The short version:
Why does a Borcea-Voisin threefold has trivial fundamental group?
Well, what is a Borcea-Voisin threefold? These are named after Borcea and Voisin, who introduced a method of constructing Calabi-Yau threefolds from elliptic curves (CY-onefolds) and K3 surfaces (CY-twofolds), and going through Voisin's proof that these threefolds are indeed Calabi-Yau, I'm stuck on why the fundamental group is trivial.
In detail, let $E$ be an elliptic curve with canonical involution $(x,y)\mapsto (x,-y)$ (we're working in characteristic $0$). Let $S$ be a K3 surface with (non-symplectic) involution $\sigma$ acting by $-1$ on $H^{2,0}(S)$. Then the (crepant) resolution of $E\times S/\iota\times\sigma$ is a Calabi-Yau threefold. Denote the resolution by $X$.
To see this, there are a few things to show (though I'm happy with showing the necessary Hodge numbers are zero, once $\pi_1(X)$ is trivial). Firstly, we can write the resolution $X$ as $\widetilde{E\times S}/\widetilde{\iota\times\sigma}$, where we blowup the fixed locus of $\iota\times\sigma$ on $E\times S$, then extend the action of $\iota\times\sigma$.
The fixed locus of an elliptic curve under the involution is four isolated points (topological Lefschetz trace) and the fixed locus of $S$ under $\sigma$ is the union of (say) $N$ disjoint curves (holomorphic and topological Lefschetz trace). Thus, the fixed locus of $E\times S$ is the union of each $P_i\times C_j$, where each $P_i$ is an isolated (fixed) point on $E$ and $C_j$ is a (fixed) curve on $S$.
If there is at least one fixed curve $C$ on $S$, there is at least one fixed point on $E\times S$, say $P$, and $\widetilde{\iota\times\sigma}$ acts on the fundamental group $\pi_1(\widetilde{E\times S},P)$ by $-1$, as $$ \pi_1(\widetilde{E\times S},P) \simeq \pi_1(E\times S,P) \simeq \pi_1(E)\times\pi_1(S) \simeq \pi_1(E), $$ and the action of $\iota$ on $E$ passes through to $\pi_1(E)$.
Voisin now claims this implies $\pi_1(X)$ is trivial, and I'm guessing it shouldn't be too difficult to justify, given there are no details, but I'm not too sure how this is done. I've also asked a few people that know far more about the fundamental group than I do, and no one had a solution (though they all say it's probably simple...)
I'd love to just say $$ \pi_1(X/G) \simeq \pi_1(X)^G, $$ for any $X$ with action by $G$, but this isn't always true...
Is there something horribly simple I've missed for this case? Thanks!