Splitting of short exact sequence of holomorphic vector bundle I'm having trouble understanding the failure of splitting of holomorphic vector bundle. Following are my thoughts on this issue:
In general, for a s.e.s. within the same category (by which I mean $E,F,G$ are objects of same category and $f,g$ are permitted morphisms)
$$0 \rightarrow E \xrightarrow{f} F\xrightarrow{g} G \rightarrow 0$$
I have an "intuitive proof" for the splitting:
1.We have a direct sum decomposition $F\cong f(E)\bigoplus F/f(E)$.
2.Since $f$ is injective by exactness, $F \cong E \bigoplus F/f(E)$.
3.We have $F/Kerg \cong G$.
4.By exactness, $Kerg\cong Imf$, thus $F \cong E \bigoplus G$.
Once I came up with this argument, I started to think what could go wrong. (I'm not pretty familiar with category theory, so please correct me if I say something wrong)

*

*The quotient object is not always defined. (i.e. sit inside the same category)

*The direct sum decomposition might not hold true. (this decomposition is valid as sets, but probably not valid as objects in the category, i.e. the projections might not be morphism)

*$F/Kerg \cong G$ minght not hold true. (again, such map might not be morphism)

Except for the above three points, I did not see where could go wrong with this argument
Now I try to apply this "intuitive" argument to the following cases:

*

*vector bundles (in smooth category). It seems to me that in this case all the criterion are valid. But all of the proofs I found for the splitting of s.e.s. of smooth vector bundles make use of a fiberwise metric on $F$ to produce an orthogonal complement to prove. I wonder why we need this extra structure to prove.


*holomorphic vector bundles. I thik in this case the problem happens for reason 1, i.e. the quotient object is not always defined as a holomorphic vector bundle, but I'm not pretty sure.
So my questions overall are:

*

*Does the intuitve argument I presented serve as the usual argument of splitting of s.e.s? If not, what is the problem with the argument?


*Why do we need a fiberwise metric to have a decomposition that is "the same" as the decomposition in intuitve proof 1 rather than just simply using the "canonical decomposition" ?


*I understand that the splitting in holomorphic vecotr bundle case fails to happen since we cannot find a fiberwise metric by piece together local ones using bump functions. But this seems a techique-wise reason, not the essential reason. Any different way to see this? (probably from category view?)
 A: $\newcommand{\Proj}{\mathbf{CP}}$tl; dr: Condition 2 fails; the first step of the intuitive argument is not true.

In the "real" situation, we can take orthogonal complements using a metric and remain in the smooth category. With complex vector bundles, by contrast, a metric needs to be conjugate-linear in one variable in order to "behave decently" from a linear-algebraic perspective, e.g., to be positive-definite. (See also the exercise below.) The orthogonal complement of a holomorphic subbundle $E \subset F$ is therefore "usually not" a holomorphic subbundle of $F$.
In the same sense, the quotient $F/f(E)$ is "usually not" isomorphic to a holomorphic subbundle of $F$. One standard justification is the principle Griffiths and Harris call curvature decreases in subbundles and increases in quotient bundles.
A basic example is the tautological line bundle $E$ over a complex projective space $\Proj^{n}$, which is a subbundle of the trivial rank $(n + 1)$ bundle $F$, but the quotient (isomorphic to $E^{*} \simeq T\Proj^{n} \otimes \mathcal{O}_{\Proj^{n}}(-1)$) is not a holomorphic subbundle of $F$.

Exercise: A complex-bilinear form $h$ on a finite-dimensional complex vector space is never positive-definite.

 Proof: For every vector $v$ we have $h(iv, iv) = -h(v, v)$.

