Is $\textrm{pr}_E^*\wedge \textrm{pr}_F^*: \Omega^p(E)\otimes \Omega^q(F)\longrightarrow \Omega^{p+q}(E\oplus F)$ an isomorphism?

Question

Let $E$ and $F$ be two vector bundles over a manifold $M$. Consider the Whitney sum $$E\oplus F\longrightarrow M.$$

We have projections $\pi_E:E\oplus F\longrightarrow E$ and $\pi_F:E\oplus F\longrightarrow F$. Is it true that $$\textrm{pr}_E^*\wedge \textrm{pr}_F^*: \Omega^p(E)\otimes \Omega^q(F)\longrightarrow \Omega^{p+q}(E\oplus F),$$ is an isomorphism of $C^\infty(M)$-modules? The map $\textrm{pr}_E^*\wedge \textrm{pr}_F^*$ is of course given by:

$$\omega_E\otimes \omega_F\longmapsto \textrm{pr}_E^*\omega_E\wedge \textrm{pr}_F^*\omega_F,$$ where $\textrm{pr}_E^*:\Omega^p(E)\longrightarrow \Omega^p(E\oplus F)$ and $\textrm{pr}_F^*:\Omega^q(F)\longrightarrow \Omega^q(E\oplus F)$ are the induce maps on forms.

Thanks.

Obs: I just messed everything up. I know that $$\Omega^k(E\oplus F)\simeq \bigoplus_{p+q=k}\Omega^p(E)\otimes \Omega^q(F),$$ so I'll rewrite the question later.

Answer 1

Take $E$ and $F$ to be trivial line bundles and consider the situation over one point. Then your question reduces to: Given a vector space $V$, is the wedge product an isomorphism $\Lambda^k V \otimes \Lambda^l V \rightarrow \Lambda^{k+l} V$? Certainly not, because of the bilinearity of $\wedge$, the map is not even injective ($ \lambda x \wedge y = x\wedge \lambda y$). And in general it is not surjective as well.

Answer 2

No, because for $M$ a point and $E,F$ of dimension $e,f$ an isomorphism would imply $$ \binom{e}{p}\cdot \binom{f}{q}\stackrel {\operatorname {sic}}{=}\binom{e+f}{p+q} \quad (\operatorname {HORROR OF HORRORS }!)$$