Let $\Sigma \subset M$ be a compact Riemann surface of genus $g\geq 2$ in a compact complex manifold. If there is a non-trivial map $M\to \Sigma$ is $M$ a direct product of $\Sigma$ and another manifold?
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5$\begingroup$ I presume u can get a counterexample by taking a non trivial extension, $V$, of $\mathcal{O}$ by itself and taking the projective bundle $\mathbb{P}(V)$. The sub $\mathcal{O}$ gives you the section. $\endgroup$– user108998Commented Sep 21, 2021 at 10:39
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2$\begingroup$ In such a counterexample, the non-trivial map is even a retraction (i.e., is identity on $\Sigma$). That is, a retract need not be a direct factor. $\endgroup$– YCorCommented Sep 21, 2021 at 13:47
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$\begingroup$ Is there an example with compact complex manifolds where there is a surjective map but no retract? $\endgroup$– jkelimCommented Sep 21, 2021 at 14:22
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As requested, an example where there is no retract:
Take $E$ an elliptic curve, and $C$ a smooth curve of genus $\geq 2$ in $E \times E$. Take $D$ a ramified double cover of $C$, and let $M$ be the quotient of $D \times E \times E$ by the product of the involution on $D$ over $C$ and translation by a point of order $2$ on the first copy of $E$. This product involution has no fixed points so the quotient is smooth (and compact).
Then $M$ maps to $C$ by projection onto the first factor. It contains $E \times E$ (the fiber of this projection onto any unramified point) and thus contains $C$.
But there is no retract/ section, since that would pull back to be a section of $D \times E \times E \to D$ invariant under the involution, which means restricting to a ramified fiber we would get a point of $E \times E$ invariant under translation by a point of order $2$, which is impossible.
answered Sep 21, 2021 at 16:52