Help with definition of "flow" My question concerns the distinction between an orbit in a flow, and a continuous one-to-one image of $\mathbb R$.  Here is a definition:

The next excerpt indicates that this shape:

is NOT an orbit of a flow.


EDIT: I need someone to explain why $X$, the one-to-one image of $\mathbb R$ in picture, is not the orbit of a flow $\pi:X\times \mathbb R \to X$
 A: The map $\pi_x$ does not have to be one-to-one at all. Consider the map $\pi: X \times \mathbb R \rightarrow X$ given by $\pi(x,t)=x$. This is a perfectly valid flow yet the image of $\mathbb R$ under the motion through $x$ is the singleton $x$. This is about as "non injective" as maps come.
A: The picture is not the orbit of a flow because if $\pi_x$ is not one-to-one, then it is periodic. Suppose $\pi_x(t_0)=\pi_x(t_0+s)$ for some $s>0$. Then using property (ii) of flows, for any $t\in\Bbb R$ we have
$$\pi_x(t+s)=\pi(x,(t_0+s)+(t-t_0))=\pi(\pi(x,t_0+s),t-t_0)=\pi(\pi(x,t_0),t-t_0)=\pi(x,t)=\pi_x(t).$$
A: I don't think it's continuous at the tripoint. Consider the map $\pi(-,t)$ (that is, fix a $t$ and vary $x$).
A: I think the simplest way to see that the depicted space is not the orbit of a flow is from the fact that it is not topologically homogeneous. (a fact 
that can be shown by a simple cut point argument)
For the same reason, it cannot be an orbit of any continuous action of a topological group on a topological space. To see this, consider first the
more general case of a group $G$ acting on a set $X$. It is not hard to
verify that $O \subset X$ is an orbit of this action if and only 


*

*$O \ne \emptyset$,

*$\pi$ can be restricted to an action $\pi_O: O \times G \to O$, in other words $\pi[O \times G] = O$, and

*$\pi_O$ is a transitive action, i.e. for all $x, y \in O$ there is a $t \in G$ such that $\pi(x, t) = y$.


With the additional assumption that $X$ and $G$ have topologies for which
$\pi$ is continuous, it is clear that $\pi_O$ is also continuous and thus we have a continuous transitive group action on $O$. Since for every $t \in G$ the mappings $x \mapsto \pi_O(x, t)$ and $x \mapsto \pi_O(x, -t)$ are continuous and they are each other's inverse, they are homeomorphisms. 
Combined with the transitivity of $\pi_O$ this implies that $O$ is 
homogeneous.
Note that $G$ does not have to be any special kind of group and the 
topology chosen on $G$ is immaterial.
On the other hand the argument does not extend to actions of monoids, because they don't generally act by isomorphisms.
