I'm currently reading J.P May's book, "A Concise Course in Algebraic Topology".
I don't understand his proof of the fundamental groupoid version of Van Kampen's theorem, particularly the part where he proves that $\tilde\eta ([f])=\tilde\eta[g]$, if $[f]=[g].$
Let $f$ & $g$ be paths form $x$ to $y$ in $X$. We wish to define $\tilde\eta:\Pi(X)\to\mathscr{C}$,given $\eta_U:\Pi(U)\to \mathscr{C}$.
The book defines $\tilde{\eta}(x)=\eta_U(x)$, and $\tilde\eta([f])=\eta([f])$ if $f$ lies within $U$.If $f$ doesn't lie within $U$, it is the composite path of paths $f_i$ all lying within some $U_i$ and $\tilde \eta([f])=\Pi\eta_{U_i}([f_i]) $. He now wishes to show that this is well defined. He says that if $[f]=[g]$, there is a homotopy, $h:f\simeq g$. You can divide the square into a grid of finitely many subsquares, such that $h$ is mapped into one of the $U$ of the cover. He then says that we can choose a grid that refines the partition of $I\times\{0\}$ for $f$ and $I\times\{1\}$ for $g$. Now he concludes that $\tilde\eta([f])=\eta([f])$, thanks to it being the consequence of finitely may relation which hold in one of the $U$. I don't understand the last step.