Let $D\subset X$ be a smooth divisor in a smooth complex variety. On $D$ we have the normal bundle $N$. Removing the zero section and retracting we get an $S^1$ bundle. Call this bundle $N'$. Now I'd like to understand the first homology of $N'$. We get from the Serre spectral sequence $$H_i(D,H_j(S^1,\mathbb{C}))\Rightarrow H_{i+j}(N',\mathbb{C}).$$ On this $E_2$ page, we get a map $E^2_{2,0}=H_2(D,\mathbb{C})=\mathbb{C}\to E^2_{0,1} =H_0(D,\mathbb{C})$. I guess this map is taking the cap product with the Euler class of $N'$, i.e. the first Chern class of $N$, but I'm not sure about this. Is this correct?

It follows that if the Chern class is non-trivial, then $E^2_{0,1}=E^\infty_{0,1}=0$. Since clearly $E^2_{1,0}=E^\infty_{1,0}=H_1(D,\mathbb{C})$, we thus find that $$H_1(N',\mathbb{C})\cong H_1(D,\mathbb{C}).$$ Now I wonder if this isomorphism is realised by the natural map? I.e. is the map just the pushforward in homology induced by the projection?

I tried reading up on the construction of the Serre spectral sequence, but it doesn't really help me with understanding the map to be honest.