What is the cokernel of the map $H^2\big(\pi_1(X), \mathbb Z\big) \longrightarrow H^2(X,\mathbb Z).$

Question

For a manifold $X$ (for simplicity, assumed to be compact), let $\pi_1(X)$ be the fundamental group of $X$. What is the cokernel of the map $$H^2\big(\pi_1(X), \mathbb Z\big) \longrightarrow H^2(X,\mathbb Z).$$ The above map is as in Interpretation of the monomorphism $H^2(\pi_1(X),\mathbb{Z}) \rightarrow H^2(X,\mathbb{Z})$.

Answer 1

Let $\tilde{X} $ be the universal covering of $X$. There is a Hochschild-Serre spectral sequence with $E^{pq}_2=H^p(\pi _1(X), H^q(\tilde{X},\mathbb{Z} ))$ converging to $H^{*}(X,\mathbb{Z})$. Since $H^1(\tilde{X},\mathbb{Z} )=0$, the cokernel you are looking for is $E^{0,2}_{\infty}$, that is, the kernel of $d_3: H^2(\tilde{X},\mathbb{Z} )^{\pi _1(X)}\rightarrow H^3(\pi _1(X),\mathbb{Z})$. I am afraid there is no simpler expression.