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Hypothesis: Let $G$ be a topological group with identity element $e$. Let $\mu$ denote the multiplication mapping in $G$.

Goal: Show that $\pi_1(G,e) = \pi(G)$ is an abelian group via the hint below.

Hint: There are two products on $\pi(G)$. The usual product $\circ$ defined for the fundamental group and the product $\ast$ induced by

$$ \ast: \pi(G) \times \pi(G) \cong \pi(G \times G) \overset{\pi(\mu)}{\rightarrow} \pi(G). $$

Show that there is a common two sided unit, $u \in \pi(G)$ for both products and there is a distributive law

$$ (f \ast g) \circ (a \ast b) = (f \circ a) \ast (g \circ b) $$

Attempt:

  1. Suppose the distributive law in the hint holds. Suppose further that $1$ -- the identity element in $\pi(G)$ with respect to $\circ$ -- serves also as the identity element in $\pi(G)$ with respect to $\ast$.

  2. Then for $a,b \in \pi(G)$, we would have the following relationship:

    $$ (1 \ast a) \circ (b \ast 1) = (1 \circ b) \ast (a \circ 1) $$

    so that

    $$ a \circ b = b \ast a $$

  3. Then if we can show that for all $f,g \in \pi(G)$ that $f \ast g \iff f \circ g$, we could complete the above relation to

    $$ a \circ b = b \ast a = b \circ a $$

    so that $\pi(G)$ is abelian as desired.

Question: Am I on the right track?

Stefan Hamcke
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user125103
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    Looks good. Note that $1\circ b = b\circ 1$. – Daniel Fischer Mar 26 '14 at 18:43
  • Good point. I think the more relevant relation is $1 \ast a = a \ast 1$ so that we can say $a \circ b = (a \ast 1) \circ (1 \ast b) = (a \circ 1) \ast (1 \circ b) = a \ast b$. – user125103 Mar 26 '14 at 19:31
  • How does one go about proving the distributive law above? – user125103 Mar 26 '14 at 19:32
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    I argue $$b\circ a = (1\circ a)\ast (b\circ 1) = a\ast b = (a\circ 1)\ast (1\circ b) = a\circ b.$$ The distributive law is a direct consequence of the definitions, $$(f\circ g)(t) = \begin{cases} f(2t) &, t \leqslant \frac12\ g(2t-1) &, t \geqslant \frac12 \end{cases}$$ and $(f\ast g)(t) = f(t)\cdot g(t)$. – Daniel Fischer Mar 26 '14 at 19:36
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    Very relevant: https://en.wikipedia.org/wiki/Eckmann%E2%80%93Hilton_argument – Najib Idrissi Mar 27 '14 at 13:49
  • @DanielFischer : Hi Daniel, hope you get this message even though your response is one year old. I am very, very interested in understanding the proof to distributive law you wrote above. Is there anyway you could give a more elaborate explanation, befitting the slow mind of a slowpoke? Thanks for your time and help. – A.Magnus Mar 05 '15 at 23:47
  • @A.Magnus I think the comments under the question and Stefan Hamcke's answer here should help. If those aren't sufficient, ping me again. (But I'm going to bed soon, so a response will take some time.) – Daniel Fischer Mar 06 '15 at 00:08
  • @DanielFischer : Hope you had slept well & woke up fresh! I guess I didn't say it right: I'd would love to understand the proof the OP's distributive law, ie., $(f \ast g) \circ (a \ast b) = (f \circ a) \ast (g \circ b).$ Especially I didn't see how your piecewise function $(f \circ g)(t)$ fit in to the proof. I tried digesting Stefan's writing & yours but (sigh!) my analogue brain just isn't a match with your digital! Thanks again for your time and generosity! – A.Magnus Mar 06 '15 at 03:01
  • @DanielFischer : I have long been lamenting the lack of rich formating features and limited space when writing comments. However, your old comment above dated 03/26/2014 shows lots of formatting -- centering, piecewise function, etc. Additionally, you were able to creat link in latest comment. How did you accomplish all that? Thanks again and again. – A.Magnus Mar 06 '15 at 03:28
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    @A.Magnus For two loops $v,w$ in a space $X$ with the same base point $x_0$, we define the composition/concatenation of these loops by $$(v\odot w)(t) = \begin{cases} v(2t) &, 0 \leqslant t \leqslant \frac{1}{2}\ w(2t-1) &, \frac{1}{2} \leqslant t \leqslant 1. \end{cases}$$ Since the loops have the same base point, $v\odot w$ is well-defined, and a loop with the same base point. Then the operation on the set $\pi_1(X,x_0)$ of homotopy classes of loops in $X$ with base point $x_0$ induced by $\odot$ is $$[v]\circ [w] := [v\odot w].$$ – Daniel Fischer Mar 06 '15 at 11:29
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    One needs to check that $\circ$ is well-defined and defines a group structure on $\pi_1(X,x_0)$. Then we look at a topological group $G$, where we can define the pointwise product of two loops with base point $e$ as $$(v\otimes w)(t) = v(t)\cdot w(t),$$ which again is a loop with base point $e$. On $\pi_1(G,e)$, the operation induced by $\otimes$ is $$[v] \ast [w] := [v\otimes w].$$ Again one checks that that is well-defined and induces a group structure on $\pi_1(G,e)$. Now one sees that $$([f]\ast [g])\circ ([a]\ast [b]) = [f\otimes g] \circ [a\otimes b] = [(f\otimes g) \odot (a\otimes b)]$$ – Daniel Fischer Mar 06 '15 at 11:30
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    and $$([f]\circ [a])\ast ([g]\circ [b]) = [f\odot a] \ast [g\odot b] = [(f\odot a)\otimes (g\odot b)],$$ and inserting the definitions yields $$(f\otimes g)\odot (a\otimes b) = (f\odot a)\otimes (g\odot b),$$ which gives the desired distributive law. Re formatting: you can use MathJax in comments, $$...$$ gives you centered displayed formulae, and [link](http://example.com) is the usual Markdown link formatting. See also the "help" link under the "Add comment" button and the extended help. – Daniel Fischer Mar 06 '15 at 11:31
  • @DanielFischer : Thanks, I am beginning to see the light now. One last question: Is there any formal name to this distributive law? Thanks again. Just to let you know the magnitude of my gratitude for your time and patience: Thanks$^n$ - "Thanks" powered to the $n$th. (But I am sure my HS English teacher won't be very amused by this term.) – A.Magnus Mar 06 '15 at 14:49
  • @DanielFischer I was wondering, how would you show that they actually share the same identity element?

