Is the following map a bijection? Let's cionsider  $$u:\mathbb{R}^n\times [0,\infty)\to\mathbb{R}^n$$ a time dependant vector field, such that it is smooth. If $X(x,t)$ is solution to
$$
\frac{d}{dt}X(x,t)=u(X(x,t),t)
$$
$$
X(x,0)=x
$$
Would that mean that for every $t_0\geq 0$, $X(x,t_0)$ is a bijection between $\mathbb{R}^n$ and $\mathbb{R}^n$.
Intuitively seems to be right, but I wouldn't know how to prove it. Any hint would be appreciated.
 A: No, this is not necessarily true. The problem is that the solutions need not exist globally.
Let's consider the function $u(y)=-y^2$ and the  corresponding initial value problem
\begin{align}
\dot{y}(t)&=u(y(t))\\
y(0)&=y_0
\end{align}
We need to distinguish two cases: If $y_0>0$, then the solution is given by $y(t)=\frac{y_0}{1+y_0t}$ and it is defined for $t\in (-\frac{1}{y_0},\infty).$ If $y_0<0$, then the solution is again given by $y(t)=\frac{y_0}{1+y_0t}$ but now it's defined for $t\in (-\infty,-\frac{1}{y_0})$. (The case $y_0=0$ will not be important, but in this case the solution is globally defined and given by $y(t)=0$)
Now we can easily check that none of these solutions satisfies $y(2)=1$, which means that the corresponding flow that you called $X$ is not bijective for $t=2$. Suppose this was true, then
\begin{align}
\frac{y_0}{1+y_0\cdot 2}&=1\\
\Longleftrightarrow y_0&=-1
\end{align}
But the solution corresponding to $y_0=-1$ is given by $y(t)=\frac{-1}{1-t}$ and this solution is only defined on the interval $(-\infty,1)$, since it goes to infinity for $t\to 1$. In particular it's not defined for $t=2$ and therefore there is no solution of the above IVP that attains the value $1$ at time $t=2$.
