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I have a question about an argument in the proof of Lemma 1.2.(1) in Quotients of K3 surfaces modulo involutions by D. Q. Zhang:

Let Let $(X, \sigma)$ be X be a smooth projective K3 surface with an involution $\sigma$ such that $\sigma^*\omega=-\omega$ for a non-zero holomorphic $2$-form $\omega$. Let $S := X/\sigma$ be the quotient space and $f : X \to S$ the quotient morphism.

Then (1) $S$ is a smooth surface with irregularity $q(S):= h^1(S, \mathcal{O}_S) = 0$.

The first part on smoothness is fine, see prev. Lemma 1.1.(1). But how do we obtain that the irregularity $q(S)$ is zero?

It is well known that irregularity of K3 surfaces is zero, and we have a dominant finite morphism $f: X \to S$.

Question: Is it always true that $q(S) \le q(X)$? How to see it? If not, what was actually Zhang's argument there to get $q(S)=0$?

Remark: Zhang works in complex setting, ie base field algebraically closed of char $0$. Does it matter for the statement on proposed inequality between irregularities?

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    $\begingroup$ Since the K3 surface is simply-connected, every finite unbranched cover of $S$ is intermediate between $S$ and its finite cover $X$. Thus the fundamental group of $S$ is finite. Therefore the irregularity of $S$ equals zero. $\endgroup$
    – Jason Starr
    Commented Nov 11 at 15:59
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    $\begingroup$ By the Universa Coefficients Theorem, the first cohomology group is the dual of the first homology group. If the irregularity is positive, so is the first Betti number. So the first homology group is infinite. Therefore the fundamental group is infinite. $\endgroup$
    – Jason Starr
    Commented Nov 11 at 16:25
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    $\begingroup$ You do not need $X$ to be unbranched over $S$. Every unbranched cover of $S$ pulls back to an unbranched cover of $X$. Since $X$ is simply connected, there is a section of the pullback unbranched cover. This equals the graph of an $S$-morphism from $X$ to the original unbranched cover. In particular the degree of that unbranched cover divides the degree of $X$ over $S$. So the fundamental group of $S$ is finite. $\endgroup$
    – Jason Starr
    Commented Nov 11 at 16:30
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    $\begingroup$ More generally, for any surjective map $f \colon Y \to X$ of smooth projective varieties, the pullback $H^i(X,\Omega_X^j) \to H^i(Y,\Omega_Y^j)$ is injective for all $i,j$, at least in characteristic $0$. It is true in any characteristic for any Weil cohomology theory (but Hodge and de Rham cohomology have torsion coefficients if $\operatorname{char} k > 0$, and there the result is false); one reference is Kleiman's paper in Dix exposés sur la cohomologie des schémas, Proposition 1.2.4. The proof is a simple consequence of Poincaré duality and the projection formula. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Nov 11 at 17:00
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    $\begingroup$ related to this $\endgroup$
    – user267839
    Commented Nov 12 at 11:43

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