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The point is that by your construction $f:\mathcal{E}_\Gamma\to Y_\Gamma$ is an elliptic surface with a section $\sigma$. So as Keerthi said in the comments, $f$ is projective. In fact the relative divisor $3\sigma$ gives the Weirstrass embedding $\mathcal{E}_\Gamma\to \mathbb{P}(f_*\mathcal{O}(3\sigma))$). (Ben's comment shows that projectivity can fail without a section.) From here, quasiprojectivity is straight forward: If $V$ is an extension of $f_*\mathcal{O}(3\sigma)$ to a vector bundle on the smooth projective closure of $ Y_\Gamma$, then the closure of $\mathcal{E}_\Gamma$ in $\mathbb{P}(V)$ is a projective variety. Therefore $\mathcal{E}_\Gamma$ is quasiprojective.

Given the endless stream of comments, perhaps I should add a few words of clarification:

1. Since $Y=Y_\Gamma$ is noncompact, any vector bundle such as $f_*\mathcal{O}(3\sigma)$ is in fact (analytically) trivial. This is definitely overkill, but you can use Grauert, "Analytichse Faserungen..." Math. Ann 1958.
2. Thus $\mathbb{P}(\mathcal{O}(3\sigma))\cong Y\times \mathbb{P}^2$. This embeds into $\bar{Y}\times \mathbb{P}^2$, where $\bar Y$ is a (the) smooth projective compactification of $Y$.
3. We can take the closure of $\mathcal{E}_\Gamma$ to get a projective variety, and the quasiprojectivity of this family follows easily.
4. Note that the fibres of the closure $\overline{\mathcal{E}_\Gamma}$ may be singular.
5. If the Hopf surface had a section, it would lift to a rational curve in $\mathbb{C}^2-\lbrace 0\rbrace$. Well, I'll let you think about why that might be a problem.

The point is that by your construction $f:\mathcal{E}_\Gamma\to Y_\Gamma$ is an elliptic surface with a section $\sigma$. So as Keerthi said in the comments, $f$ is projective. In fact the relative divisor $3\sigma$ gives the Weirstrass embedding $\mathcal{E}_\Gamma\to \mathbb{P}(f_*\mathcal{O}(3\sigma))$). (Ben's comment shows that projectivity can fail without a section.) From here, quasiprojectivity is straight forward: If $V$ is an extension of $f_*\mathcal{O}(3\sigma)$ to a vector bundle on the smooth projective closure of $ Y_\Gamma$, then the closure of $\mathcal{E}_\Gamma$ in $\mathbb{P}(V)$ is a projective variety. Therefore $\mathcal{E}_\Gamma$ is quasiprojective.

Given the endless stream of comments, perhaps I should add a few words of clarification:

1. Since $Y=Y_\Gamma$ is noncompact, any vector bundle such as $f_*\mathcal{O}(3\sigma)$ is in fact (analytically) trivial. This is definitely overkill, but you can use Grauert, "Analytichse Faserungen..." Math. Ann 1958.
2. Thus $\mathbb{P}(\mathcal{O}(3\sigma))\cong Y\times \mathbb{P}^2$. This embeds into $\bar{Y}\times \mathbb{P}^2$, where $\bar Y$ is a (the) smooth projective compactification of $Y$.
3. We can take the closure of $\mathcal{E}_\Gamma$ to get a projective variety, and the quasiprojectivity of this family follows easily.
4. Note that the fibres of the closure $\overline{\mathcal{E}_\Gamma}$ may be singular.
5. If the Hopf surface had a section, it would lift to a rational curve in $\mathbb{C}^2-\lbrace 0\rbrace$. Well, I'll let you think about why that might be a problem.

OK, I guess there some issues.... I'm converting this to CW. Anyone, who wants to fix this is welcome to. I've got to finishing my refereeing....

