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If $M$ is a complex manifold. we can write $d = \partial + \overline{\partial}$. Does pullback commute with either $\partial$ or $\overline{\partial}$?

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I think you can argue yes due to linearity of the pull-back and independence of $\partial$ and $\bar{\partial}$. In other words, begin with the fact the whole $d$ commutes with the pull-back and try to isolate the same result for the partials. –  James S. Cook Oct 12 '12 at 3:41
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If $M$ and $N$ are complex manifolds and $f:M\to N$ is a smooth map, then $f^*$ commutes with $\partial$ and $\bar\partial$ if and only if $f$ is holomorphic. To prove the "only if" implication, write $f$ in local holomorphic coordinates as $(w^1,\dots,w^n) = (f^1(z),\dots,f^n(z))$, and note that the equation $\bar\partial(f^*w^j)= f^*(\bar\partial f^j)$ reduces to $\bar\partial f^j=0$, which is exactly the Cauchy-Riemann equations for $f^j$. The converse is a straightforward computation in local holomorphic coordinates.

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