Dolbeault coholomogy induced by pull-back of forms

Question

Ive been racking my brain and I cant seem to get it right. I have proven that the pull back induces a linear map between the de rham cohomologies of groups but I still cant prove that a pull-back $f* : C^\infty(Y, \Omega^k) \rightarrow C^\infty(X, \Omega^k)$ induces a linear map between the Dolbeault coholomogy $f* : H^{p,q}(Y) \rightarrow H^{p,q}(X)$ ? And yes $f: X \rightarrow Y$ is holomorphic. I think i need to show that pull back maps forms of type (p,q) to type (p,q) before I can show that the induced map is linear or well-defined.

Answer 1

Yes, you do need to know that, so see this recent post. And then it will follow by type considerations that $\bar\partial$ commutes with pullback, so pullback by a holomorphic function respects the Dolbeault cohomology.

Answer 2

In the case of the De Rham cohomology, the pullback of a smooth map induces a linear map in cohomology since it commutes with the exterior derivative. For the Dolbeault cohomology, this is the exact same thing with $\overline{\partial}$ and holomorphic functions, namely:

Let $X$ and $Y$ be two complex manifolds and let $f\colon X\rightarrow Y$ be a holomorphic function, then one has: $$\forall u\in C^\infty(X,\Lambda^{p,q}),f^*\overline{\partial}u=\overline{\partial}f^*u.$$