Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This is my first post on math.stackexchange.com - please excuse me if I have overseen some relevant part of the FAQ or this question among those already answered or if I do misbehave in any other way, it maybe takes a bit of use to understand how to find out whether one can ask a question or not.

I am supposed to teach myself the content "Sheaves, Cohomology and the de Rham Theorem" using Warner's Foundations of Differentiable Manifolds and Lie Groups. I have an issue with a very early supposedly easy problem and this suggest that I do not get any grip on the objects I am dealing with. Could somebody maybe give me a hint or a point of view which stays in Warner's framework (no ringed spaces a.s.o.)? Before I delve into the questions, I fix a bit of vocabulary (as this might vary from author to author):

  • $M$ is a manifold
  • $K$ is a principal ideal domain
  • A sheaf $\mathcal{S}$ of $K$-modules over $M$ is a tuple $(\mathcal{S},\pi)$ where $\mathcal{S}$ is a topological space and $\pi:\mathcal{S}\to M$ is a local homeomorphism such that for all $m\in M$ the preimage $\pi^{-1}(m)$ is a $K$-module and the composition laws (on the modules in) $\mathcal{S}$ are continuous.

Now for the question: Let $(\mathcal{S},\pi)$ be a sheaf of $K$-modules over a manifold $M$. Let $U\subset M$ open and $f:U\to\mathcal{S}$ be a section of $\mathcal{S}$ over $U$, i.e. $f$ is continuous and $\pi\circ f=\operatorname{id}_{U}$. Assume $m\in U$ such that $f(m)=0$ in $\pi^{-1}(m)$. Then there exists a neighbourhood $m\in V\subset U$ such that $f$ vanishes on all of $V$, i.e. $f(v)=0\in\pi^{-1}(v)$.

I assume that this must be true as of the following: "[...] if sections $f$ and $g$ agree at $m\in M$, then they must agree on a neighbourhood of $m$".

As I had no idea how to answer this question, I decided to first work on the first problem in the exercise section, i.e. that the null-section $g:M\to\mathcal{S}$ defined by $g(m)=0\in\pi^{-1}(m)$ is continuous. I haven't even managed to prove this. That's why I am asking for help. I have tried to prove it pretty directly, i.e. I chose a neighbourhood $g(m)=0\in W\subseteq\mathcal{S}$ such that $\pi\big|_{W}$ is a homeomorphism and then started looking for a neigbourhood $m\in U\subseteq M$ which satisfies $g(U)\subseteq W$. I used some neighbourhood $m\in V\subseteq\pi(W)$ and $\pi\circ g\big|_{V}=\operatorname{id}_{V}$ to deduce that $g(V)\subseteq \pi^{-1}(\pi(W))$. The latter simply is: $$ \pi^{-1}(\pi(W))=\bigcup_{w\in W}(\{0\}\cup\pi^{-1}(\pi(w))\setminus\{0\}) $$ which is what I did stare at but this did not help. I feel that I really do not get a grip on these objects, so that is why I am turning to you.

  • Maybe somebody could give me a hint on what to do
  • More usefully maybe somebody could give me a hint on how to interprete these things in the context of Warner's definition.
share|improve this question
add comment

1 Answer

up vote 1 down vote accepted

You can cosider such a sheaf as a collection of modules bunched together in a continuous way, 'over the points of $M$'. Note that the fibres $\pi^{-1}(m)$ are discrete because $\pi$ is a local homeomorphism, hence the topology inherited to the modules is discrete.

Then, try to prove that if $f:X\to Y$ continuous and is a map above the local homeomorphisms $\xi:X\to Z$ and $\eta:Y\to Z$ (that is, $\eta\circ f=\xi$), then $f$ itself is a local homeomorphism, too.

As the module operations are continuous, we also have that the map $\mathcal S\to \mathcal S$ mapping $s_m\mapsto (s_m-s_m)$ i.e. $s_m\mapsto 0_m$ over $m\in M$ is continuous, so, by the above it follows that the subset $\{ 0_m\mid m\in M\}$ is open, and also that the corresponding null section $m\mapsto 0_m$ is continuous.

About the case when sections $f,g$ coincide at a point: $f(m)=g(m)$, use the fact that $\pi$ is a local homeomorphism, that is the sections are local inverses of $\pi$, i.e. there is a neighborhood $U\subset \mathcal S$ of $f(m)=g(m)$, such that $\pi|_U:U\to \pi(U)$ is a homeomorphism, so that $$f|_{\pi(U)}=(\pi|_U)^{-1}=g|_{\pi(U)}. $$

share|improve this answer
    
Thank you so much. I did not realize that the fibres inherit the discrete topology, which is an important piece of information. This also makes it a lot easier to imagine a sheaf. I was always bewildered by the fact that I have very little information on the topology of these bunched together modules. It turns out that I have a lot! I was looking at cases where discreteness was true when thinking about a proof and I expect this to be very helpful. The fact that 'as a germ' the inversion of $\pi$ is unique is very helpful of course. –  M. Luethi Nov 11 '12 at 22:02
    
I did go through everything discussed as above and it seems that the discreteness of the fibers is indeed what I needed in order to match what I hoped sheaves looked like and what they indeed look like. I hope now I can finally get started properly. Thank you once again. –  M. Luethi Nov 12 '12 at 9:04
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.