Topological group closed path

Question

A topological group $G$ is a group that is also a topological space in which the maps $u:G×G → G$， $v:G→G$ defined by $u(g_1, g_2)=g_1g_2$ and $v(g)=g^{-1}$ are continuous. Let $f,h$ be closed paths in $G$ based at the Identity element $e$ of $G$. Define $f.h$ by $(f.h)(t) = u(f(t),h(t))$ for all $t\in I$. Show that the homomorphism $v_*:\pi(G,e)→\pi(G,e)$ satisfies $v_*[f]=[f]^{-1}$.

Here is what I think, by definition of the induced homomorphism we know $v_*[f]=[vf]$. now by the definition of $v_*$ we have $v(f)=f^{-1}$. hence, $v_*[f]=[vf]=[f^{-1}]$. so all I need to show is $[f^{-1}]=[f]^{-1}$ if what I have done is correct, but I am having trouble about this.

This exercise is proposed in Kosniowski's "A first course in Algebraic Topology" chapter 15. Any help would be appreciated!

Answer 1

Since $f * f^{-1} \sim f\cdot f^{-1} = c_e$ and $f^{-1} * f \sim f^{-1}\cdot f = c_e$, $[f]^{-1} = [f^{-1}]$.

Answer 2

More on this situation is available in this paper on covering groups of non-connected topological groups.