result, every loop will be homotopic to $* \xrightarrow{g} *$ for some $g$.

John Palmieri
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  • Thanks! (Sorry for the late reply) I am understanding what you're trying to say, but... then does it mean that every element of $BG$ is a... loop? – May Aug 15 '23 at 09:21
  • Well, $G$ as a category has only one object, so $BG$ has only a single $0$-simplex. Therefore every $1$-simplex starts and ends at that $0$-simplex, so every $1$-simplex is a loop, which means that every concatenation of $1$-simplices is a loop. – John Palmieri Aug 15 '23 at 15:05
  • I would not say, though, that every element of $BG$ is a loop. The elements of $B_nG$ are $n$-simplices, and topologically those aren't loops but higher-dimensional things. They all have just one vertex, though. – John Palmieri Aug 15 '23 at 15:07
  • I see. Thanks again :). – May Aug 17 '23 at 04:17