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The classifying space $BG=|Nerve(G)|$ of an arbitrary topological group $G$ does not necessarily have the homotopy type of a CW-complex but the fundamental group should still be accessible. What is $\pi_{1}(BG)$? A reference on this would be great. My initial guess: $\pi_{1}(BG)$ is the quotient group $\pi_{0}(G)$ for arbitrary $G$

Motivation: There is a natural way to make $\pi_1$ a functor to topological groups. I am interested in relating the topologies of $G$ and $\pi_{1}(BG)$ but the topology on $\pi_{1}(X)$ is boring (discrete) when $X$ is a CW-complex.

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    $\begingroup$ Should that be $\pi_0(G)$? Since the universal bundle $EG$ is contractible, then the long exact sequence of homotopy groups break into [1=\pi_1(EG)\to\pi_1(BG)\to \pi_0{G}\to\pi_0(EG)=1.] $\endgroup$
    – Fei YE
    Commented Oct 9, 2010 at 20:19
  • $\begingroup$ Is it obvious that the function $\pi_{1}(BG)\rightarrow \pi_{0}(G)$ from the LES is a homomorphism? $\endgroup$
    – Jeremy Brazas
    Commented Oct 9, 2010 at 21:05
  • $\begingroup$ @Jeremy, that it's a homomorphism looks fairly obvious to me --- but don't ask me why... @Fei, following on Jeremy's remark, the proofs (e.g. of long exact sequence) are usually presented for topological groups of the homotopy type of a CW complex, which is a specialization explicitly excluded by the question as posed. Put another way, it's not clear to me even that the sequence is exact. Jeremy, just how arbitrary can these groups be? Is a real algebraic group with the Zariski topology a case you want to consider? $\endgroup$
    – some guy on the street
    Commented Oct 9, 2010 at 21:42
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    $\begingroup$ In general, there is no map $\pi_1(BG) \to \pi_0(G)$. See my post below. $\endgroup$
    – André Henriques
    Commented Oct 9, 2010 at 22:27

3 Answers 3

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If $G$ is homeomorphic to a Cantor set (e.g. $G=\mathbb Z_p$), then $BG$ contains a copy of the Hawaiian earrings in it. To see this, take a sequence of points of $G$ that converges to the identity element: you'll get a corresponding sequence of 1-cells in $BG$ that converge to the the degenerate 1-cell. The fundamental group of the Hawaiian earrings is a rather wild object, and looks nothing like the free group that you might naively expect.

If, on the other hand, if you agreed to redefine $BG$ to be the fat geometric realization of the simplicial space $NG$, then you would get $\pi_1(BG)\cong\pi_0(G)$, as desired. I would even bet that the above isomorphism respects the natural topologies that are present on both sides.

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  • $\begingroup$ Could you explain the subtelty between fat and ordinary geometric realization ? $\endgroup$
    – BS.
    Commented Oct 10, 2010 at 9:34
  • $\begingroup$ Thanks very much Andre! The business about the geometric realization makes sense. I am not that familiar with fat geometric realizations but I believe the difference comes from restricting the identifications to only strictly increasing maps in $\Delta$ (correct me if I am wrong). Is there somewhere I can find a proof of the assertion about the group isomorphism? $\endgroup$
    – Jeremy Brazas
    Commented Oct 10, 2010 at 15:03
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    $\begingroup$ To get a better homotopy type, one can alternatively replace $G$ by the realization $G' :=|S.G|$, which is a topological group in the compactly generated topology ($S.$ = total singular complex, $| \quad |$ = realization). Then the map of topological groups $G' \to G$ is a weak homotopy equivalence. Then the fat and thin realizations of $N.G'$ coincide up to homotopy. $\endgroup$
    – John Klein
    Commented Nov 25, 2011 at 19:53
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    $\begingroup$ @jdc: One advantage of the thin realization is that it commutes with products (Milnor's theorem) so long as you work in the category of compactly generated spaces. Another is that level-wise constructions, like level-wise loop spaces, give the right homotopy type (this is in May's Geometry of Iterated Loop Spaces). I would agree with Segal (see the Appendix to Categories and Cohomology Theories) that the thick realization gives the correct homotopy type in general, and the thin version is a useful tool (e.g. for the previous reasons) when it's homotopy equivalent to the thick version. $\endgroup$
    – Dan Ramras
    Commented Jul 11, 2018 at 3:00
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    $\begingroup$ @JeremyBrazas I never found a proof of Andre's general statement, so I gave a short proof in the appendix to my preprint. (I prove the more general statement for arbitrary topological monoids with $\pi_0 G$ is replaced by its Grothendieck group.) In addition to using the thick realization, I needed to use the compactly generated topology on $G^n$ when forming the nerve $NG$; I don't know if this matters. I haven't thought about whether my argument says anything about the topologized fundamental group. $\endgroup$
    – Dan Ramras
    Commented Jul 11, 2018 at 3:17
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The first reference in this general area was:

N. E. Steenrod, "Milgram's classifying space of a topological group", Topology 7 (1968) 349–368.

Working in the category of compactly generated Hausdorff spaces, and combining his Theorems 8.1 and 8.3 we have:

Theorem: Let $G$ be a topological group with unit $e$, such that $(G, e)$ is an NDR. Then the canonical map $EG \to BG$ is a quasi-fibration and a principal $G$-bundle.

The NDR-condition asks that the inclusion of $e$ in $G$ is a cofibration. That is of course weaker than asking for $G$ to be a CW-space, but not as general as how the question was posed.

Since $EG$ is contractible, the long exact sequence in homotopy for a quasi-fibration, at the usual base point of $BG$, gives a bijection $$ \partial : \pi_1(BG) \cong \pi_0(G) $$ of $\pi_1(BG)$-sets. This at least gives you an isomorphism between $\pi_1(BG)$ and some group structure on $\pi_0(G)$.

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Readers might also like to know a result on the first k-invariant of BG contained in

R. Brown and C.B. Spencer, $\cal G$-groupoids, crossed modules and the fundamental groupoid of a topological group'', Proc. Kon. Ned. Akad. v. Wet. 7 (1976) 296-302.

This shows that the associated crossed module to fundamental groupoid of $G$ has $k$-invariant which is exactly the $k$-invariant of $BG$.

Since this query is about non-connected topological groups, another related ressult is on universal covers of non-connected topological groups

R. Brown and O. Mucuk, ``Covering groups of non-connected topological groups revisited'', Math. Proc. Camb. Phil. Soc, 115 (1994) 97-110.

which relates the question of the existence of topological group universal covers of $G$ to the theory of ostructions to extensions of abstract groups.

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  • $\begingroup$ I should have added that the $k$-invariant referred to is exactly the obstruction to there being a `universal covering' topological group $\tilede G$ of $G$ (in the sense that there is a morphism $p: \tilde{ G} \to G$ of topological groups which on each component is a universal cover of spaces). $\endgroup$
    – Ronnie Brown
    Commented Nov 25, 2011 at 19:06
  • $\begingroup$ Misprint: $\tilede{ G}$ should be $\tilde{G}$. $\endgroup$
    – Ronnie Brown
    Commented Nov 25, 2011 at 19:07

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