Consider a holomorphic bundle $F$ over a complex manifold $M$. I have read on a comment here on mathstack that given a holomorphic subbundle $E$ of $F$ there are many non-isomorphic holomorphic bundles $G$ that complete the sort exact sequence
$$0\to E \xrightarrow{h} F \to G \to 0\,.$$
I really can not understand how is this possible. By defintion of the short exact sequence it should be $F/im(h) = G$, where $im(h) =E$. Hence any two holomorphic bundles $G_1$, $G_2$ that complete the short exact sequnce above should be isomorphic to $F/E$. Note that by convention $h$ should have constant rank along the fibers.
Best Answer
I think you may have been the victim of a game of telephone. What you say is true; given a holomorphic subbundle $S \subset E$ there is exactly one holomorphic bundle $Q$ that fits into a short exact sequence $$ 0 \to S \to E \to Q \to 0. $$ The proof is just by general algebra: Suppose there are two such bundles $Q$ and $Q'$. Then we can define a map $Q \to Q'$ by lifting an element to $E$ and projecting it down. This is well-defined for the usual reasons and holomorphic, and has a trivial kernel so it is an isomorphism.
However, given two holomorphic vector bundles $S$ and $Q$ there may be many bundles $E$ such that there is a short exact sequence $$ 0 \to S \to E \to Q \to 0. $$ The different bundles $E$ are called extensions of $S$ by $Q$. There is always one obvious extension, $E = S \oplus Q$, but there may be other ones. Their isomorphism classes are classified by the cohomology group $H^1(X, \operatorname{Hom}(Q, S))$, where $X$ is the underlying complex manifold.
For example, if $X$ is a curve of genus $g$ and $S = Q = \mathcal{O}_X$ are the trivial line bundles, then $$ H^1(X, \operatorname{Hom}(Q, S)) = H^1(X, \mathcal O_X) $$ is a $g$-dimensional vector space. When $X$ is the projective line there are thus no nontrivial extensions of the trivial line bundle by itself, but for higher-genus curves there are such nontrivial extensions.
You can read about this in Demailly's book (Chapter V.14) or look into Ext functors.