# Sections of pullback bundle are generated by pullback of sections

Let $$M,N$$ be smooth manifolds, $$\pi:E\to M$$ a smooth rank $$r$$ vector bundle, $$f:N\to M$$ a smooth map. We define $$f^*E=\{(p,v)\in N\times E: f(p)=\pi(v)\},$$ which has a natural structure of a vector bundle over $$N$$. Consider a section $$\sigma\in\Gamma(E)$$. Then $$f^*\sigma:p\mapsto (p,\sigma\circ f(p))$$ defines a smooth section of $$f^*E$$. I want to show that $$\Gamma(f^*E)$$, the space of smooth sections of $$f^*$$E, is generated as a $$C^\infty(M)$$-module by sections of the form $$f^*\sigma$$ with $$\sigma\in\Gamma(E)$$ (in other words: $$\Gamma(f^*E)=f^*\Gamma(E)\otimes_{f^*C^\infty(M)}C^\infty(N)$$).

Let's first show this locally. Consider a local trivilisation $$\phi:E\vert_U\tilde\to U\times\mathbb R^r$$. Let $$\sigma_i\in\Gamma(E\vert_U)$$ correspond to the constant section $$p\mapsto (p,e_i)$$ of $$U\times\mathbb R^n$$, where $$e_i$$ is the $$i$$-th standard basis vector of $$\mathbb R^r$$. In other words: $$\sigma_i:p\mapsto\phi^{-1}(p,e_i)$$. Since $$f^*E\vert_{f^{-1}(U)}\cong f^{-1}(U)\times\mathbb R^k$$, we see that $$f^*\sigma_i$$ corresponds under that identification to $$p\mapsto (p,e_i)$$. It's clear that any section of $$f^*E\vert_{f^{-1}(U)}$$ is a linear combination of $$f^*\sigma_i$$.

For the global case, I am thinking of invoking the result that there exists a finite cover $$\{U_\alpha\}$$ of $$M$$ and trivialisations $$\phi_\alpha: E\vert_{U_\alpha}\tilde\to U_\alpha\times\mathbb R^r$$. However, what I would like to do first is extend the previously defined $$f^*\sigma_i$$ to a global pull-back of a section. I do know it's possible to extend sections locally using a partition of unity argument, but I don't know how to achieve this in such a way that the global extension is still the pull-back of a finite number of sections...

So my question is: given a local section $$\sigma\in\Gamma(E\vert_U)$$ ($$U\subset M$$ open), we have a section $$f^*\sigma\in\Gamma(f^*E\vert_{f^{-1}(U)})$$; how can this section be extended to a section of the form $$\sum_{i=1}^k f_i f^*\sigma_i$$, $$f_i\in C^\infty(M),\sigma_i\in\Gamma(E)$$?

Could someone help me out?

I've been given the following hint in the chat: if $$U_i$$ is an open cover, then there exists(*) an open cover $$V_i$$ over the same index set such that $$\overline{V_i}\subset U_i$$. I think I can work it out with this result; I'll post my solution tomorrow.

See Ng Chikeung's answer for a nice short argument.

Before I write my answer, I want to point out that of course my initial question whether a local section can be extended globally is not true (for the trivial bundle $$M\times\mathbb R$$ this is equivalent to whether any smooth function defined on $$U\subset M$$ can be extended globally, and we know this is not true, e.g. $$1/x$$ on $$(0,1)$$ can't be extended to $$[0,1)$$). However, we are not in this situation; we are in the situation where we have a local section that only needs to be extended on a closed subset (closed w.r.t. $$M$$) of $$U$$, and this can be done.
Using the same notation as in the question post, let $$\sigma\in\Gamma(E)$$ and let $$U_\alpha$$ be a finite trivialising cover for $$E$$. As already remarked, we have a finite sum $$\sigma\vert_{f^{-1}(U_\alpha)}=\sum_i f_i f^*\sigma_i$$ with $$f_i\in C^\infty(f^{-1}(U_\alpha))$$, $$\sigma_i\in\Gamma(E\vert_{U_\alpha})$$. Following the hint, we know that there exists an open cover $$V_\alpha$$ of $$M$$ (with the same index set as $$U_\alpha$$) such that $$\overline{V_\alpha}\subset U_\alpha$$. By the Extension Lemma for Vector Bundle (Lemma 10.12 in Lee's Smooth Manifolds) we can extend $$\sigma_i\vert_{\overline{V_\alpha}}$$ to a global section $$\tilde\sigma_i\in\Gamma(E)$$. We now choose a partition of unity $$\psi_\alpha$$ subordinate to the cover $$f^{-1}(V_\alpha)$$ of $$N$$. Then we have a finite sum $$\sigma=\sum_{\alpha}\psi_\alpha\sigma$$. Note that $$\psi_\alpha\sigma\vert_{f^{-1}(U_\alpha)}=\sum_i\psi_\alpha f_i f^*\sigma_i$$. We can extend $$f_i\vert_{\text{supp}(\psi_\alpha)}$$ to $$\tilde f_i\in C^\infty(N)$$. Then $$\psi_\alpha\sigma=\sum_i \psi_\alpha\tilde f_i f^*\tilde\sigma_i$$.