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