MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).

3 ,; added 6 characters in body

Let me show that the answer to your last question is yes, by explaining how to construct an Enriques surface inside a simply connected threefold.

Let $\psi \colon V \to \mathbb{P}^3$ be a double covering ramified branched on a smooth quartic surface and let $\tau \in \textrm{Aut}(V)$ be an involution fixing eight points. Set $X=V/ \tau$. Then one proves that $X$ is a rational threefold with eight singular points of type $\frac{1}{2}(1,1,1)$. The anti-canonical divisor $-K_X$ is not Cartier but we have $-K_X \equiv_{Q} H$, where $H$ is an ample divisor on $X$ with $H^3=8$ such that the general element of $|H|$ is a smooth Enriques surface.

Now consider any desingularization $\pi \colon Y \to X$. Since $X$ is rational, $Y$ is a smooth rational variety, hence $\pi_1(Y)=0$. Moreover $X$ has only isolated singularities, so the general element of $|\pi^*H|$ is again a smooth Enriques surface.

For

Example. Let $V \subset \mathbb{P}(1^3, \; 2)$ defined by the detailsequation $$u^2=x^4+y^4+z^4+t^4,$$ where $x,\; y, \; z,\; t$ have weight $1$ and $u$ has weight $2$, and consider the involution $\tau \colon \mathbb{P}(1^3, \; 2) \to \mathbb{P}(1^3, \;2)$ given by $$\tau(x:y:z:t:u)=(x:y:-z:-t:-u).$$ Then $V$ is naturally a double cover of $\mathbb{P}^3$ branched on a Fermat quartic surface and moreover $V$ is $\tau$-invariant. The restricted involution $\tau|_V$ fixes the eight points of $V$ given by $u=z=t=0$ and $u=x=y=0$.

For further details see I. Cheltsov, Rationality of Enriques-Fano threefold of genus five, Isvestiya Math. 68 (2004) and the references given therein.

2 added 3 characters in body

Let me show that the answer to your last question is yes, by explaining how to construct an Enriques surface inside a simply connected threefold.

let

Let $\psi \colon V \to \mathbb{P}^3$ be a double covering ramified on a smooth quadric quartic surface and let $\tau \in \textrm{Aut}(V)$ be an involution fixing eight points. Set $X=V/ \tau$. Then one proves that $X$ is a rational threefold with eight singular points of type $\frac{1}{2}(1,1,1)$. The anti-canonical divisor $-K_X$ is not Cartier but we have $-K_X \equiv_{Q} H$, where $H$ is an ample divisor on $X$ with $H^3=8$ such that the general element of $|H|$ is a smooth Enriques surface.

Now consider any desingularization $\pi \colon Y \to X$. Since $X$ is rational, $Y$ is a smooth rational variety, hence $\pi_1(Y)=0$. Moreover $X$ has only isolated singularities, so the general element of $|\pi^*H|$ is a smooth Enriques surface.

For the details, see I. Cheltsov, Rationality of Enriques-Fano threefold of genus five, Isvestiya Math. 68 (2004) and the references given therein.

1

Let me show that the answer to your last question is yes, by explaining how to construct an Enriques surface inside a simply connected threefold.

let $\psi \colon V \to \mathbb{P}^3$ be a double covering ramified on a smooth quadric surface and let $\tau \in \textrm{Aut}(V)$ be an involution fixing eight points. Set $X=V/ \tau$. Then one proves that $X$ is a rational threefold with eight singular points of type $\frac{1}{2}(1,1,1)$. The anti-canonical divisor $-K_X$ is not Cartier but we have $-K_X \equiv_{Q} H$, where $H$ is an ample divisor on $X$ with $H^3=8$ such that the general element of $|H|$ is a smooth Enriques surface.

Now consider any desingularization $\pi \colon Y \to X$. Since $X$ is rational, $Y$ is a smooth rational variety, hence $\pi_1(Y)=0$. Moreover $X$ has only isolated singularities, so the general element of $|\pi^*H|$ is a smooth Enriques surface.

For the details, see I. Cheltsov, Rationality of Enriques-Fano threefold of genus five, Isvestiya Math. 68 (2004) and the references given therein.