k-form: sum of wedge products of 1-forms? Let M be a smooth manifold. Can every k-form $\omega$ on M be written as a sum of k-forms, that are wedge products of 1-forms, i.e. $\omega = \sum_{i=0}^n \alpha_1^{(i)} \wedge \ldots \wedge \alpha_k^{(i)} $, where $\alpha_l^{(i)} \in \Omega^1(M) $ ? If M is compact, one can cover M be finitely many charts and use a partition of unity to see that this holds...but how about the general case?      
 A: Johannes' answer can be upgraded to the following statement:
Let $M$ be a second countable and Hausdorff manifold (who cares about others?) and $\pi_i\colon E_i \longrightarrow M$ vector bundles for $i = 1, \ldots, N$. Then the canonical map
\begin{equation}
   \Gamma^\infty(E_1) \otimes_{C^\infty(M)} \cdots \otimes_{C^\infty(M)}
   \Gamma^\infty(E_N)
   \longrightarrow
   \Gamma^\infty(E_1 \otimes \cdots \otimes E_N)
\end{equation}
is an isomorphism of $C^\infty(M)$-modules. The same holds if you do some symmetrization/antisymmetrization in tensor powers of a single vector bundle in addition.
My favorite argument uses the smooth version of Serre-Swan's theorem: the sections $\Gamma^\infty(E)$ are a finitely generated and projective module over $C^\infty(M)$ and any finitely generated projective module is (up to iso) of that form. This can e.g. be proved along the same lines as Johannes' argument by embedding the tangent bundle of $E$ into some big $\mathbb{R}^n$ and observe that $E$ is naturally identifiable with the vertical subbundle of $TE$ viewed as bundle over $M$.
Then the above statement is shown by noting that the map is clearly injective (and well-defined). The surjectivity is the harder part. But the tensor product on the left hand side is again finitely generated and projective, so it has to be the space of smooth sections of  some vector bundle. Then the injectivity gives that this vector bundle can be included as a subbundle of $E_1 \otimes \cdots \otimes E_N$. Counting fibre dimensions shows that they coincide...
The main point is that Serre Swan also holds in the non-compact case. An alternative proof of this can be found e.g. in Well's book in complex differential geometry.
A: The same argument as in comapct case works if you use in addition Ostrand's theorem on colored dimension
A: The answer is yes and you need transversality in some form. Here we use Whitneys embedding theorem, but a weaker statement would suffice. Embed $M \subset \mathbb{R}^m$. Thus there is a monomorphism $TM \to M \times \mathbb{R}^m$ of vector bundles, dualizing to an epimorphism $M \times \mathbb{R}^m \to T^{\ast} M$. Let $a_1,\ldots ,a_m$ be the images of the basis vectors. You can write any $k$-form on $M$ as a linear combination with $C^{\infty} (M)$-coefficients of the forms $a_{i_1} \wedge \ldots a_{i_k}$. Done.
