TITLE: Compact complex surface that admits a Kodaira fibration is Kahler
QUESTION [5 upvotes]: A Kodaira fibration is a compact complex surface X endowed with a holomorphic submersion onto a Riemann surface $\pi: X\to\Sigma$ which has connected fibers and is not isotrivial.
Is there an easy way to see why a compact complex surface that admits a Kodaira fibration is Kahler? I know for a complex compact surface is Kahler if and only if its first Betti number is even. I wonder it's possible to deduce that a compact complex surface that admits a Kodaira fibration has even Betti number?

REPLY [2 votes]: Let $f \colon S \longrightarrow B$ be a Kodaira fibration, and let $F$ be a general fibre. Then by [Kas68, Thm. 1.1] we have $g(B) \geq 2$ and $g(F) \geq 3$. 
In particular, $S$ contains no rational or elliptic curves: in fact, such curves cannot neither dominate the base (because $g(B) \geq 2$) nor be contained in fibres (because the fibration is by assumption smooth).
So every Kodaira fibred surface $S$ is minimal and, by the superadditivity of the Kodaira dimension, it is of general type. 
In particular, it is not only Kähler  but actually algebraic (i.e., projective).
References.
[Kas68] A. Kas: On deformations of a certain type of irregular algebraic surface, American J. Math. 90, 789-804 (1968). ZBL0202.51702.