Homotopy between $a^{-1}$ and $a^*$ in topological group.

Question

In a topological group $G$ , indicate with $g^*$ the inverse of $g\in G$ and with $a^*: I\to G$ the path that is obtained from $a: I \to G$ by setting $a^*(s) = a(s)^*$. Prove that for every closed path $a:I \to G$, based on the neutral element of $G$, we have $a^*\simeq a^{-1}$, where $a^{-1}(s) = a(1-s)$. Anybody can help me?

Answer 1

This follows from the Eckamnn-Hilton argument : let's look at $\pi_1(G,e)$, denote by $ab$ the usual concatenation of paths, and by $a\otimes b$ the "groupwise" muliplication, that is (by abusing notation and mistaking the class of a path with the path itself) $(a\otimes b)(t) = a(t)b(t)$.

Then clearly both $\pi_1(G,e)$ and $(\pi_1(G,e),\otimes)$ are groups. But also since multiplication (let's denote it by $m$) in $G$ is continuous, $\otimes: \pi_1(G,e) \times \pi_1(G,e) \to \pi_1(G,e)$ is actually induced by $m: G\times G\to G$ (identifying $\pi_1(G)\times \pi_1(G)$ and $\pi_1(G\times G)$ by the canonical isomorphism).

Hence, being induced by a continuous map, it is a group morphism ! By the Eckmann-Hilton argument, it suffices to show that $\otimes$ and standard multiplication are equal; in particular the two inversions ($*$ and $^{-1}$) are equal as well. But equal in $\pi_1$ means homotopic "in real life" : that's what we wanted to prove.

Granted, this is a bit abstract but it's the real reason why things work. If you want to unwrap what's happening you can see that $(ac)\otimes (bd) = m_* ((ac, bd)) = m_*((a,b)(c,d)) = m_*(a,b)m_*(c,d) = (a\otimes b)(c\otimes d)$ which is what makes the Eckmann-Hilton argument work.

One may also ask for a more down-to-earth, explicit homotopy. Here's an idea of how it would work : you want to prove that $\otimes$ is standard multiplication; consider two paths $a,b$. Up to a homotopy, you can make $a$ move quicker, so that it rund twice as fast up to $\frac{1}{2}$, and then stays at $e$. Similarly, you can make $b$ run twice as fast, but after a while where it stayed at $e$: that is $e$ up to $\frac{1}{2}$ then $b$ twice as fast. This gives you $a', b'$. But now $a\otimes b$ is homotopic to $a'\otimes b'$, and it's quite clear that $a'\otimes b'$ is the concatenation of $a$ and $b$. Once again, you see that you have equality between two group operations, so equality between their inverses.

But let me just say that the Eckmann-Hilton argument is important, in that it helps for this exercise, but it can also be used to show : $\pi_1(G,e)$ is abelian and for any topological space $X$ and $x\in X$, if $n\geq 2$, $\pi_n(X,x)$ is abelian

(Edit: maybe the point I'm making is not clear: showing that $a^* $ and $a^{-1}$ are homotopic is the same as showing that their classes are equal in $\pi_1(G,e)$. What I'm essentially doing here is showing in various ways that the two group structures on $\pi_1(G,e)$ -the one coming from $G$, that is multiplication at each moment, and the one coming from $\pi_1$, that is concatenation- are the same, hence they must have the same inversion: since one of these inversions is $a^*$ and the other $a^{-1}$, we get what we wanted. Thus it suffices to show that these group structures are the same, which is what I'm doing here)