Let $X$ be a complex manifold. Let $$0\rightarrow E\rightarrow F\rightarrow G\rightarrow 0$$ be an exact sequence of holomorphic bundles over $X$. Show that $\det F\cong\det E\otimes\det G$.
I thought to try by associating the corresponding sheaves of $\mathcal{O}_X$-modules. If the sequence splits (like it is the case in the real setting), it is simple. In general it does not split, however we have splitting at the level of stalks i.e. $F_x\cong E_x\oplus G_x$ for every $x\in X$. If $\phi$, $\psi$ and $\tau$ are corresponding transition maps that is e.g. $\phi (U\cap V):\mathcal{O}(U\cap V) ^r\rightarrow E(U\cap V)\rightarrow \mathcal{O}(U\cap V) ^r$, then at every point we shall have matrix for $\psi _x$ is block diagonal with blocks $\phi _x$ and something else that should give determinant isomorphic to that of $G$. How we can relate the transition functions of bundles forming an exact sequence and use it to prove the claim? To answer the last question, for simplicity one can assume $G$ is subbundle of $E$. (Then the first question becomes why $\det G \cong \det F/E$ )
The transition functions for $F$ are a block upper triangular matrix with top left block the transition functions for $E$ and bottom right block the transition functions for $G$. (The top right is some garbage we don't care about.) It is a linear algebra fact that the determinant of such a matrix is the product of the determinants of its diagonal blocks. Since the transition function of a tensor product of line bundles is the product of their transition functions, this proves the claim.