$P_1=UD$ and $(P_2, P_3, \ldots, P_j) =
                                       (R_1, R_2, \ldots, R_m)$. It follows that $t=k+1$ and $j=m+1$. Thus we can recover $(P_1, P_2, \ldots, P_j)$. For the case $w_1=3$, we have $P_1=R_1UD$ and $t=k$. Also, we can recover $(P_2, \cdots, P_j)$ from $(R_2,  \ldots, R_m)$. For the case $w_1=4$, we have $t=k$, $P_1=UR_1D$, and $(P_2,
\ldots, P_j)=(R_2, \ldots, R_m)$.

It remains to deal with the situation $w_1=1$, which involves Cases 1 and 4 of the bijection. If $k=1$, then we have $t=1$, $P_1=UD$ and $(P_2,  \ldots, P_j)=(R_1,\ldots, R_m)$. If $k=2$, then we have $t=1$, $P_1=R_1UR_2D$ and $(P_2,  \ldots, P_j)= (R_3,
 \ldots, R_m)$. If $k>2$, then we have $t=k-1$, $P_1=R_1UR_2D$, and $(P_2,  \ldots, P_j)=(R_3,  \ldots, R_m)$. Thus, we have shown that $\phi$ is a bijection.

An example of the bijection $\phi$ is given in Figure \[dyck\], where normal vertices are drawn with black points, return points that cut off the Dyck paths into segments are drawn with white points, and the distinguished segment is marked with a $*$ on its last return step.

(120,25)

(0,15)[(1,1)[4]{}]{} (4,19)[(1,-1)[3.7]{}]{} (8.3,15.3)[(1,1)[8]{}]{} (16,23)[(1,-1)[8]{}]{} (24,15)[(1,1)[4]{}]{} (28,19)[(1,-1)[3.7]{}]{}(29.5,16.9)[$\ast$]{}(32.3,15.3)[(1,1)[4]{}]{} (36,19)[(1,-1)[3.7]{}]{} (0,15) (4,19) (8,15) (12,19) (16,23) (20,19) (24,15)(28,19) (32,15) (36,19) (40,15)

(43,19)[$\longleftrightarrow$]{} (54,18)[$2$]{}

(58,15)[(1,1)[8]{}]{} (66,23)[(1,-1)[8]{}]{} (74,15)[(1,1)[4]{}]{} (78,19)[(1,-1)[3.7]{}]{}(79.5,16.9)[$\ast$]{}(82.3,15.3)[(1,1)[4]{}]{} (86,19)[(1,-1)[3.7]{}]{} (58,15) (62,19) (66,23) 