can I ask a question: what is the cokernel of the map $$H^2\big(\pi_1(X), \mathbb Z\big) \longrightarrow H^2(X,\mathbb Z).$$

@MaartenDerickx, for all unbranched covers.

@MaartenDerickx thanks, and you are right. should be commutative algebra.

@J.Martel Acturally, what you asked is very difficult to me. As far as I know, there is no systematic way to determine whether $C$ is of CM type or not.

@abx: $C$ has complex multiplication means the Jacobian $Jac(C)$ of $C$ has complex multiplication. The Jacobian $Jac(C)$ is an abelian variety. an abelian variety $A$ of dimension $g$ is said to have complex multiplication if the Endmorphism algebra $End_{\mathbb Q}(A)$ contains a field $K$ such that $[K:Q]=2g$ (See Mumford's <abelian variety>)

Thanks. I wonder how could you get the formula? $$\kappa(\phi_i)=\frac{cds_i}{s_i(s_i-1)}.$$

I have a question about your answer. I think $C_{g,[n]}$ is not the moduli space for curves with one marked point+level structure. Let me denote by $M_{g,[n],1}$ the later space. The reason is the following. If a curve $C$ has a non-trivial automorphism group $G$. Then $[C,p,\alpha]$ and $[C,\tau(p),\alpha]$ represent the same point of $M_{g,[n],1}$ for any $\tau \in G$. This implies that the fibre of $M_{g,[n],1} \to M_{g,[n]}$ over $[C,\alpha]\in M_{g,[n]}$ is isomorphic to $C/G$. But the universe family $C_{g,[n]} \to M_{g,[n]}$ requires that the fibre over $[C,\alpha]$ isomorphicto C