Equivalent bundle atlases

Let $\pi:E\to M$ be a smooth fiber bundle with typical fiber $F$. Two bundle atlases $\mathcal A$ and $\mathcal B$ for $\pi$ are equivalent if their union $\mathcal A\cup \mathcal B$ is again a bundle atlas; equivalently, every local trivialization from $\mathcal A$ is compatible with every local trivialization from $\mathcal B$ in the sense that the induced changes of trivialization on overlaps are smooth and fiberwise diffeomorphisms.

A common rephrasing is that $\mathcal A$ and $\mathcal B$ admit a common refinement: there exists a third atlas $\mathcal C$ such that each chart of $\mathcal C$ is a restriction of a chart from $\mathcal A$ and also of a chart from $\mathcal B$. In this way, a smooth fiber bundle structure can be viewed as an equivalence class of atlases.

On overlaps between a chart from $\mathcal A$ and a chart from $\mathcal B$, the compatibility is expressed by smooth transition functions taking values in $\mathrm{Diff}(F)$.

Examples

1. Refinement by shrinking: if $\{(U_i,\Phi_i)\}$ is an atlas and $V_i\subset U_i$ is an open cover, then $\{(V_i,\Phi_i|_{\pi^{-1}(V_i)})\}$ is an equivalent atlas.
2. Different gauges on a trivial bundle: on $M\times F$, changing trivializations by $(x,f)\mapsto(x,h(x)(f))$ for a smooth $h:M\to \mathrm{Diff}(F)$ yields an equivalent atlas.
3. Tangent bundle: atlases for $TM$ coming from different smooth atlases of $M$ are equivalent because coordinate changes on $M$ induce smooth fiberwise linear transition maps on $TM$.