Let $M$ be a smooth manifold. Denote $C_M^\infty$ to the sheaf of real-valued functions on $M$. For a smooth vector bundle $\pi:E\to M$, denote $\Sigma_E$ to the sheaf of smooth sections of $\pi$ (it is a $C_M^\infty$-module). I am wondering the following:
If $E\to F\to G$ is an exact sequence of smooth vector bundles over $M$ (by definition, $E|_p\to F|_p\to G|_p$ is exact for every $p\in M$), is then $\Sigma_E\to\Sigma_F\to\Sigma_G$ an exact sequence of $C_M^\infty$-modules?
If the sequence were short exact then the answer is yes, for (i) we have s.e.s. on sections (see here) and (ii) filtered colimits are exact, so we get short exactness on stalks.
However, I don't know what happens in the general case. If the category $\mathsf{VB}(M)$ of vector bundles over $M$ were abelian, then we would be done: for a functor between abelian categories, it is the same to preserve s.e.s. than to preserve exact sequences. But to my knowledge, $\mathsf{VB}(M)$ is not abelian (I think the image and kernel of a morphism of vector bundles $f$ exist iff $f$ has constant rank).
Further thoughts to tackle the question: first, it is clear that since $E\to F\to G$ is a complex, $\Sigma_E\to\Sigma_F\to\Sigma_G$ is a complex (the composite vanishes). On the other hand, since the question of $\Sigma_E\to\Sigma_F\to\Sigma_G$ being exact is local on $M$, we can assume all $E,F,G$ are trivial vector bundles (at this point, I don't think the situation is much different than studying exactness at stalks). Thus, we have to show that if $\sigma$ is a local smooth section of $F$ that is sent to the zero section of $G$, then it comes from a smooth section of $E$. But how one could do that? If $F\to G$ had constant rank, then its kernel would be a smooth subbundle of $F$ (and $\Sigma_F\to\Sigma_G$ would have a kernel that is an $C^\infty_M$-submodule of $\Sigma_F$). Hence, $E\to F$ would induce a submersion onto the kernel of $F\to G$, so I guess one could use the rank theorem to lift $\sigma$ to a smooth section of $E$. But what happens in the case that $F\to G$ doesn't have constant rank? (Maybe there are counterexamples on this case?)
Best Answer
$\def\rank{\operatorname{rank}} \def\null{\operatorname{null}} \def\ker{\operatorname{Ker}} \def\im{\operatorname{Im}}$The answer is yes: exactness in the bundles implies exactness in the sheaves of sections. Actually, we have exactness on sections, see Proposition 8 below.
Proof. Consider the matrix of $\varphi$ with respect to two local frames of $E$ and $F$ and apply this. $\square$
Proof. Call $\null\psi,\rank\varphi:M\to\mathbb{N}$ to the functions $p\mapsto\null\psi_p,\rank\varphi_p$. Since $\rank\varphi=\null\psi=\rank F-\rank\psi$, it suffices to show that one of $\varphi$ or $\psi$ has locally constant rank. Let $p_0\in M$. Since $\rank\varphi,\rank\psi$ are lower semicontinuous functions (Lemma 2), there is an open neighborhood $U\subset M$ of $p_0$ such that for all $p\in U$, \begin{align*} \rank\varphi_p&\geq\rank\varphi_{p_0}\\ &=\null\psi_{p_0}\\ &=\rank F-\rank\psi_{p_0}\\ &\geq\rank F-\rank\psi_p\\ &=\null\psi_p\\ &=\rank\varphi_p. \end{align*} Hence, all “$\geq$” are actual equalities, so $\varphi|_U$ has constant rank. $\square$
Lemma 4 is Theorem 10.34 of J.M.Lee, Introduction to Smooth Manifolds, 2nd ed., so we redirect the reader there to see the proof.
Given a morphism of vector bundles $\varphi:E\to F$ over $M$, we denote $\Sigma_\varphi$ to the induced morphism on sheaves of sections $\Sigma_E\to\Sigma_F$.
Proof. The equality $\ker\Sigma_\varphi=\Sigma_{\ker\varphi}$ is easy to show.
