I am basically a beginner in the manifold course. We have seen that if $X$ is a vector field given over a manifold $M$ and $x\in M$ then there exists a curve passing through $x$ such that locally the velocity vectors of the curve agree with the vector field or more concretely,
There exists a (unique) curve $\gamma:(-\epsilon,\epsilon)\to \mathbb R$ such that $X(\gamma(t))=\gamma'(t)$ with $\gamma(0)=x$.This solution is obtained by Picard's existence and uniqueness theorem on the above Initial Value Problem in a local chart and then pulling everything back to the Manifold neighbourhood and this is known as the integral curve to $X$ through $x$ and is denoted by $\gamma^{x, X}$. We can thus define a function $F:(-\epsilon,\epsilon)\times U\to \mathbb R$ (Where $U\subset M$ is an open neighbourhood of $x$ in $M$) as follows:
$F^X(t,x)=F^X_t(x)=\gamma^{x, X}(t)$ which is sometimes written as $\gamma_t^{x, X}$.Now we have an immediate result(Although intuitively clear, it does not seem to be that immediate to me formally):
The result is that $F_t\circ F_s=F_{t+s}$ whenever the two sides make sense. But I am not able to prove this. Perhaps because I have some weakness handling two variables $t,s$ simultaneously.But I know what I should do. If I show that $T_t\circ T_s$ is an integral curve of $X$ through $x$ then we are through. But the differentiation computation seems to be awful.
Can somebody provide a little help?
Fix a vector field $X,$ so we can omit it from our notation, and fix $s \in \mathbb{R}$. Take an arbitrary point $x$ so that $F_{t}(F_{s}(x))$ and $F_{t+s}(x)$ are both defined. By definition $F_{t+s}(x) = \gamma^{x}(t+s),$ where $\gamma^{x}$ is the unique maximal integral curve for $X$ with $\gamma^{x}(0) = x.$ On the other hand, letting $y = F_{s}(x) = \gamma^{x}(s),$ we have $F_{t}(y) = \gamma^{y}(t),$ where $\gamma^{y}$ is the unique maximal integral curve for $X$ with $\gamma^{y}(0) = y.$
We want to show that $\gamma^{y}(t) = \gamma^{x}(t+s).$
By uniqueness of maximal integral curves for $X$, if two maximal integral curves for $X$ agree at $0$, then they must be the same curve. How can we use that here? Observe that if we define $\alpha(t) = \gamma^{x}(t+s),$ then $\alpha(t)$ is a maximal integral curve for $X$ (check this by writing $\alpha'$ in terms of ${\gamma^{x}}'$), and moreover $\alpha(0) = \gamma^{x}(s) = y = \gamma^{y}(0).$ By uniqueness, we conclude that $\gamma^{y}(t) = \alpha(t) = \gamma^{x}(t+s)$, and we are done.