    In one case, we have the identity $$[f]\circ[e]=[f]$$ and in this case $[e]$ is the constant map because we are talking about loops in the fundamental group.

    On the other hand, we have$$[f]\ast[1]=[f\otimes1]=f(t)\cdot1$$ where $\cdot$ is the group operation and 1 is the group identity, so [1] is the homotopy class where the group identity belongs. How do I know that this homotopy class is the same as the homotopy class [e] of the constant map?

     – Mike Mar 29 '16 at 02:55
  • If I suppose that $[e]$ is the identity for the $\circ$ operation and try to apply it to $\ast$ I get:

    $$[f]\ast[e]=[f\otimes e]$$ which gives $$f(t)\cdot e(t)=f(t)\cdot C$$ where $C$ is the base point of the fundamental group, and there is no guarantee that the base point is the group identity.

     – Mike Mar 29 '16 at 03:00
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    @Craig We are looking specifically at $\pi_1(G,e)$, the base point is the identity element of the group $G$. The identity for $\circ$ is the homotopy class of the constant loop at the base point, and with base point $e$, that's also the identity for $\ast$. (For a different base point, the pointwise multiplication of loops would not [directly] induce an operation on $\pi_1(G,p)$. One could push it over via the isomorphism $\pi_1(G,p) \cong \pi_1(G,e)$ [since $\pi_1(G,e)$ is abelian the isomorphism is independent of the choice of path connecting $e$ and $p$], but it's not immediate.) – Daniel Fischer Mar 29 '16 at 10:43

2 Answers2

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Or a one-line proof: "the fundamental group functor preserves products, hence it sends group objects to group objects". Note that the group objects in ${\bf Top}$ are precisely the topological groups, and group objects in ${\bf Grp}$ correspond to abelian groups.

Edoardo Lanari
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One can get more out of this situation,which especially useful if the topological group $G$ is not connected. For more details, see this paper.

Let $\tilde{G}$ be the set of homotopy classes rel end points of paths in $G$ which start at $e$. The group structure of $G$ induces a group structure on $\tilde{G}$. The final point map defines $t: \tilde{G} \to G$, which under appropriate local conditions is the universal cover of $G$ at $e$. This morphism may also be given the structure of crossed module, using the conjugation operation of $G$ on $\tilde{G}$.

Recall that a crossed module $\mu: M \to P$ is a morphism of groups together with an action of $P$ on $M$ written $(m,p) \mapsto m^p$ with the properties

  1. $\mu(m^p)= p^{-1}\mu(m) p$;

  2. $m^{-1}nm=n^{\mu m}$

for all $m,n \in M, p \in P$. Such a crossed module determines a $k$-invariant $k \in H^3(Cok\; \mu, Ker\; \mu)$.

For the crossed module coming from a topological group as above, the $k$-invariant is trivial if and only if the topological group $G$ has a universal cover of all components such that the totality can be given the structure of topological group with the covering map a morphism of topological groups.

Ronnie Brown
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