I am reading the fundamental theorem on time-dependent flows from the book Introduction to Smooth Manifolds, written by John Lee (see page 237). The last line of the theorem (i.e., equation $9.18$) says that if $p\in M_{t_1,t_0}$ and $\psi_{t_1,t_0}(p)\in M_{t_2,t_1}$, then $p\in M_{t_2,t_0}$ and $$\psi_{t_2,t_1}\circ \psi_{t_1,t_0}(p)=\psi_{t_2,t_0}(p).$$
Notice that the left-hand side of this equation depends on $t_1$, but the right-hand side doesn't. Should it not be $$\psi_{t_2,t_1}\circ \psi_{t_0,t_1}(p)=\psi_{t_2+t_0,t_1}(p)?$$ The question arises while checking the group laws (i.e., equation $9.3$ on page 209) of the flow of each vector field $V_t\colon M\to TM$ defined by $V_t(p)=V(t,p)$.
Best Answer
You're mixing the notations. With $\psi_t$ you'd be fine to add them. However this is $\psi_{t_0,t_1}$ which is actually the curve $\beta_{t_1-t_0}$ as explained on page 238. Note that $$\psi_{t_0,t_1} \circ \psi_{t_1,t_2} = \beta_{t_1-t_0} \circ \beta_{t_2-t_1} = \beta_{t_1-t_0+t_2-t_1} =\beta_{t_2-t_0} = \psi_{t_0,t_2}$$ as claimed. You can see how the group law was ultimately used to prove the result.