Why is a Kummer surface simply-connected? By a kummer surface, I mean a quotient of two-dimensional complex torus by multiplication by $-1$. It is a K3 surface and known to be simply connected.
But it is not clear to me why it is simply-connected from its construction, which is a quotient of a surface with infinite fundamental group.
Is there a direct explanation or any references? 
Thanks.   
 A: I haven't checked details but the following might work. Coverings of the Kummer surface will lift to coverings of the abelian surface (whose fundamental group is clearly $\mathbb{Z}^4$), so the fundamental group of the Kummer surface is abelian. Now it's enough to show that the first homology group is trivial. The latter I expect you will find in textbooks, e.g. as a computation of the Hodge diamond of a K3 surface.
A: In the paper "The homology of Kummer manifolds" E. Spanier proves among other things that the fundental group of the resolution $K$ of singularities of the quotient
$$(S^1\times S^1)^{\times n}/(\mathbb{Z}/2)$$
is simply connected (theorem 1). The proof is elementary and can be easily adapted to the singular quotient itself.
In your case $n=2$, $K$ is a K3-surface and the quotient is the Kummer surface.
Paper available here:
http://www.ams.org/journals/proc/1956-007-01/S0002-9939-1956-0087188-3/S0002-9939-1956-0087188-3.pdf
Edit : let me give some more details, let us consider
$$\mathcal{K}=(S^1\times S^1 \times S^1\times S^1)/(\mathbb{Z}/2)$$
it has 16 singular points $\{s_1,\dots,s_{16}\}$, the link of each singular point is homeomorphic to $\mathbb{R}P^3=S^3/(\mathbb{Z}/2)$. Let $$\mathcal{K}_{r}=\mathcal{K}-\bigcup_{i=1}^{16} U_i$$
where $U_i$ is a "nice" neighbourhood of the singular point $s_i$ i.e.
$$U_i\cong c \mathbb{R}P^3$$
it is homeomorphic to the open cone of the link $\mathbb{R}P^3$. 
Now you notice that $\mathcal{K}_r$ is a manifold with boundary, it has $16$ boundary components and the main point is to prove that the morphism
$$\mathcal{I}:\coprod_{i=1}^{16}\pi_1(\mathbb{R}P^3)\rightarrow \pi_1(\mathcal{K}_r)$$ induced by the inclusions of each boundary components into $\mathcal{K}_r$ is surjective. 
Using the fact that each neighbourhood $U_i$ is contractible you get that 
$$\pi_1(\mathcal{K})\cong\pi_1 (\mathcal{K}_r)/Im(\mathcal{I})=\{e\}.$$
