I'm looking at the proof in Griffith Harris of the following fact:let $S$ be a compact algebraic complex surface such that its Betti numbers are equal to that of $\mathbb{P}^2$ and whose canonical bundle is not positive.
Then $S$ is biholomorphic to $\mathbb{P}^2$. At the beginning of the proof, they say that $H^1(S,\mathcal{O}_S)$ vanish(which depends on the cohomological condition) and then they say so $Pic(M) \cong H^2(S,\mathbb{Z}) \cong \mathbb{Z}$.
The fact I'm struggling to understand is the last isomorphism: from the cohomological conditions we just know that the rank of the $H^2(S;\mathbb{Z})$ is one, but it's looking like there obviously should be no torsion (which I do not see why).
It doesn't seem important for G-H's argument that $H^2(S,\mathbb{Z})$ is torsion-free, but here is a proof anyway:
By the universal coefficient theorem, there is a short exact sequence $$0 \rightarrow \text{Ext}^1(H_1(S,\mathbb{Z}), \mathbb{Z}) \rightarrow H^2(S,\mathbb{Z}) \rightarrow \text{Hom}(H_2(S,\mathbb{Z}),\mathbb{Z}) \rightarrow 0$$ To see that $H^2(S,\mathbb{Z})$ is torsion-free it is therefore sufficient to show that $H_1(S,\mathbb{Z})=0$; since $H_1(S,\mathbb{Z})$ is a quotient of $\pi_1(S)$, it suffices to show that $S$ has no nontrivial finite covering spaces $\tilde{S}$.
Such an $\tilde{S}$ would satisfy $$\chi(\mathcal{O}_{\tilde{S}}) = d\chi(\mathcal{O}_{S}) = d$$
where $d$ denotes the degree of $f:\tilde{S} \rightarrow S$. The equality $d=1$ then follows from the equalities $h^{0,1}=h^{0,2}=0$; in turn, these latter equalities follow from Kodaira vanishing if $-K_{\tilde{S}}$ is ample. But the argument of Lazzaro Campeotti in the comments shows that $-K_S$ is ample, and so $-K_{\tilde{S}}=-f^*K_S$ is also ample.