(3.33) Is this well-defined? Yes because $X$ is simply-connected (pre-class question: prove this).

(4.04) Is $F$ surjective? Given a loop $\delta$ in $X/G$ based at $[x]$, path-lifting gives us a lift $\tilde{\delta}$ in $X$ starting at $x$. The endpoint $\tilde{\delta}(1)$ is in $q^{-1}([x]$, so there exists $g\in G$ such that $\tilde{\delta}(1)=\rho(g)(x)$. This implies that $F(g)=[\delta]$.

(5.51) Is $F$ a homomorphism? Given $g,h\in G$, pick paths $\gamma_g$ (from $x$ to $\rho(g)(x)$ in $X$) and $\gamma_h$ (from $x$ to $\rho(h)(x)$ in $X$). The path $\rho(g)\gamma_h$ connects $\rho(g)(x)$ to $\rho(g)\rho(h)(x)=\rho(gh)(x)$. The concatenation $(\rho(g)\gamma_h)\cdot\gamma_g$ makes sense and provides a path from $x$ to $\rho(gh)(x)$. The projection $q((\rho(g)\gamma_h)\cdot\gamma_g)$ is then a loop in $X/G$ whose lift connects $x$ with $\rho(gh)(x)$, so

(9.44) Therefore $F$ is an antihomomorphism. This is because I defined the concatenation of $a$ followed by $b$ to be $b\cdot a$. This had the advantage of making monodromy into a homomorphism. To make the current $F$ into a homomorphism, we should use right actions of the group, $x\mapsto xg$, satisfying $(xg)h=x(gh)$. Note that any homomorphism $f\colon G\to H$ can be turned into an antihomomorphism $\bar{f}\colon G\to H$ by $\bar{f}(g)=f(g^{-1})$, so homomorphisms and antihomomorphisms are equally useful.

(11.14) Is $F$ injective? I claim that $\ker(F)$ is trivial. If $g\in\ker(F)$ then the loop $q(\gamma_g)$ is nullhomotopic (where $\gamma_g$ is a path from $x$ to $\rho(g)(x)$ in $X$). By homotopy lifting, this implies that $\gamma_g$ is homotopic to the constant path rel endpoints, which implies that $\rho(g)(x)=x$. But the action of the group is properly discontinuous, so $\rho(g)x=x$ implies $g=1$. Therefore the kernel of $F$ is trivial.