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I have tried to prove that the equivalence of loops is preserved by the $\ast$ product, i.e. if $\alpha \sim_{x_0} \alpha'$ and $\beta \sim_{x_0} \beta'$, then $\alpha \ast \beta \sim_{x_0} \alpha' \ast \beta'$, where $\alpha$, $\alpha'$, $\beta$, and $\beta'$ are loops based at $x_0$, and $\ast$ is the usual product of paths given by

$(\alpha \ast \beta)(t)= \begin{cases} \alpha(2t) & 0 \le t \le \frac{1}{2} \\ \beta(2t+1) & \frac{1}{2} \le t \le 1 \\ \end{cases}$.

Since $\alpha \sim_{x_0} \alpha'$ and $\beta \sim_{x_0} \beta'$, there are homotopies $H(t,s)$ and $J(t,s)$ for which $H(t,0)=\alpha(t)$, $H(t,1)=\alpha'(t)$, $H(0,s)=\alpha(0)=\alpha'(0)$, and $H(1,s)=\alpha(1)=\alpha'(1)$, and similar for $J(t,s)$.

It makes sense that if we need a homotopy of a product then we should take the product of the already known homotopies. So, let $K(t,s):=H(t,s) \ast J(t,s)$. I have verified that the requirements of a homotopy (like those shown for $H$ above) hold for $K(t,s)$, so I think it is a homotopy from $\alpha \ast \beta$ to $\alpha' \ast \beta'$.

I would just like to know if it makes sense to take the product of homotopies in this way, or if there's something I'm not seeing that prevents this from working.


share|cite|improve this question
Yes, it makes sense. How do you define that? – Berci Feb 8 '13 at 19:25
@Berci: Define what, exactly? $K(t,s):=H(t,s) \ast J(t,s)$? – Alex Petzke Feb 9 '13 at 0:03

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