The theorem is: Let be f, g : X→Y two continuous maps between topogolical spaces and
H : X$\times$[0, 1] a homotopy such that H(x,0)=f(x) and H(x,1)=g(x).
Given $x_0\in X$ let be $y_0=f(x_0)$, $y_1=g(x_0)$ and $\tau(t)=H(x_0,t)$ a path in Y from $y_0$ and $y_1$.
Then, $g_*=\varphi_{[\tau]}\circ f_*$
i.e. $\require{AMScd}$ \begin{CD} \pi_1(X,x_0) @>f_*>> \pi_1(Y,y_0)\\ @V V V @VV \varphi_{[\tau]} V\\ @>>g_*> \pi_1(Y,y_1) \end{CD}
where
$f_*:\pi_1(X,x_0) \rightarrow \pi_1(Y,y_0),\space f_*([\gamma])=[f\circ \gamma]$,
$g_*:\pi_1(X,x_0) \rightarrow \pi_1(Y,y_1),\space g_*([\gamma])=[g\circ \gamma]$ and
$\varphi_{[\tau]}:\pi_1(Y,y_0) \rightarrow \pi_1(Y,y_1),\space\varphi_{[\tau]}([\gamma])=[\tau^{-1}]·[\gamma]·[\tau]$ In the diagram it's supposed that $\pi_1(X,x_0)$ goes directly to $\pi_1(Y,y_1)$ by $g_*$
$f_*$ and $g_*$ are group homomorphism and $\varphi_{[\tau]}$ is a group isomorphism.
I began supposing
$g_*([\gamma])=\varphi_{[\tau]}\circ f_*([\gamma])=[\tau^{-1}]·[f\circ\gamma]·[\tau] \iff [\tau]·g_*([\gamma])·[\tau^{-1}]·f_*([\gamma]^{-1})=e_{y_0}$
and from here I don't know how to continue, I think there's something with four components which can help me to complete the proof.
Thanks.
