# Koszul-Malgrange Holomorphic structure on a pullback bundle

I'm finding myself a little confused about Koszul-Malgrange holomorphic structures in a certain context.

Suppose $$M$$ is a complex manifold, $$N$$ is a smooth manifold with a smooth complex vector bundle $$V$$ and bundle connection $$\nabla$$, and $$f:M\to N$$ is a smooth map. We form the pullback bundle $$f^*V$$ over $$M$$ and give it the pullback connection $$f^*\nabla$$. Let $$P: f^*V \to V$$ be the associated smooth projection map. Assuming $$f^*\nabla^{0,1}\cdot f^*\nabla^{0,1}=0$$, a theorem of Koszul-Malgrange says there is a holomorphic structure on $$f^*V$$ that turns it into a holomorphic vector bundle over $$M$$.

My question is about the following. Suppose we have a (possibly locally defined) holomorphic section $$s$$ of $$f^*V$$. In a suitable trivialization over some open set $$\Omega$$, so that $$f^*V$$ splits as $$\Omega\times \mathbb{C}^n$$, we can write $$s(z) = (z, h(z))$$ where $$h:\Omega\to \mathbb{C}^n$$ is holomorphic.

Now, consider the map $$P\circ s: \Omega \to V$$ I'm basically wondering if it makes sense to write $$P\circ s =(f(z), h(z))$$ for some choice of trivialization of $$V$$. The sort-of roadblock that I'm hitting is that the trivialization of the pullback we used should come from the Koszul-Malgrange theorem, rather than be pullback of a trivialization of $$V$$. So, I'm finding it hard to fiddle the definition of the pullback bundle.

I'm mainly interested in the case where $$\dim M \leq \dim N$$ and $$f$$ is regular everywhere. Then in my trivizalization over $$\Omega$$, the map $$f_*$$ should be an isomorphism of $$f^*V|_{\Omega}\to V|_{f(\Omega)}$$ that acts by $$(z,v)\mapsto (f(z),v)$$. Of course, this induces a trivialization over $$f(\Omega)$$. Then we can find an open set $$U$$ containing $$f(\Omega)$$ in which we have a smooth trivialization that extends our trivialization over $$f(\Omega)$$. I think that in this trivialization, $$P\circ s$$ should assume the desired form.

Okay I thought about it a bit more, and here is what I can say in general. We can find a neighbourhood $$U\subset N$$ such that $$V|_U$$ splits diffeomorphically as $$U\times \mathbb{C}^n$$ and the section $$s$$ takes the form $$s(z) = (f(z), A(z)\circ h(z))$$, where $$A(z)$$ is a smoothly varying family of invertible $$n\times n$$ complex matrices.
Let $$(U,\varphi)$$ be a trivialization of $$V$$ over some open set $$U\subset N$$ and $$\Omega\subset M$$ such that $$f(\Omega) \subset U$$. In $$\Omega$$ we have the pullback trivialization $$\psi$$, and in these trivializations the map $$P$$ takes $$\Omega\times \mathbb{C}^n\to U\times \mathbb{C}^n$$ via $$(z,v)\mapsto (f(z),v)$$. Meanwhile, according to Koszul-Malgrange (by the way, the easiest exposition I've found is in here http://gtnmk.droppages.com/2019/250B-2019-set.pdf), then for $$p\in \Omega$$ we can find $$\Omega_p\subset \Omega$$ and a holomorphic trivialization $$k:f^*V|_{\Omega_p}\to \Omega_p \times \mathbb{C}^n$$. Let us just restrict $$\Omega$$ so that $$\Omega_p=\Omega$$. This should make the notation easier.
If $$s(z) = (z,b(z))$$ with respect to the trivialization $$\psi$$, then $$P\circ s= (f(z),b(z))$$ with respect to the trivializations $$\psi,\varphi$$. Under $$k$$, $$s$$ assumes the form $$s(z) = (z,h(z))$$, with $$h$$ holomorphic. The relation between $$b$$ and $$h$$ is as follows. $$\varphi$$ and $$k$$ are trivializations, so in coordinates, we can write $$\varphi\circ k^{-1}(z,v) = (z, A(z)v)$$ for some smoothly varying family of invertible matrices $$A(z)$$. $$A(z)$$ then satisfies $$b(z) = A(z) \circ h(z)$$ which gives the result.