Definition of the flow of an ODE and its inverse At lesson, the teacher considers a flow $\Phi$ given by the solutions of the ode system for $t\in[0, T]$ and $x\in\mathbb R^d$,
$$
\begin{cases}
y'(s)=b(y(s), s),&s\leq T\\
y(t)=x
\end{cases},\qquad(\star)
$$
that is $\Phi(x, t, s)=y(s)$ solving $(\star)$. He said that we will be mostly concerned with $\Phi(\cdot, 0, \cdot)$. The field $b$ is assumed to be Lipschitz continuous in both variables and bounded.
Then, he intoduces the inverse $\Psi$ of the above flow as follows: $\Psi(x, 0, s)=y(s)$ satisfying
$$
\begin{cases}
y'(s)=-b(y(s), t-s),&s<t\leq T\\
y(0)=x
\end{cases},\qquad(\triangle)
$$
and he said that $\Psi$ is such that
$$
\Phi(\Psi(x, 0, s), 0, s)=x,\quad \Psi(\Phi(x, 0, s), 0, s)=x.\qquad (\star\star)
$$
I do not understand $(\star\star)$. Can someone help me? Maybe is the definition of the inverse wrong?
Thank you
 A: After @lutz-lehmann pointed out that my arguments would apply only to an autonomous system, I will try to make an analogous argument for the related autonomous system.

I will stick to the $t = 0$ case.

Let $s\mapsto\mathbf{u}(s) = [y(s),~s]^{\mathsf{T}}$ be the solution of
\begin{eqnarray}
\frac{d}{ds}\left[\begin{array}{c}y(s)\\s\end{array}\right] &=& \left[\begin{array}{c}b(y(s),s)\\1\end{array}\right] ~=~ \left[\begin{array}{c}b(\mathbf{u}(s))\\1\end{array}\right],\\
\mathbf{u}(0) &=& \left[\begin{array}{c}x\\0\end{array}\right].
\end{eqnarray}
Since $b$'s argument is $\mathbf{u}(s)$, this system is autonomous.
Let $U$ be the flow for this autonomous system. Then $U(\cdot,\cdot,s)$ will map this example's initial conditions as follows:
\begin{equation}
U(\cdot,\cdot,s):\left[\begin{array}{c}x\\0\end{array}\right]\mapsto\left[\begin{array}{c}y(s)\\s\end{array}\right].
\end{equation}
In other words,
\begin{equation}
U(x,0,s) = \left[\begin{array}{c}\Phi(x,0,s)\\s\end{array}\right].
\end{equation}

Let $s\mapsto\mathbf{v}(s) = [w(s),-s]^{\mathsf{T}}$ be the solution of
\begin{eqnarray}
\frac{d}{ds}\left[\begin{array}{c}w(s)\\-s\end{array}\right] &=& \left[\begin{array}{c}-b(w(s),-s)\\-1\end{array}\right] ~=~ \left[\begin{array}{c}-b(\mathbf{v}(s))\\-1\end{array}\right],\\
\mathbf{v}(0) &=& \left[\begin{array}{c}x\\0\end{array}\right].
\end{eqnarray}
This system is also autonomous.
Let $V$ be the flow of this system. Then $V(\cdot,\cdot,s)$ maps this example's initial conditions as follows:
\begin{equation}
V(\cdot,\cdot,s):\left[\begin{array}{c}x\\0\end{array}\right]\mapsto\left[\begin{array}{c}w(s)\\-s\end{array}\right].
\end{equation}
In other words,
\begin{equation}
V(x,0,s) = \left[\begin{array}{c}\Psi(x,0,s)\\-s\end{array}\right].
\end{equation}

Since $U$ is the flow of an autonomous system, $U(\cdot,\cdot,s)$ pushes a solution $\mathbf{u}$ from its value at any time (due to the unchaning vector field) to its value $s$ time units later. The complication here is that $w(s)$ is not a fixed point.
Note, however, that at each point $(x,t)$, the vector-valued right-hand sides of the two autonomous systems differ only in sign: one is $[b(x,t),~1]^{\mathsf{T}}$ and the other is $[-b(x,t),-1]^{\mathsf{T}}$. That means a solution of one is a solution of the other run backward in time.
This implies that $V(\cdot,\cdot,s) = U(\cdot,\cdot,-s)$. This, in turn, implies that, since $[y(s),~s]^{\mathsf{T}}$ and $[w(s),-s]^{\mathsf{T}}$ started at the same point ($[x,~0]^{\mathsf{T}}$) at time $0$ and evolved according to opposite-pointing vector fields, $w(s) = y(-s)$.
\begin{eqnarray}
\Phi(\Psi(x,0,s),0,s) &=& \Phi(w(s),0,s)\\
&=& \Phi(y(-s),0,s)\\
&=& \Phi(\Phi(x,0,-s),0,s)\\
&=& x
\end{eqnarray}
I'm sure this can be argued with much more sophistication, but I think the rough shape of the argument is clear. I apologize for the completely incorrect previous answer.
