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. 



Why is a Kummer surface simply-connected?
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 
$$\pi\_1(\mathcal{K})\cong\pi\_1 (\mathcal{K}\_r)/Im(\mathcal{I})=\{e\}.$$


