The fundamental group of quotient space of 3-folds

Question

Let $S$ be a K3 surface with an involution $\iota_S$, $E$ an elliptic curve with an involution $\iota_E$. Assume the fixed locus of $S$ under $\iota_S$ contains $N>0$ disjoint curves. Note the fixed locus of $E$ under $\iota_E$ contains 4 points. $\iota_S\times \iota_E$ acts on $S\times E$ as an involution, and the fixed locus, say $L$, contains $4N$ disjoint curves. Blow up $S\times E$ on $L$, to get $\widetilde{S\times E}$ and extend $\iota_S\times \iota_E$ to $\widetilde{S\times E}$ to get an involution $\widetilde{\iota_S\times\iota_E}$.

In a paper of Voisin, p.280, Lemma 1.3's proof: since $\widetilde{\iota_S\times\iota_E}$ acts as $-1$ on $\pi_1(\widetilde{S\times E})\cong\mathbb Z^2$, $\pi_1(\widetilde{S\times E}/\widetilde{\iota_S\times\iota_E})=1$. But I wonder why? I don't think in general $\pi_1(X/G)=\pi_1(X)^G$.

Answer 1

The point is that the involution under consideration has at least one fixed point.

Let $Y$ denote your 3-fold, $X$ be its quotient by the involution, $y\in Y$ be a fixed point, and $x$ be its image in $X$. Then you can lift any loop on $X$ with basepoint $x$ to a loop on $Y$ with basepoint $y$, so the homomorphism $\pi_1(Y)\to \pi_1(X)$ is surjective.

Now pick a fixed point $s$ of the involution on $S$ and consider the inclusion $E\to Y$ from the strict transform of $\{s\}\times E$. This induces an isomorphism on $\pi_1$ (K3 surfaces are simply connected), so the previous result shows that $\pi_1(E)\to\pi_1(X)$ is surjective. However, $E\to X$ factors through the quotient $\mathbb{P}^1$ of $E$, so it induces $0$ on $\pi_1$. Therefore we get $\pi_1(X)=0$.