I meant, why the k-form has to be defined over the cotangent bundle? I know that the cotangent space is the dual of the tangent space, but this not explain why is defined there
@Fargle I know, but I do not want to assume anything, actually I am trying to build piece by piece a hausdorff, second-countable mtrizability manifold.
@Fargle I see. Then, if we suppose that a Hausdorff space has paritions of unity, how should I proceed in order to build second countable for that space?
Right, but partition of unity allow us to extend local properties to global ones, this help us to contrust topological spaces, but no every topological spaces need to be second countable.
Reading the definition of partition of unity:
Let $A\subset \Bbb R^n$ and let $\mathcal{O}$ be an open cover of $A$. Then there is a collection $\Phi$ of $C^\infty$ functions $\varphi$ defined in an open set containing $A$, with the following properties:
For each $x \in A$ we have $0 ...
@PaulPlummer Thanks you. I was asking because later it says that every Banach space which is not isomorphic to a Hilbert space, has closed subspace that has no complement. And since every finite-dimensional Banach space has complement for every closed subspace, I was on doubt, so thanks!
Reading the definition of partition of unity:
Let $A\subset \Bbb R^n$ and let $\mathcal{O}$ be an open cover of $A$. Then there is a collection $\Phi$ of $C^\infty$ functions $\varphi$ defined in an open set containing $A$, with the following properties:
For each $x \in A$ we have $0 ...