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Let $S$ be an smooth algebraic surface and $C$ be a smooth curve, both over $\mathbb{C}$. Assume that a finite group $G$ acts on $S$ and $C$ respectively and the induced acting of $G$ on $S\times C$ has no fixed point. I now want to compute $NS((S\times C)/G)$. Is it true that $NS((S\times C)/G)\cong NS(S\times C)^G$ as abelian groups?

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No, since the torsion parts may differ. You can create a counterexample by letting $S$ be a K3 surface with $G=\mathbb{Z}/2$ acting on $S$ by a fixed point involution and trivially on $C$. If you tensor $NS$ by $\mathbb{Q}$, then this should be OK. – Donu Arapura Oct 21 at 13:02

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