Differential forms and behaviour respect product

Question

Let $M$ a smooth $n$-manifold and $E = M \times \mathbf{R}^m$ the product manifold. Let $\pi: E \to M$ the projection onto the first factor. Let $k \leq n$. It's known that $$ \{dx_{i_1} \wedge \cdots \wedge dx_{i_k} : i_1 < \cdots < i_k\} $$ is a local frame of $\Omega^k(M)$. Let $\omega \in \Omega^k(M \times \mathbf{R}^n)$. Locally $$ \omega = f(x_1,\dots,x_n,y_1,\dots,y_m)dx_{i_1} \wedge \cdots \wedge dx_{i_r} \wedge dy_{j_1} \wedge \cdots \wedge dy_{j_s} $$ where $r+s = k$ and $y$ is coordinates in $\mathbf{R}^m$.

Therefore, can we deduce that $$ \Omega^{k}(M \times \mathbf{R}^m) = \sum_{p =1}^{k} \Omega^{p}(M) \wedge \Omega^{k-p}(\mathbf{R}^m) $$ ?

I'm very confused because it's true that $\Lambda(V \times W) = \Lambda V \otimes \Lambda W$ and here appears the wedge product. In particular, I want to prove that every $k$-form in $M \times \mathbf{R}^m$ is a linear combination of $$ \pi^* (\phi) f(x,y) dy_{j_1} \wedge \cdots \wedge dy_{j_s} $$ where $\phi \in \Omega^{*}(M)$. Notice that the last formula is global I don't need charts

Answer 1

Your last equation gives you the answer: while you indeed have the wedge products of the two coframe, the "function" part may mix the coordinates. In fact you write $f(x,y)$ and not every such $f$ can be written as a product of a function on $M$ with a function on $R^n$, for example $\sin(xy)$. If you prefer, $C^\infty (M\times N) \neq C^\infty(M) C^\infty(N)$.

Answer 2

What is true is what you stated. If $M$ and $N$ are manifolds, then $\Lambda^k\left(M\times N\right) = \bigoplus_{r+s=k} \Lambda^r(M)\otimes\Lambda^s(N)$. The projections $\pi_M : M\times N \to M$ and $\pi_N : M\times N \to N$ give a morphism $\bigoplus_{r+s=k}\Omega^r(M)\otimes \Omega^s(N) \to \Omega^{k}(M\times N)$. There is no reason that it will be surjective.

For example, if $k=0$, you have a morphism \begin{align} \Omega^0(M)\otimes \Omega^0(N) &\longrightarrow \Omega^0(M\times N) \\ (f,g) &\longmapsto (p,q)\mapsto f(p)g(q) \end{align}but it is clearly not surjetive in many examples. The fact is that in the fibers, everything works fine, but when you take sections, that are global notions, you cannot act like you were working pointwisely.