Prove the pullback bundle is a vector bundle

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.

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 $$\mathbb{R}^k.$$

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