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.



$S^1$ normal bundle on divisor and Serre spectral sequence
The answer to both your questions is yes - this differential is cap with the Euler class, and the isomorphism is induced by the projection. The first is in any good reference on the homological Gysin sequence, e.g. Spanier's "Algebraic Topology" text, Chapter 5, Section 7 (Theorem 11 and Formulas 15). 


This already gives that the projection is an isomorphism on $H\_1$ when the Euler class is non-trivial. That the projection agrees with the edge homomorphism in the Leray spectral sequence is shown in many books which cover spectral sequences, e.g. Switzer's "Algebraic Topology: Homotopy and Homology", 15.29. 


