I'am working on a shorter proof of a theorem but to manage it I need to know if a lemma is true.
Conjecture: Given a manifold $M$ and an short exact sequence of vector bundles $$ 0 \rightarrow E' \rightarrow E \rightarrow E'' \rightarrow 0 $$ then $\Lambda^\cdot E' \otimes \Lambda^\cdot E'' \simeq \Lambda^\cdot E $, meaning by $\Lambda(\cdot)$ the vector bundle of external algebra. Is also a isomorphism of algebras?
This is not true unless the sequence splits. The only thing which is true in general is $\Lambda^{m}E'\otimes\Lambda^{n}E"\cong\Lambda^{m+n}E$ where $m,n$ are ranks of $E',E"$. Just as an illustration, let $m=n=1$. Then $\Lambda^. E'=\mathcal{O}\oplus E'$ and similarly for $E"$. So, the left side in your isomorphism is $\mathcal{O}\oplus (E'\oplus E")\oplus E'\otimes E"$, while the right side is $\mathcal{O}\oplus E\oplus \Lambda^2 E$, the last term as I said is just $E'\otimes E"$. In general these are not even isomorphic as vector bundles, unless $E\cong E'\oplus E"$.