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Let us prove that any non-simply connected surface without curves of negative self-intersection provides a counterexample

This answer is completely rewritten. This includes Abelian surfaces and generic products of curves of higher genusis not an actual answer but a thought related to the question.

ProofI decided to leave it hear since it is short. Suppose

Note first that $X$ if there is a regular map from a surface with isolated singularities in $X$ to $\mathbb CP^3$. Then the sinuglarities of the surfaces are normal. Hence the minimal resolution of these P^3$whose image has only isolated singularitiesmust have , then$X$has curves with negative self-intersections (that project to singularity)self-intersection. So In particular, if the resolution does not have$X$has no such curves, then its image in$X$\mathbb P^3$ is non-singularsmooth.It follows that

Now, suppose we have a surface $X$ has trivial fundamental group.

Correction. I see now that with isolated singularities in order for this reasoning to work one has to guarantee that on $\mathbb CP^3$,say of general type and consider the question:

Question. Let $X'$ be the minimal resolution of singularities of on $X$ there are no rational X$.Can we say something about$-1$-curves. I wonder X$ if this holds indeed provided the degree of $X$ is high enough. Or maybe one can "classify" surfaces in X'$contains rational$\mathbb CP^3$for which this statement fails. So I still believe this idea can work out...-1$ curves?

Proof. Suppose that $X$ is a surface with isolated singularities in $\mathbb CP^3$. Then the sinuglarities of the surfaces are normal. Hence the minimal resolution of these singularities must have curves with negative self-intersections (that project to singularity). So if the resolution does not have such curves, $X$ is non-singular. It follows that $X$ has trivial fundamental group.
Correction. I see now that in order for this reasoning to work one has to guarantee that on the minimal resolution of singularities of $X$ there are no rational $-1$-curves. I wonder if this holds indeed provided the degree of $X$ is high enough. Or maybe one can "classify" surfaces in $\mathbb CP^3$ for which this statement fails. So I still believe this idea can work out...
Proof. Suppose that $X$ is a surface with isolated singularities in $\mathbb CP^3$. Then the sinuglarities of the surfaces are normal. Hence the minimal resolution of these singularities must have curves with negative self-intersections (that project to singularity). So if the resolution does not have such curves, $X$ is non-singular. It follows that $X$ has trivial fundamental group.