I'd like to understand the proof given to me that if $f : X \longmapsto Y$ is an homotopy equivalence then the induced map $f_{*}$ on the $\pi_{1}$ is an isomorphism.
I saw there are other methods, but the one my professor gave, was slightly different from anything I try to found on Stack (freely link them otherwise).
Since that $g \circ f \sim Id_{X}$ we have the same induced map of the identity on the $\pi_{1}$, which "coincide with some $\gamma_{\#}$, and then is an isomorphism". (The same argument is applied to $f \circ g \sim Id_{Y}$)
Where $\gamma_{\#} : \pi_{1}(X,a) \longmapsto \pi_{1}(X,b)$ defined as $\gamma_{\#}([\alpha]) = [\bar{\gamma}\star \alpha \star \gamma]$, with $\gamma$ a path from $a \to b$. Which result to be an isomorphism if $X$ is path connected.
I don't understand the sentence between the quote marks. Why every isomorphism of $\pi_{1}$ should be induced from such a $\gamma$ ?
The idea is that homotopic map induces the same map on the fondamental group, the only problem is that the base point can change but you fix it with the isomorphism $\gamma_\#$. Let $F:X\times I \to Y$ be a homotopy between $f$ and $g$ and $a\in X$ a base point. Let $\gamma$ the path from $f(a)$ to $g(a)$, given by $F(a,t)$, then you can check that the diagram $$\require{AMScd} \require{cancel} \def\diaguparrow#1{\smash{\raise.6em\rlap{\ \ \scriptstyle #1} \lower.6em{\cancelto{}{\Space{2em}{1.7em}{0px}}}}} \begin{CD} && \pi_1(Y,f(a))\\ & \diaguparrow{f_*} @VV\gamma_\#V \\ \pi_1(X,a) @>>g_*> \pi_1(Y,g(a)) \end{CD}$$ is commuting (this is just one of the classic and tedious verification one have to do before working on fondamental group), so $\gamma_\#\circ f_*=g_*$. If $f:X \to X$ and $ g$ is the identity you get the result you need.