Let $X$ be a K3 surface, show that $H_1(X,\mathbb Z)=0$ Let $X$ be a(n algebraic) K3 surface, i.e., $X$ is a smooth algebraic surface with trivial canonical bundle and $H^1(X,\mathcal{O})=0$. This assumption directly implies that $H^1(X,\mathbb C)=0$, so $H_1(X,\mathbb C)=0$ by Poincaré duality. In particular $H_1(X,\mathbb Z)$ is a torsion group.
Next, to show that $H_1(X,\mathbb Z)$ is actually $0$, the book says otherwise there is a nontrivial torsion element in fundamental group which allows us to pass to a finite covering $p:\tilde{X}\to X$, but $\tilde{X}$ is still a K3 surface, and considering any K3 surface has Euler characteristic 24, so $p$ has to be an identity map, which is a contradiction.
Here is my question: First homology group is the Abelianzation of fundamental group $H_1=\pi_1/[\pi_1,\pi_1]$, but why a nontrivial torsion element in $H_1$ has a nontrivial torsion representative in $\pi_1$?
 A: First proof.
By GAGA principle, $X$ is a (connected compact) complex surface; using the Serre duality, one has
\begin{equation}
H^2(X,\mathcal{O}_X)\cong H^0(X,\omega_X)^{\vee}\cong\mathbb{C},
\end{equation}
and by a dimensional argument $H^3(X,\mathcal{O}_X)=0,\,H^4(X,\mathcal{O}_X)=0$.
Using the exponential sequence
\begin{equation}
0\to\underline{\mathbb{Z}}\to\mathcal{O}_X\to\mathcal{O}_X^{\times}\to0
\end{equation}
and passing to long exact sequence in cohomology, one has:
\begin{gather}
0\to H^0\left(X,\underline{\mathbb{Z}}\right)\cong\mathbb{Z}\to H^0(X,\mathcal{O}_X)\cong\mathbb{C}\xrightarrow{\exp}H^0\left(X,\mathcal{O}_X^{\times}\right)\cong\mathbb{C}^{\times}\to H^1\left(X,\underline{\mathbb{Z}}\right)\to H^1(X,\mathcal{O}_X)=0,\\
0\to H^1\left(X,\mathcal{O}_X^{\times}\right)\cong Pic(X)\to H^2\left(X,\underline{\mathbb{Z}}\right)\to H^2(X,\mathcal{O}_X)\cong\mathbb{C}\to H^2\left(X,\mathcal{O}_X^{\times}\right)\to H^3\left(X,\underline{\mathbb{Z}}\right)\to H^3(X,\mathcal{O}_X)=0,\\
0\to H^3\left(X,\mathcal{O}_X^{\times}\right)\to H^4\left(X,\underline{\mathbb{Z}}\right)\to H^4(X,\mathcal{O}_X)=0,\\
0\to H^4\left(X,\mathcal{O}_X^{\times}\right)\to0;
\end{gather}
so $H^3\left(X,\mathcal{O}_X^{\times}\right)\cong H^4\left(X,\underline{\mathbb{Z}}\right)$ and $H^4\left(X,\mathcal{O}_X^{\times}\right)=0$.
By exactness of
\begin{equation}
0\to\mathbb{Z}\to\mathbb{C}\xrightarrow{\exp}\mathbb{C}^{\times}\to0
\end{equation}
the previous sequence splits in
\begin{equation}
0\to H^1\left(X,\underline{\mathbb{Z}}\right)\to H^1(X,\mathcal{O}_X)=0
\end{equation}
so $H^1\left(X,\underline{\mathbb{Z}}\right)\cong H^1(X,\mathbb{Z})=0$.
By the Universal Coefficient Theorem for Cohomology, there exists a surjective morphism of Abelian groups
\begin{equation}
0=H^1(X,\mathbb{Z})\to\hom(H_1(X,\mathbb{Z}),\mathbb{Z})\to0,
\end{equation}
that is $H_1(X,\mathbb{Z})$ is either a non trivial torsion Abelian group or $0$.
Because $Pic(X)$ and $\mathbb{C}$ are torsion free groups (see remark 1.2.5), then also $H^2\left(X,\underline{\mathbb{Z}}\right)\cong H^2(X,\mathbb{Z})$ is a torsion free group; as consequence $H^{\dim_{\mathbb{R}}X-2+1}(X,\mathbb{Z})=H^3(X,\mathbb{Z})$ is a torsion free group; by Poincaré duality, $H_1(X,\mathbb{Z})$ is a torsion free group, and by previous reasoning $H_1(X,\mathbb{Z})=0$.
Remarks.

*

*Let $X$ be a compact, without boundary, oriented (real) manifold  of dimension $d$. The torsion subgroup of $H^p(X,\mathbb{Z})$ is isomorphic to torsion subgroup of $H^{d-p+1}(X,\mathbb{Z})$; where $p\in\{1,\dots,d\}$. (cfr. exercise 54)

*By this proof, one has the following (partial) result: $\pi_1(X)$ is a perfect group. For exact, $\pi_1(X)=0$; see the second proof.

*By Poincaré duality, $H^3\left(X,\mathcal{O}_X^{\times}\right)\cong H^4\left(X,\underline{\mathbb{Z}}\right)\cong H^4(X,\mathbb{Z})\cong H_0(X,\mathbb{Z})\cong\mathbb{Z}$ and $H^3\left(X,\underline{\mathbb{Z}}\right)\cong H^3(X,\mathbb{Z})\cong H_1(X,\mathbb{Z})=0$; so the non-trivial part of previous long exact sequence in cohomology is
\begin{equation}
0\to Pic(X)\to H^2\left(X,\underline{\mathbb{Z}}\right)\to\mathbb{C}\to H^2\left(X,\mathcal{O}_X^{\times}\right)\to0.
\end{equation}
Second proof.
Any K3 surface is diffeomorphic to a Fermat quartic surface (see example 1.1.3.i, remark 1.3.6.i and theorem 7.1.1), in particular they are simply connected (corollary 7.1.4); that is $\pi_1(X)=0$ and the relevant Abelianization $H_1(X,\mathbb{Z})$ is trivial.
References

*

*Davis J. F., Kirk P. (2001) Lectures Notes in Algebraic Topology, Graduate Studies in Mathematics 35 American Mathematical Society

*Huybrechts D. (2016) Lectures on K3-surfaces, Cambridge University Press

A: What book is this? The argument is incorrect. The existence of a nontrivial finite cover corresponds to the existence of a nontrivial subgroup of finite index, not the existence of a nontrivial torsion element. And this is guaranteed by the existence of a nontrivial finite quotient of $\pi_1$, which is in turn guaranteed by the existence of a nontrivial finite quotient of $H_1$.
