I'm struggling with the proof of the following result:
Proposition
Let $-\infty \leq a < b \leq \infty$. A mapping $\phi: \mathbb{R} \times (a,b) \rightarrow (a,b)$ defines a continuous flow if there exists a finite or countable set of disjoint intervals $(a_n,b_n) \subset (a,b)$, $ \left( n \in J \right)$ and for every $n \in J$ there exists an homomorphism $r_n$ from $(a_n,b_n)$ onto $\mathbb{R}$ such that,
$$ \phi(t,x) = \begin{cases} x , \; &\text{if} \; x \notin \cup_{n \in J} (a_n,b_n) \\ r_n^{-1}(r_n(x)+t) , \; &\text{if} \; x \in (a_n,b_n), \, n \in J \end{cases} $$ for all $t \in \mathbb{R}$.
The proof goes as follows. The set $K := \{x \in (a,b): \phi(t,x) = x, \; \;\forall t \in \mathbb{R}\}$ is closed in $(a,b)$, so $(a,b)\setminus K$ is open and is the union of a finite or countable set of disjoint intervals $(a_n,b_n)$, $n \in J$. Take $x_n \in (a_n,b_n)$ and define $\alpha_n(t) := \phi(t,x_n)$. This is an injective mapping from $\mathbb{R}$ into $(a_n,b_n)$.
Question: Why are those maps injective?
I have tried the following. Take $t_2 > t_1 \in \mathbb{R}$ such that $\alpha_n(t_1) = \alpha_n(t_2)$. Therefore, as $t_2 = t_1 + \tau$ for some $\tau >0$ we have that $$ \phi(t_1,x_n) = \phi(\tau,\phi(t_1,x_n)), $$ so $y_n := \phi(t_1,x_n) \in (a_n,b_n)$ verifies that $y_n = \phi(\tau,y_n)$. Then my idea is to use this new point to find a contradiction with the fact that $y_n \notin K$. But so far I haven't been able to do it.
Any suggestions?