The Gysin sequence associated to (2.13) yields

(2.14)

$\displaystyle 0\rightarrow H^{1}(\mathbb{T}^{2})\xrightarrow{\pi^{*}}H^{1}(\Nil^{3}_{b})\rightarrow H^{0}(\mathbb{T}^{2})\xrightarrow{\cup e}H^{2}(\mathbb{T}^{2})\rightarrow\cdots$

Since the Euler class $e$ of the bundle is $b$ times a generator of $H^{2}(\mathbb{T}^{2})$, the mapping $\cup e:\mathbb{R}\cong H^{0}(\mathbb{T}^{2})\rightarrow H^{2}(\mathbb{T}^{2})\cong\mathbb{R}$ is just multiplication by $b$, so this mapping is an isomorphism. Consequently, $\pi^{*}:H^{1}(\mathbb{T}^{2})\rightarrow H^{1}(\Nil^{3}_{b})$ is also an isomorphism. Since $\Nil^{3}_{b}$ is a compact orientable $3$-manifold, Poincaré duality implies that $b_{1}=b_{2}$. ∎