The point is that by your construction $f:\mathcal{E}_\Gamma\to Y_\Gamma$ is an elliptic surface with a section $\sigma$. So as Keerthi said in the comments, $f$ is projective. In fact the relative divisor $3\sigma$ gives the Weirstrass embedding $\mathcal{E}_\Gamma\to \mathbb{P}(f_*\mathcal{O}(3\sigma))$). (Ben's comment shows that projectivity can fail without a section.) From here, quasiprojectivity is straight forward: If $V$ is an extension of $f_*\mathcal{O}(3\sigma)$ to a vector bundle on the smooth projective closure of $ Y_\Gamma$, then the closure of $\mathcal{E}_\Gamma$ in $\mathbb{P}(V)$ is a projective variety. Therefore $\mathcal{E}_\Gamma$ is quasiprojective.

The point is that by your construction $f:\mathcal{E}_\Gamma\to Y_\Gamma$ is an elliptic surface with a section $\sigma$. So as Keerthi said in the comments, $f$ is projective. In fact the relative divisor $3\sigma$ gives the Weirstrass embedding $\mathcal{E}_\Gamma\to \mathbb{P}(f_*\mathcal{O}(3\sigma))$). (Ben's comment shows that projectivity can fail without a section.) From here, quasiprojectivity is straight forward: If $V$ is an extension of $f_*\mathcal{O}(3\sigma)$ to a vector bundle on the smooth projective closure of $ Y_\Gamma$, then the closure of $\mathcal{E}_\Gamma$ in $\mathbb{P}(V)$ is a projective variety. Therefore $\mathcal{E}_\Gamma$ is quasiprojective.

Given the endless stream of comments, perhaps I should add a few words of clarification:

1. Since $Y=Y_\Gamma$ is noncompact, any vector bundle such as $f_*\mathcal{O}(3\sigma)$ is in fact (analytically) trivial. This is definitely overkill, but you can use Grauert, "Analytichse Faserungen..." Math. Ann 1958.
2. Thus $\mathbb{P}(\mathcal{O}(3\sigma))\cong Y\times \mathbb{P}^2$. This embeds into $\bar{Y}\times \mathbb{P}^2$, where $\bar Y$ is a (the) smooth projective compactification of $Y$.
3. We can take the closure of $\mathcal{E}_\Gamma$ to get a projective variety, and the quasiprojectivity of this family follows easily.
4. Note that the fibres of the closure $\overline{\mathcal{E}_\Gamma}$ may be singular.
5. If the Hopf surface had a section, it would lift to a rational curve in $\mathbb{C}^2-\lbrace 0\rbrace$. Well, I'll let you think about why that might be a problem.

The point is that by your construction $f:\mathcal{E}_\Gamma\to Y_\Gamma$ is an elliptic surface with a section $\sigma$. So as Keerthi said in the comments, $f$ is projective. In fact the relative divisor $3\sigma$ gives the Weirstrass embedding $\mathcal{E}_\Gamma\to \mathbb{P}(f_*\mathcal{O}(\sigma))$). (Ben's comment shows that projectivity can fail without a section.) From here, quasiprojectivity is straight forward: If $V$ is an extension of $f_*\mathcal{O}(\sigma))$ to the smooth projective closure  $ Y_\Gamma$, then the closure of $\mathcal{E}_\Gamma$ in $\mathbb{P}(V)$ is projective.

The point is that by your construction $f:\mathcal{E}_\Gamma\to Y_\Gamma$ is an elliptic surface with a section $\sigma$. So as Keerthi said in the comments, $f$ is projective. In fact the relative divisor $3\sigma$ gives the Weirstrass embedding $\mathcal{E}_\Gamma\to \mathbb{P}(f_*\mathcal{O}(3\sigma))$). (Ben's comment shows that projectivity can fail without a section.) From here, quasiprojectivity is straight forward: If $V$ is an extension of $f_*\mathcal{O}(3\sigma)$ to a vector bundle on the smooth projective closure of $ Y_\Gamma$, then the closure of $\mathcal{E}_\Gamma$ in $\mathbb{P}(V)$ is a projective variety. Therefore $\mathcal{E}_\Gamma$ is quasiprojective.