Let $X$ and $Y$ be two topological spaces with basic points $x_0$ and $y_0$, respectively, and give $X\times Y$ with the product topology. Prove that $\pi_1(X\times Y, (x_0,y_0))$ is isomorphic to $\pi_1(X,x_0)\times \pi_1(Y,y_0)$.
I have found an internet proof of this but I do not understand the part of $\phi ^{-1}$ that is well defined, could someone explain to me please? How is $\phi^{-1}$ defined? Would not it be better to prove that $\phi$ is bijective? Thanks
The inverse $\phi^{-1}$ is defined by $\phi^{-1}([f],[g])=[(f,g)]$, where $(f,g):[0,1]\to X\times Y$ is the map defined by $(f,g)(t)=(f(t),g(t))$. So, to check that this is well-defined, you have to check that if $[f]=[\tilde{f}]$ and $[g]=[\tilde{g}]$ then $[(f,g)]=[(\tilde{f},\tilde{g})]$, and this is exactly what the provided argument is doing. Once you know this map is well-defined, it is obvious that it is the inverse of $\phi$, since $(f,g)$ is the unique map $h$ such that $p_1\circ h=f$ and $p_2\circ g=f$.
(In other words, if you were just dealing with maps rather than homotopy classes of maps, $\phi^{-1}(f,g)=(f,g)$ would obviously be the inverse of $\phi$. To get an inverse when talking about homotopy classes of maps, you just have to check that this inverse operation is well-defined with respect to homotopies.)