I arrived at the following question about vector-field-induced flows on a smooth manifold.
Let $M$ be a smooth manifold, $X \in \Gamma(TM)$ a smooth vector field and $\Phi_X: \{(p,I_p)\ |\, p \in M\} \to M$ its maximal local flow, i.e. such that an integral curve $\gamma$ of $X$ with $\gamma(0)=p$ can be maximally extended to the open interval $0 \in I_p$.
I read a claim in a book I'm currently studying, that for any such flow $\Phi_X$ the map $\hat \phi : (\mathbb{R},+) \to (\mathrm{Diff}(M),\circ),\, t \mapsto \Phi_X(\cdot,t)$ is a local group homomorphism. (In the book it is also mentioned that any such flow is a diffeomorphism for some open neighbourhood of $t=0$, that I also don't understand in first place and intuitively would not agree with.)
Clearly, if $X$ (and thus also $\Phi_X$) is complete, i.e. $\forall p \in M: I_p = \mathbb{R}$, then $\hat \phi$ is even a group homomorphism, the one-parameter group of diffeomorphisms generated by the flow of $X$.
I, however, don't see how the local group homomorphism property is satisfied for a non-complete vector field.
(May someone provide me with a pretty definition of a local group homomorphism between topological groups, btw?)
As a counterexample I had $M = (-1,1)$ with the identity chart $(M,id)$ and the vector field $X = \frac{\partial}{\partial id^1}$ in mind. since in that case there clearly doesn't even exist any $t \neq 0$ such that $\Phi_X(\cdot,t)$ is a diffeomorphism, as it is not even defined on the entire $M$?
Am I disregarding something here?