Here is the problem,
Let $\phi$ be the flow of the differential equation $\dot x = f(x)$, $x(0)=x_{0}$, where $f(x)$ is a $C^1$ function $f:W\subset \rm I\!R^n \rightarrow \rm I\!R^n $, $W$ open. Show that the $\phi$ has the property that $\phi_{s+t}(x)=\phi_{s}(\phi_t(x))$, in the sense that if onde side is defined, so the other, and they are equal.
So, we now there exists a maximal interval $(\alpha,\beta)$ where there is a solution to $\dot x = f(x)$, $x(0)=x_{0}$, and consider first $s,t >0$ and suppose $\phi_{s}(\phi_{t}(x))$ is defined.
The book of Hirsch and Smale have a little portion of the proof by defining the function $y:(\alpha,s+t] \rightarrow W $
$$ y(r) = \Bigg\{ \quad \quad \phi_{r}(x) \quad \qquad \quad \quad \text{if} \quad \alpha <r \leq t, \\ \qquad \qquad \qquad \phi_{r-t}(\phi_{t}(x)) \quad \quad \quad \quad \text{if} \quad t\leq r\leq t+s. \\ $$
(Sorry for the equation, but I couldn't find my mystake, I tried using cases but I don't know why it didn't work)
Since $y(0)=x$, then by the Existence and uniqueness theorem it must be the case that $y(s+t)=\phi_{s+t}(x)$ and the first part is proved. Now when $s,t <0 $ it is very similar now considering the interval $[s+t,\beta)$. The idea in this two is that you can paste two functions, one in terms of "$s$" so that $y$ is just the same as $\phi_{t}(x)$. But when you consider $s<0, s+t>0$ then I can't see how can I paste the functions, because is like going forward and then backwards. Hope you can help.
I have already seen it, consider $y:[-t,0]\rightarrow W$,$\quad y(r)=\phi_{t+r}(x)$, so $y(0)=\phi_{0}(\phi_{t}(x))$ therefore the function $u:[s,0]\rightarrow W$, $u(r)=\phi_{s}(\phi_{t}(x))$ have the same time as $y$, hence $y(s)=u(s)$. And the rest cases of signs of $s,t$ must be similar.
I'm very sorry but it came up to me at this moment.