Let $\alpha$ be a complex $k$-form on $X$ and $f$ a smooth map between complex manifolds. Is then $$\overline{f^* \alpha} = f^* \overline \alpha$$
This is a local question, so we can assume $\alpha = g \,dz_{i_1} \wedge \dots \wedge d\bar z_{j_q}$ and $f=(f_1,\dots,f_n)$. So it suffices to check this on functions and on $dz_{i_1}$ and $d\bar z_{j_1}$
I think it is true on the coordinate 1-forms, because $\overline{f^* d z_i} = \overline {d f_i} = \bar d \overline{ f_i} = d \overline{ f_i}.$
But how is it for functions? Is $\overline{f^* g} = \overline{g\circ f}$ the same as $f^* \bar g = \bar g \circ f$?
I thought, this was obvious, because $\bar f (z) = \overline {f(z)}$.
But then I saw this question and got confused.
Edit: I saw that $f(x,y) = u(x,y) + i v(x,y)$ with $u,v$ real, so $\bar f = u - iv$. I still think, it is true, but I begin to wonder, as I don't see it written in any of the books.