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The answer in no, because of the following result:

Theorem 1. Let $X$ be a non-ruled minimal surface. Then there exists a finite ramified covering $S \to X$ of degree $>1$, such that $S$ is minimal of general type with $K_S$ very ample, $\pi_1(S) \cong \pi_1(X)$ and $S$ is not birationally equivalent to $X$. We can moreover assume that $S$ has negative index, i.e. $K_S^2 - 8 \chi(\mathcal{O}_S) <0$.

So the fundamental group $\pi_1(X)$ alone does not determine the birational type of $X$, and in general not even its diffeomorphism type. However, when

When $X$ is the product of two curves, however, something more can be said, if provided that one also knows the topological Euler number. More precisely one proves the following

Theorem 2. Let $C_1$, $C_2$ be smooth curves of genus $g_1$, $g_2$, with $g_i \geq 2$, and let $X=C_1 \times C_2$. Then any surface $S$ such that $\pi_1(S) \cong \pi_1(X)$ and $e(S)=e(X)$ is isomorphic to a product of two curves of the same genera.

Theorems 1 and 2 were proven by F. Catanese in his paper Fibred surfaces, varieties isogenous to a product and related moduli spaces, which considers the more general situation $X=(C_1 \times C_2)/G$, where $G$ is a finite group acting freely on the product $C_1 \times C_2$.

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The answer in no, because of the following result:

Theorem 1. Let $X$ be a non-ruled minimal surface. Then there exists a finite ramified covering $S \to X$ of degree $>1$, such that $S$ is minimal of general type with $K_S$ very ample, $\pi_1(S) \cong \pi_1(X)$ and $S$ is not birationally equivalent to $X$. We can moreover assume that $S$ has negative index, i.e. $K_S^2 - 8 \chi(\mathcal{O}_S) <0$.

So the fundamental group alone does not determine the birational type. However, when $X$ is the product of two curves something more can be said, in fact the fundamental group and if one also knows the topological Euler numbertogether do actually determine the isomorphism type. More precisely one proves the following

Theorem 2. Let $C_1$, $C_2$ be smooth curves of genus $g_1$, $g_2$, with $g_i \geq 2$, and let $X=C_1 \times C_2$. Then any surface $S$ such that $\pi_1(S) \cong \pi_1(X)$ and $e(S)=e(X)$ is isomorphic to $X$. a product of two curves of the same genera.

Theorems 1 and 2 were proven by F. Catanese in his paper Fibred surfaces, varieties isogenous to a product and related moduli spaces, which actually considers the more general situation $X=(C_1 \times C_2)/G$, where $G$ is a finite group acting freely on the product $C_1 \times C_2$.

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The answer in no, because of the following result:

Theorem 1. Let $X$ be a non-ruled minimal surface. Then there exists a finite ramified covering $S \to X$ of degree $>1$, such that $S$ is minimal of general type with $K_S$ very ample, $\pi_1(S) \cong \pi_1(X)$ and $S$ is not birationally equivalent to $X$. We can moreover assume that $S$ has negative index, i.e. $K_S^2 - 8 \chi(\mathcal{O}_S) <0$.

So the fundamental group alone does not determine the birational type. However, for when $X$ is the product of two curves something more can be said, in fact the fundamental group and the topological Euler number together do actually determine the isomorphism type. More precisely one proves the following

Theorem 2. Let $C_1$, $C_2$ be smooth curves of genus $g_1$, $g_2$, with $g_i \geq 2$, and let $X=C_1 \times C_2$. Then any surface $S$ such that $\pi_1(S) \cong \pi_1(X)$ and $e(S)=e(X)$ is isomorphic to $X$.

Theorems 1 and 2 were proven by F. Catanese in his paper Fibred surfaces, varieties isogenous to a product and related moduli spaces, which actually considers the more general situation $X=(C_1 \times C_2)/G$, where $G$ is a finite group acting freely on the product $C_1 \times C_2$.

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