I am having trouble proving a statement for my differential geometry class. The problem is as follows:
Let $(E,B,\pi)$ be a vector bundle and let $\Psi : W \longrightarrow W' \times F'$ be a vector bundle chart. Show that $W = \pi^{-1}(W \cap B)$.
Although this statement seems rather simple (and probably is), I am not too sure if it even holds true (see Point 2 below). I tried to show equality by showing $W \subseteq \pi^{-1}(W \cap B)$ as well as $W \supseteq \pi^{-1}(W \cap B)$.
0.) First note that the set $\Psi^{-1}(W',0)$ coincides with the set $B \cap W$ (right?).
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1.) $\enspace \subseteq \enspace : \quad$ For a point $e \in E$ the projection $\pi$ is defined as $\pi(e) = \Psi^{-1}(w',0)$ where $\Psi(e) = (w',f')$ for some $(w',f')$. Since one only considers a single vector-bundle-chart this definition implies
$$\pi(W) \enspace = \enspace \Psi^{-1}(W',0) \quad \Longrightarrow \quad W \enspace \subseteq \enspace \pi^{-1} \Big( \Psi^{-1}(W',0) \Big) \enspace = \enspace \pi^{-1}( W \cap B )$$
(Note that the "equal"-sign is replaced by a "subset"-sign because $\pi$ is not injective.)
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2.) $\enspace \supseteq \enspace : \quad$The other direction I find problematic. Because if $b \in W \cap B$ is a point in $W$ and in $B$, then $\pi^{-1}$ should be the whole fiber over $b$. But this fiber is not restricted to the subset $W$. So how to show this direction?
E D I T 1:
Since there seem to be different definition of the basis $B$, the set $B$ is assumed to be
$$ B \enspace := \enspace \big\{ \; e \in E \; \big| \; \text{$\exists$ vector-bundle-chart $(\Psi,W)$ s. t. $e = \Psi^{-1}(w',0)$ for some $w' \in W'$} \; \big\}$$