I'm working with Viro's textbook on topology, and I stuck on this exercise on induced homomorphisms:
Where $ T_s : \pi_1(X,x_0) \to \pi_1(X,x_1) : [\alpha] \mapsto [s^{-1}\alpha s] $ -- a translation map.
I picked an element $ [u] \in \pi_1(X,x_0) $ and applied given compositions in order to get the same result, but that's what I got: $$ (f_* \circ T_s)[u] = [f \circ (s^{-1}us)] = [(f\circ s^{-1})(f\circ u)(f\circ s)] $$ $$ (T_{f\circ s} \circ f_*)[u] = T_{f\circ s}([f\circ u]) = [(f\circ s)^{-1} (f\circ u) (f\circ s)] $$
And I can't get further: either I made a mistake somewhere or I need to show that $ f\circ s^{-1} = (f\circ s)^{-1} $. I feel like this question is a bit silly and all, but I can't figure this out.
Note that the inverse here denotes inverse in the sense of paths, i.e. $s^{-1}(t) = s(-t)$. With this in mind, we have
$$(f\circ s)^{-1}(t) = (f\circ s)(-t) = f(s(-t)) = f(s^{-1}(t)) = (f\circ s^{-1})(t)$$
and hence $(f\circ s)^{-1} = f\circ s^{-1}$.