If $\tau>0$ and $v:[0,\tau]\times\mathbb R\to\mathbb R$ is continuous in the first argument and satisfies $$\sup_{t\in[0,\:\tau]}\left|v(t,x)-v(t,y)\right|\le c|x-y|\tag1$$ for some $c\ge0$, we know that, for all $(s,x)\in[0,\tau]\times\mathbb R$, there is an unique $X^{s,\:x}\in C^0([0,\tau])$ with $$T_{s,\:t}(x):=X^{s,\:x}(t)=x+\int_s^tv(r,X^{s,\:x}(r))\:{\rm d}r\;\;\;\text{for all }t\in[s,\tau]\tag2.$$
Let $0\le s\le t\le\tau$. I am able to show that $T_{s,\:t}$ is bijective, but can we give an explicit formula$^1$ for $T_{s,\:t}^{-1}$?
Let me elaborate on what I've tried: Assume for the moment that $s=0$ and let $$w(r,y):=-v(t-r,y)\;\;\;\text{for }(r,y)\in[0,\tau]\times\mathbb R.$$ By the same argument as before, for all $y\in\mathbb R$, there is a unique $Y^y\in C^0([0,t])$ with $$Y^y(r)=y+\int_0^rw(q,Y^y(q))\:{\rm d}q\;\;\;\text{for all }r\in[0,t]\tag3.$$ We easily see that $$Y^{X^{0,\:x}(t)}=X^{0,\:x}(t-\;\cdot\;)\;\;\;\text{for all }x\in\mathbb R\tag4$$ and $$X^{0,\:Y^y(t)}=Y^y(t-\;\cdot\;)\;\;\;\text{for all }y\in\mathbb R\tag5.$$ If we now define $$S_r(y):=Y^y(s)\;\;\;\text{for }(r,y)\in[0,t]\times\mathbb R,$$ we obtain $S_t\circ T_{0,\:t}=\operatorname{id}_\mathbb R$ and $T_{0,\:t}\circ S_t=\operatorname{id}_\mathbb R$ from $(4)$ and $(5)$, respectively, i.e. $$T_{0,\:t}^{-1}=S_t\tag6.$$
The crucial thing is that the family $(S_r)_{r\in[0,\:t]}$ depends on $t$. Can we somehow show that it also holds $T_{0,\:r}^{-1}=S_r$ for all $r\in[0,t]$? What I would really like to obtain is a family $(S_{s,\:t}:0\le s\le t\le\tau)$ such that $S_{s,\:t}=T_{s,\:t}^{-1}$.