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I'm really struggling with some exercise my professor left me about fundamental group so I think I need some clarification.
During one of them I found that any loop $\omega$, where $\omega$ belongs to the fundamental group of a connected space, it is homotopic to its inverse $\omega^{-1}$, but what does that means?
In my opinion the only way that could happen is that they are both contractible. Am I missing any other possibility?

Barbamento
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    This might help: https://math.stackexchange.com/questions/2459811/is-a-loop-in-x-based-at-a-in-x-homotopic-to-its-reverse-loop – Aniruddha Deshmukh Apr 08 '21 at 07:57
  • I spent more than one hour searching if that question had already been answered but I didn't find it. That was exactly what I was lookin for, thanks a lot – Barbamento Apr 08 '21 at 08:05
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    You should also be careful because a loop can be homotopic to its own inverse and not be contractible. – memerson Apr 08 '21 at 08:06
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    $\omega=\omega^{-1}$ in $\pi_1(X)$ simply means that $\omega$ is of order $2$. It does not have to be contractible. This is true for example when $\pi_1(X)=\mathbb{Z}_2$ which happens when $X$ is the real projective plane $\mathbb{R}P^n$ for $n>1$. – freakish Apr 08 '21 at 08:08

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No they don't need to be (I guess you mean) null homotopic.

Recall that the reverse loop is just $\omega^{-1}(t)=\omega(1-t)$.

Actually it implies that the fundamental group has an exponent of $2$: we have $\omega\cong\omega^{-1}\implies\omega^2\cong e$. Put differently, each $\omega$ would have to be involutive. In particular, $\pi_1(X)$ would then be abelian.

So the space doesn't have to be simply connected, but spaces like the circle and torus do not have this property. It is true, however, if the space is simply connected, as you surmised, since then all the loops are null homotopic. Take $S^n,n\ge2$, for example.

  • The question doesn't mention an incorrect statement made by the professor. – jMdA Apr 08 '21 at 08:42
  • @jMdA yes it does: it says the professor stated that in a connected space a loop is homotopic to its reverse. –  Apr 08 '21 at 08:44
  • No, It's not a statement from the professor. It says: "During one of them I found that..." so it's likely the professor had nothing to do with the statement that "in a connected space any $\omega$ is homotopic to $\omega^{-1}$". Maybe it's simply a mistake by the person who asked the question, or perhaps there's missing context that made the statement make sense in the exercise. – jMdA Apr 08 '21 at 09:11
  • @jMdA oh, ok, i suppose I could have read it more carefully –  Apr 08 '21 at 09:13
  • Yes it wasn't a statement from my professor. Since he want us to understand what we do there's no point in asking here the solutions of my exercises. I use it to check if the conclusion I got could be correct or to have a better understanding of some notions.

    And I'm almost sure that he spend some times in here.

     – Barbamento Apr 09 '21 at 07:41