I'm just trying to work through and understand if this really simple example is actually a fiber bundle, as I'm not sure I fully understand the definition.
We have the manifold $B= (0,1) \subset \mathbb{R}$ , with fiber $V=\mathbb{R}^2$ and an atlas $(k_1:(0,\frac{1}{2}) \rightarrow (0,\frac{2}{3})) , (k_2:(\frac{1}{2},1) \rightarrow (\frac{1}{3},1))$ where $k_1(t)=\frac{4}{3}t$ and $k_2(t)=\frac{4}{3}t-\frac{1}{3}$.
Then if E is the entire space define its atlas by $k'_1(t,(x,y))=(\frac{4}{3}t,(tx,y))$ and $k'_2(t,(x,y))=(\frac{4}{3}t-\frac{1}{3},(x,y))$
so let $\pi : E\rightarrow B$ be such that $\pi (t,v)= t$.
My question is just is this a smooth bundle and is it trivial? I know this is a really basic question but I just want to check my understanding is right.