Let $M$ be a closed oriented real manifold with a free smooth circle action. Denote $BS^1$ to be the classifying space of principal circle bundles and $ES^1\rightarrow BS^1$ to be the universal principal circle bundle.


Now we consider the fibre product $M\_{S^1}:=M\times\_{S^1}ES^1$. When the fundamental group $\pi\_1(M)=0$, one can show that $$H^2(M\_{S^1}, \mathbb{Z})\cong H^2(M, \mathbb{Z})\oplus \mathbb{Z}.\ \ \ \ \ \ \ \ \ \ \ \ \ (\star)$$


My question is: Is it possible to loosen the condition "$\pi\_1(M)=0$"? For example, does $(\star)$ still hold true if we only assume $H^1(M, \mathbb{Z})=0$?



on second cohomology of $S^1$-manifold
Yes. This follows from the Leray-Serre spectral sequence of the fibre bundle
$$
M\to M\_{S^1} \to BS^1
$$
which has $E\_2^{p,q}=H^p(BS^1;H^q(M;\mathbb{Z}))$ and converges to (the associated graded of the filtration on) $H^\*(M\_{S^1};\mathbb{Z})$. Note that $BS^1\simeq \mathbb{C}P^\infty$ has cohomology concentrated in even degrees. So the only possible differentials into or out of the diagonal $p+q=2$ are $$d\_2:H^0(BS^1;H^2(M;\mathbb{Z}))\to H^2(BS^1;H^1(M;\mathbb{Z}))$$ and $$d\_2:H^0(BS^1;H^1(M;\mathbb{Z}))\to H^2(BS^1;H^0(M;\mathbb{Z})),$$ both of which must be zero if we assume that $H^1(M;\mathbb{Z})=0$.


