Is there a local flow that is diffeomorphic at any time? This question is regarding p.223-224 of Loring Tu's Introduction to Manifolds (Second edition).
Without proof, the author previously assumed the following.
Theorem. Let $M$ be a manifold and $X$ be a smooth vector field on it. For any $p \in M$ and a chart $(U, x^{1}, \cdots, x^{n})$ there is a neighborhood $W \ni p$ in $U$ and a smooth map (called a local flow) $F : (-\epsilon, \epsilon) \times W \rightarrow U$ such that for each $q \in W$, the mapping $F_{q} : W \rightarrow U$ given by $t \mapsto F(t, q)$ defines an integral curve of $X$ starting at $q$ (i.e., $F(0, q) = q$ and $d/dt|_{t = t_{0}}F(t, q) = X_{F(t_{0}, q)}$ for any $t_{0} \in (-\epsilon, \epsilon)$).
In p.223, the author says that for $p \in M$, we may choose a neighborhood $U \ni p$ in $M$ with a local flow $\varphi : (-\epsilon, \epsilon) \times U \rightarrow M$ such that $(\partial/\partial t) \varphi_{t}(q) = X_{\varphi_{t}(q)}$ and $\varphi_{0}(q)$ for all $q \in U$, where $\varphi_{t}(q) := \varphi(t, q)$.
In one of the earlier sections, the book explained that $\varphi_{s} \circ \varphi_{t} = \varphi_{s + t}$, whenever both sides make sense by uniqueness of integral curves of a vector field. So far, I am comfortable understanding what the book is trying to say.
Here is what I don't understand.
Where I need help. In p.224, the author says that the formula $\varphi_{s} \circ \varphi_{t} = \varphi_{s + t}$ implies that $\varphi_{t} : U \rightarrow \varphi_{t}(U)$ is a diffeomorphism because $\varphi_{-t} \circ \varphi_{t} = \varphi_{0} = id$ and $\varphi_{t} \circ \varphi_{-t} = \varphi_{0} = id$. I am concerned with the domain of $\varphi_{-t}$ because it does not look trivial to me that it is defined on $\varphi_{t}(U)$. It is also strange that in the line "$\varphi_{-t} \circ \varphi_{t} = \varphi_{0} = id$ and $\varphi_{t} \circ \varphi_{-t} = \varphi_{0} = id$," the identity must be the same ones, which seems to imply that $\varphi_{t}(U) = U$, and again, I do not see any reason that this is true.
I suppose I have this difficulty since I do not understand bolts and nuts of the existence of local flow, and at this moment, it seems to take a long roundabout from studying manifolds to read the proof of it. Thus, I ask you for a generous help in explaining why $\varphi_{t} : U \rightarrow \varphi_{t}(U)$ is a diffeomorphism with inverse $\varphi_{-t}$. Thank you in advance.
 A: 
bolts and nuts of the existence of local flow

The proof  may  be pretty long and boring (it belongs to ODE books), but the result you need is easy to  state: 

The existence set $ E=\{(p,t)\in X\times \mathbb R :  \varphi_t(p) \text{ is defined}\}$ is open in $X\times \mathbb R$. Also, the map $(p,t)\mapsto \varphi_t(p)$ is jointly continuous on $E$. 

Let's see how the above helps. Let $p\in \varphi_t(U)$ be any point. Pick $q\in U$ such that $\varphi_t(q)=p$. The time-shifted map $ s\mapsto \varphi_{t+s}(q)$, $-t\le s\le 0$ is also a trajectory of $X$ (since  $X$ is time-independent). We can denote this map $\varphi_s(p)$, since $\varphi_0(p)=p$. So, $(p,-t)\in E$. 
By the openness of $E$, $\varphi_{-t}$ is defined in a neighborhood of $p$. By the continuity, it maps a small neighborhood of $p$ into $U$.   Hence, this neighborhood lies in $\varphi_t(U)$, showing that $\varphi_t(U)$ is open. 
To summarize: we've seen that $\varphi_t$ maps $U$ onto an open set $\varphi_t(U)$, then $\varphi_{-t}$ is defined on $\varphi_t(U)$. That the compositions  $\varphi_t \circ \varphi_{-t} $ and $\varphi_{-t} \circ \varphi_{t} $ bring us back to the point we started with is a direct consequence of the definition: we follow the same curve back and forth.  
