There are a lot of different definitions of a Calabi-Yau manifold. Roughly, we can divide them in two sets, see Wikipedia https://en.wikipedia.org/wiki/Calabi%E2%80%93Yau_manifold . I will refer to this link for notations and everything else.
Wikipedia says that when $M$ is simply-connected, these two definitions are equivalent. But if $M$ is only connected, then we have that the first one implies the second one but not the vice versa. I agree with both the statements.
My problem is the following. Suppose that $M$ is not simply-connected and the holonomy group $Hol(M)$ is contained in $SU(n)$ (here $n$ is the complex dimension of $M$). By the holomy principle there exists a parallel $n$-form, i.e. a parallel section of the canonical bundle $K_M=\Omega_M^n$. In particular this implies that $K_M$ is flat and since the curvature of $K_M$ is proportional to the Ricci tensor of $M$, we get that $M$ is Ricci-flat. According to Wikipedia, $M$ is Ricci-flat if and only if $c_1(M)=0$. So we are leading to say that $Hol(M)\subset SU(n)$ if and only if $c_1(M)=0$. But Wikipedia itself says that this double implication is not true when $M$ is not simply-connected. I am stuck in front of this absurd.
Can someone please point out where my argument fails and why?
Thank you in advance!
EDIT: I think to have solved my problem(s) now.
1) first of all I was wrong because Wikipedia talks about local holonomy and not restricted holonomy (for which the holonomy principle doesn't hold);
2) if $Hol(M)\subset SU(n)$ I conclude that $M$ is Ricci-flat, but the vice versa is not true in general because of the following: if $K_M$ is flat I claimed that there exists a parallel section; but what happens if the bundle has no sections? We cannot go the other direction. This is for example the case of an Enrique surface.
3) in the non simply-connected case, I can run my argument on the universal cover of $M$, which is simply-connected and hence everything works well. After that I can push-down the result and see what happens. In the case the holonomy is contained inside $SU(n)$ we can still state that the canonical bundle is trivial, but it can have non trivial curvature.