Prove the pullback bundle is a vector bundle

Question

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 $f^{*}E = \{(e,n) \in E \times N \mid \pi(e) = f(n)\}$ 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.

Answer 1

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 $\mathbb{R}^k.$

Answer 2

This is not so obvious if $f$ is not a diffeomorphism. You can't just say that $$ F^*E = \bigsqcup_{p\in N} E_{F(p)}, $$ because that's not enough to specify uniquely what the local trivializations are. You have to provide the definition of the local trivializations as part of the definition of the pullback bundle.

I think this is essentially the same answer as @Creo's, but worded a little differently.

$\newcommand\R{\mathbb{R}}$ For each trivialization $$T: O\times\R^k \rightarrow \left.E\right|_O,$$ let $U = f^{-1}(O)$, define the trivialization of $f^*E$ over $U$ to be \begin{align*} U\times \R^k &\rightarrow \left.f^*E\right|_U\\ (p,a) &\mapsto T(f(p),a). \end{align*}

I find this easier to understand using frames. A trivialization of $E$ over $O$ is equivalent to a frame of sections $$ s_j: O \rightarrow E,\ 1 \le j \le k, $$ where for each $q \in O$, $(s_1(q), \dots, s_k(q))$ is a basis of $E_q$. The transition map from one trivialization to another is the same as the corresponding change of frame map.

To define the vector bundle $f^*E$, it suffices to define a collection of mutually compatible local frames that cover the bundle. Given any frame $(s_1, \dots, s_k)$ for $E$ over an open $O \subset M$, we declare $(\bar s_1, \dots, \bar s_k)$ of $f^*E$ over $U = f^{-1}(O)$, where $$ \bar{s}_j = s_j\circ f\text{, }1 \le j \le k, $$ to be a smooth frame of $f^*E$ over $U$. It is straightforward to verify that the transition maps for this collection of local frames satisfy the necessary compatibility conditions.

Another way to say this is to specify which maps $N \rightarrow F^*E$ are smooth sections of $F^*E$. This can be done as follows: A map $\bar{s}: N \rightarrow F^*E$ is a smooth section if there exist smooth sections $s_1, \dots, s_m$ of $E$ over $M$ and smooth functions $a^1, \dots, a^m \in C^\infty(N)$ such that $$ \bar{s} = a^1s_1 + \cdots + a_ms^m. $$

A fancy way to say this is to say that the space of smooth sections of $f^*E$ is the $C^\infty(N)$-module generated by the space $$ \{ s\circ f:\ \forall\text{ sections }s\text{ of }E \}. $$

A terse description of this can be found here.