Homotopy between inverse path

Question

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?

Answer 1

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.