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Show $\otimes$ and $*$ are the same operation on $\pi_1(G, x_0)$ where $(f\otimes g)(s) = f(s) \cdot g(s)$ where $\cdot$ is the group operation on the topological group $G. $

This is a question from the Munkres text which is really frustrating me. It's number 7 on page 335 for anyone with the book. I've already proved the first two parts.

There is a hint and that is to compute $(f*$ $e_x)$ $\otimes$ $(e_x$ $*g)$.

When I do this it is pretty easy to see that this is the same as $f \otimes g$ but am at a loss as to how this is the same as $f*g$.

Stefan Hamcke
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EgoKilla
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    Show that $(a\ast b)\otimes (c\ast d) = (a\otimes c)\ast(b\otimes d)$. – Daniel Fischer Apr 05 '14 at 21:44
  • Is it just as simple as rearranging parentheses since both operations are associative? – EgoKilla Apr 05 '14 at 23:39
  • It is as simple, but it's not the associativity. It's directly the definition once you pick representatives of the homotopy classes. – Daniel Fischer Apr 05 '14 at 23:42
  • So for example $(ab) \otimes (cd) = [(ab) \otimes (cd)] = [a(b \otimes c)d]$? – EgoKilla Apr 06 '14 at 00:04
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    No. $$(a\ast b)\otimes(c\ast d) = (a\ast b)(s)\cdot(c\ast d)(s) = \begin{cases}a(2s)\cdot c(2s) &, s \leqslant \frac12\ b(2s-1)\cdot d(2s-1) &, \frac12 \leqslant s,\end{cases}$$ if we don't distinguish the homotopy classes from their representatives notationally. – Daniel Fischer Apr 06 '14 at 00:08
  • I see how to solve it from there but I don't understand how you got that. – EgoKilla Apr 06 '14 at 00:12
  • It's the definition of $\otimes$ and of $\ast$, there's nothing more to it than just writing down how the things are defined. – Daniel Fischer Apr 06 '14 at 00:13
  • Oh I see it now thanks, I wasn't thinking of path concatenation in the proper way. I was thinking you'd have to travel all the way to whatever point on ab before you could multiply by cd but really you're evaluating a and c together and b and d together. Since that's where you are on both paths. – EgoKilla Apr 06 '14 at 00:18

1 Answers1

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The operation $⊗$ can be defined on classes as the composition $$\pi_1(X,x_0)\times\pi_1(X,x_0)\xrightarrow{\cong} \pi_1(X×X,(x_0,x_0)) \xrightarrow{\pi_0(\cdot)} \pi_1(X,x_0^2)$$ which send two classes $[g]$ and $[f]$ to the class $[g⊗f]$. Since we want this to be an operation on $\pi_1(X,x_0)$, we must choose $x_0$ to be the identity $e$.

We see that both $(f⊗g)*(k⊗l)$ and $(f*k)⊗(g*l)$ evaluate to $$s\mapsto \begin{cases} f(2s)\cdot g(2s) &\text{ if }s\le\frac12\\ k(2s-1)\cdot l(2s-1) &\text{ if }s\ge\frac12 \end{cases}$$ The constant loop $x_0$ is the identity for $⊗$.

Now $[f]⊗[g] = ([f]*x_0)⊗(x_0*[g]) = ([f]⊗x_0)*(x_0⊗[g]) = [f]*[g]$

Stefan Hamcke
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