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I'm stuck at proving that the pullback bundle is a vector bundle. The question is basically we have a smooth function $F$ from the smooth manifold $N$ to the smooth manifold $M$. And $(E, \pi, M)$ is a vector bundle. Now we want to prove that $f^{*}E = \{(e,n) \in E \times N \mid \pi(e) = f(e)\}$ is the total space over $N$. This is what I tried:

Define$$\rho: f^{*}E \to N: (e,n) \mapsto n$$.

First notice that $\rho^{-1}(\{n\}) = (f^{*}E)_{f(n)}$ consists of all pairs $(e,n)$ such that $\pi(e) =f(n)$, which is the fiber of $f(n)$ attached to point $n$, which is again simply identified with $E_{f(n)}$, which is a $k$-dimensional vector space (so of course $\rho$ is surjective), and the fibers of $N$ under $\rho$ are the $k$-dimensional vector spaces that are 'attached' to $M$. Now take $p \in N$. There exist an open neighbourhood $U$ around $f(p) \in M$ such that there exists an diffeomorphism $\Phi: U \times R^k \to pi^{-1}(U)$, such that $\pi \circ \Phi = \pi_1$ and that $\Phi: \{f(p)\} \times R^k \to E_{f(p)}$ is a vector space isomorphism. Now let's take a look at $V = F^{-1}(U)$, which is an open neighbourhood of $p$, since $F$ is smooth. Note that $\rho^{-1}(V) = \{ (e,n) \in f^{*}E \mid \rho((e,n)) = n \in V \}$.

Now I don't actually know what to do to prove the local trivialization. I think I'm overseeing a simple step.

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As you wrote, let $f^*E := \{(e,n) \in E \times N| \ \pi(e)=f(n) \}$. I like to to it the other way around, so let our trivialisations for $E$ be given by $$\phi: \pi^{-1}(U) \to U \times \mathbb{R}^k, \ e \mapsto (\pi(e), pr_2(\phi(e)))$$ where $pr_2: U \times \mathbb{R}^k \to \mathbb{R}^k, (x,v) \mapsto v$ (why can we write it that way?). Now, let (as you wrote) $$\rho: f^*E \to N, (e,n) \mapsto n$$ wherefore: $$\rho^{-1}(V) = \{(e,n) \in E \times V| \ \pi(e) = f(n) \}.$$ Thus we can define the trivialisation (why?): $$\psi: \rho^{-1}(f^{-1}(U)) \to f^{-1}(U) \times \mathbb{R}^k, (e,n) \mapsto (n, pr_2(\phi(e)))$$ again, $pr_2$ ist just the projection onto $\mathbb{R}^k.$

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  • $\begingroup$ How does it follow that $\psi$ is a diffeomorhpism. $\endgroup$ Nov 5 '18 at 12:09
  • $\begingroup$ what's your given differentiable structure on $f^*E$? $\endgroup$
    – Creo
    Nov 5 '18 at 12:25
  • $\begingroup$ if you're asking $why$ the $\psi$ do define a differentiable structure: that should follow from the properties of $\phi$ $\endgroup$
    – Creo
    Nov 5 '18 at 12:28
  • $\begingroup$ is $pr_{2}(\phi)$ a diffeomorphism? $\endgroup$ Nov 5 '18 at 12:50
  • $\begingroup$ no. But $\phi$ is. Did you write down the map $\psi^{-1} \circ \psi'$ and compared it with $\phi^{-1} \circ \phi'$ ? (On $U \cap U'$) $\endgroup$
    – Creo
    Nov 5 '18 at 12:58

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