$G_t$ vanishes. So, $G_{t^r}$ vanishes a some point $x_0\in \R$. If $t>t^r$, then $\trans(F_t)>p/q$, thus $G_t$ does not vanish. Since $G_t(x_0)>G_{t^l}(x_0)=0$, we have $G_t>0$ for $t>t^r$. Passing to the limit as $t\to t^r$ shows that $G_{t^r}\geq 0$. Conversely, if $G_t$ vanishes at some point $x_0\in \R$, then $\trans(G_t)=p/q$ and if in addition $G_t\geq 0$, then $G_{t'}>0$ for $t'>t$ and so, $\trans(F_{t'})>p/q$ for $t'>t$. This shows that $t=t^r$.

The characterization of $t^l$ follows similarly.

Finally, if $G_t$ takes both positive and negative values, then $\trans(F_t)=p/q$ and we cannot be in one of the previous cases, so $t\in (t^l,t^r)$. Conversely, if $t\in (t^l,t^r)$, then $G_t$ vanishes and according to the previous cases, the sign of $G_t$ cannot be constant. Thus, $G_t$ takes both positive and negative values.

In particular, we see that when $t=t^l$ or $t=t^r$, any point where $F_t^{\circ q}-{\rm Id}-p$ vanishes is a local extremum of the function. So, if $F_t$ is of class $C^1$ and $t=t^l$ or $t=t^r$, then there is a point $x_0$ such that $$F_{t}^{\circ q}(x_0)=x_0+p \quad \text{and}\quad (F_{t}^{\circ q})'(x_0)=1.$$ The induced map $f_t:\T\to \T$ has a periodic cycle whose rotation number is $p/q$ and whose multiplier is $1$.

A fixed point $z_0$ of an analytic map 