Cohomological criterion for $G$-equivariant maps - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T07:28:40Z http://mathoverflow.net/feeds/question/108498 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/108498/cohomological-criterion-for-g-equivariant-maps Cohomological criterion for $G$-equivariant maps Koopa 2012-09-30T22:10:07Z 2012-10-06T16:27:50Z <p>Let $f:M\rightarrow N$ be a surjective map from a fourmanifold $M$ to a surface $N$, with connected fibers (each fiber is connected). Assume that $f$ admits a multiple section $s:N\rightarrow M$. </p> <p>Suppose now that we are given a finite group $G$ acting freely on $M$ in such a way that the fiber class and the multiple section class in $H^2(M,\mathbb{Z})$ are invariant or multiplied by $-1$. Can we conclude that the map $f$ is $G$-equivariant and descends to $f':M/G\rightarrow N/G$ as topological spaces (as $G$ action may not be free on $N$)? </p> <p>A point I am confused by is that the fiber class and the section class are preserved (invariant or multiplied by -1) as classes, not as cycles. </p> <p><strong>Edit</strong> My question turns out to be non-sense in its original form above. So let me change my question. I am reading <a href="http://link.springer.com/article/10.1007/s002080050125" rel="nofollow">this paper</a> by M. Gross and P.H.M. Wilson, where they study SYZ conjecture for certain CY3s. I now try to understand the proof of Theorem 2.1. where certain holomorphic map $f$ is claimed to be anti-holomorphic $C_2$-action equivariant. Here the assumption in my question is satisfied. I now understand the (anti-)holomorphicity of the map $f$ ($C_2$-action) is crucial. But it is not yet clear to me why $f$ is $C_2$-action equivariant. </p>