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They are similar. Both contain a projection map and one can define sections, moreover the fiber of the fiber bundle is just like the stalk of the sheaf.

But what are the differences between them?

Maybe a sheaf is more abstract and can break down, while a fibre bundle is more geometric and must keep itself continuous. Any other differences?

Strongart
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    Fibre bundles look the same locally at any point of your space. This does not have to be true for sheaves. – the L Mar 12 '11 at 10:54
  • I have been told (but do not understand) that there is an adjunction between the category of presheaves on a topological space and the category of bundles over the space, which restricts to an equivalence of categories between the category of sheaves and the category of étale bundles. – Zhen Lin Mar 12 '11 at 15:40
  • a fiber bundle is a local product, and a sheaf of (say rings) is a way of associating rings to the open sets of a space so that inclusions of open sets induce homomorphisms of the respective rings. theyre just different things – yoyo Mar 12 '11 at 16:13
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    @yoyo: There are many kinds of sheaves other than sheaves of rings. A fibre bundle gives rise to a sheaf of sections, from which (in reasonable circumstances, and when endowed with the appropriate extra structure) the bundle can be recovered. So, while they are different things, it is not a matter of them being just different things. – Matt E Mar 12 '11 at 21:11
  • Sheaves and fiber bundles are different in general. For example the projection $ \mathbb{R}^2\rightarrow\mathbb{R} $ on the first factor is a (trivial) fiber bundle, but not a sheaf. The intersection of sheaves and fiber bundles is locally constant sheaves or equivalently, covering spaces. – ARA Feb 06 '19 at 18:27

4 Answers4

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If $(X,\mathcal{O}_X)$ is a ringed topological space, you can look at locally free sheaves of $\mathcal{O}_X$-modules on $X$.

If $\mathcal{O}_X$ is the sheaf of continuous functions on a topological manifold (=Hausdorff and locally homeomorphic to $\mathbb{R}^n$), or the sheaf of smooth functions on a smooth manifold, you get fiber bundles (the sheaf associated to a fiber bundle is the sheaf of "regular" (=continuous or smooth here) sections).

Plop
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    ... and conversely any fiber bundle has a natural sheaf associated to it, namely the sheaf of sections of the bundle. – Matt Mar 12 '11 at 19:53
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First remark, there is the definition of sheaf from wikipedia (which by the way talks about étalé spaces and that adjunction business) and the équivalent one 1.2. p. 3 of Bredon, Glen E. (1997), "Sheaf theory" which looks much more like that of a bundle (the A in that definition is the étalé space).

The second remark (from this p.2-3) is that a bundle is locally homeomorphic to a cartesian product, whereas a sheaf is locally homeomorphic to the "base space" itself!

Other difference is that a manifold (which a bundle is) is Hausforff, not the étalé space.

Noix07
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Roughly speaking, sheaf of sections of a bundle is 'equivalent' to the bundle. From a bundle to the sheaf of its sections you can pass easily. In the opposite direction you have to construct germs from sections around each point of the base manifold and then 'topologize' such space to get the bundle. 'Equivalence' means that there are adjoint functors between category of bundles and sheaves of sections of bundles.

Panoramic view of this construction can be found in Chapter 8 of Daniel Rosiak's "Sheaf theory through examples" introductory book. Details, as always, can be found in S. Mac Lane, I. Moerdijk, "Sheaves in Geometry and Logic". This construction shows motivation for usage of germs.

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A fiber bundle is a quadruple $$(\mathcal{M},\pi,\mathcal{E},\mathcal{F})$$ (where $\mathcal{M},\mathcal{E}$ and $\mathcal{F}$ are topological spaces) with $$\pi:\mathcal{E}\to\mathcal{M}$$ such that for every $x\in\mathcal{M}$, there exists a $U\subset\mathcal{M}$ with $x\in U$ and an homeomorphism $$\chi:\pi^{-1}(U)\to U\times\mathcal{F}$$ such that: $$\chi\circ pr_1=\pi$$ (where $pr_1:\mathcal{M}\times\mathcal{F}\to\mathcal{M}$ is the projection onto the first component, i.e. $\forall (m,f)\in\mathcal{M}\times\mathcal{F}:pr_1(m,f)=m$)


In other words in a fiber bundle what is required is a homeomorphism with a cartesian product (locally). Which means that the total space $\mathcal{E}$ has to be locally homeomorphic to $\mathcal{M}\times\mathcal{F}$.
In a sheaf, the only requirement is that a topological space $\mathcal{E}$ is locally homeomorphic to another topological space $\mathcal{X}$, in fact a sheaf is a triple: $$(\mathcal{E},\pi,\mathcal{X})$$ such that $\pi$ is a surjective local homeomorphism.