We are going to show that the image presheaf of $\Sigma_\varphi$ equals $\Sigma_{\im\varphi}$. That is, for $U\subset M$ be open, we will show that $\im(\Sigma_{\varphi,U})=\Sigma_{\im\varphi}(U)$. The containment $(\subset)$ is easy, so we show the converse. Let $\sigma\in\Sigma_{\im\varphi}(U)$. By the proof of the aforementioned theorem 10.34 of Lee's book, there is an open cover $U=\bigcup_iU_i$ and local sections $\tau_i\in\Sigma_E(U_i)$ such that $\varphi\circ\tau_i=\sigma|_{U_i}$ for all $i$. Let $\{\rho_i\}$ be a partition of unity on $U$ subordinated to the open cover $\{U_i\}$. Then we can define $\tau=\sum\rho_i\tau_i$, which is a local section of $E$ over $U$. This way, \begin{align*} \varphi\circ\tau &=\varphi\circ\left(\sum\rho_i\tau_i\right)\\ &=\sum\varphi\circ(\rho_i\tau_i)\\ &=\sum\rho_i(\varphi\circ\tau_i)\\ &=\sum\rho_i\sigma|_{U_i}\\ &=\sum\rho_i\sigma\\ &=\left(\sum\rho_i\right)\sigma\\ &=\sigma. \end{align*}
Hence, $\sigma\in\im(\Sigma_{\varphi,U})$. $\square$
Proof. The image presheaf is a separated presheaf for it is a subpresheaf of a sheaf (namely, $\Sigma_F$). It is left to verify the gluing axiom. Let $U\subset M$ be open and let $U=\bigcup_{i\in I}U_i$ be an open cover. Let $\tau_i\in\Sigma_E(U_i)$ and suppose that the sections $\sigma_i=\varphi\circ \tau_i$ satisfy $\sigma_i|_{U_i\cap U_j}=\sigma_j|_{U_i\cap U_j}$. Let $\rho_i$ be a partition of unity on $U$ subordinated to $\{U_i\}$. Then $\tau=\sum\rho_i\sigma_i$ is a smooth local section of $E$ and $\varphi\circ\tau=\sum_i\rho_i\sigma_i$. Let $j\in I$ and pick $p\in U_j$. We have \begin{align*} (\varphi\circ\tau)(p) &=\sum_i(\rho_i\sigma_i)(p)\\ &=\sum_{\substack{i\in I\\p\in U_i}}(\rho_i\sigma_i)(p)\\ &=\sum_{\substack{i\in I\\p\in U_i}}\rho_i(p)\sigma_i(p)\\ &=\sum_{\substack{i\in I\\p\in U_i}}\rho_i(p)\sigma_j(p)\\ &=\sum_{i\in I}\rho_i(p)\sigma_j(p)\\ &=\left(\sum_{i\in I}\rho_i(p)\right)\sigma_j(p)\\ &=\sigma_j(p). \end{align*} Hence, $\varphi\circ\tau|_{U_j}=\sigma_j$. $\square$
Remark 7. Behind Lemma 6 there is a more general phenomenon. To explain it, we make the following definitions: a sheaf of rings $\mathcal{O}$ over a topological space $X$ is said to be to be fine if it has partitions of unity. This is, for all open subsets $U\subset X$ and all open covers $U=\bigcup_{i\in I} U_i$, there are sections $\rho_i\in\mathcal{O}(U)$, $i\in I$, such that $\{\operatorname{supp}\rho_i\}_{i\in I}$ is a locally finite family of subsets of $U$ and $\sum_{i\in I}\rho_i=1$. The previous lemma generalizes to: if $\varphi:\mathcal{F}\to\mathcal{G}$ is a morphism of $\mathcal{O}$-modules and $\mathcal{O}$ is fine, then the image presheaf of $\varphi$ is already a sheaf. The proof is formally the same, but now instead of evaluating at $p$, one takes stalks at $x$.
Proof. The converse is shown not to hold in the comment I wrote to my question. Suppose first that $E\to F\to G$ is exact. By Lemma 3, both $\varphi$ and $\psi$ have locally constant rank. By Lemma 4, $\im\varphi$ and $\ker\psi$ are smooth subbundles, and they are equal by hypothesis. By Lemma 5, we have $\ker\Sigma_\psi=\Sigma_{\ker\psi}=\Sigma_{\im\varphi}=\im\Sigma_\varphi$. This means exactness of the sequence of sheaves $\Sigma_E\to\Sigma_F\to\Sigma_G$.
The fact that “$\Sigma_E(U)\to\Sigma_F(U)\to\Sigma_G(U)$ is exact for all open $U\subset M$” implies “$\Sigma_E\to\Sigma_F\to\Sigma_G$ is exact” is a general property of sheaves of abelian groups over a topological space: since filtered colimits are exact, we get that $\Sigma_{E,p}\to\Sigma_{F,p}\to\Sigma_{G,p}$ is exact. Conversely, suppose $\Sigma_E\to\Sigma_F\to\Sigma_G$ is exact, i.e., it holds $\ker\Sigma_\varphi=\im\Sigma_\varphi$. Then, for $U\subset M$ we have $\ker\Sigma_{\varphi,U}=\ker\Sigma_\varphi(U)=\im\Sigma_\varphi(U)=\im\Sigma_{\varphi,U}$, where in the last step we have applied Lemma 6. This means that $\Sigma_E(U)\to\Sigma_F(U)\to\Sigma_G(U)$ is exact. $\square$
With the definitions of Remark 7, the second part of Proposition 8 generalizes to: Let $(X,\mathcal{O}_X)$ be a ringed space such that $\mathcal{O}_X$ is fine, then the global sections functor $\Gamma(X,-):\mathsf{Mod}(\mathcal{O}_X)\to\mathsf{Mod}(\Gamma(X,\mathcal{O}_X))$ is exact (in other words, all sheaves of $\mathcal{O}_X$-modules are acyclic). The proof is done in the same way, now with the generalization of Lemma 6 explained in Remark 7.