I'm studying Michele Audin's book - Torus Actions on Symplectic Manifolds and stumbled across an exercise I can't prove.
Exercise I.13 Prove that the Euler class of the Seifert manifold with invariants $(g|\beta, (m_1 \beta_1), \ldots, (m_r,\beta_r))$ is
$$e = \beta - \sum_{i=1}^r \frac{\beta_i}{m_i}. \tag{1}$$
Attempt: In her construction of the Classification of oriented Seifert manifolds, she considers, in the case of existence of exceptional orbits with Seifert invariants $(m_1, \beta_1) ,\ldots, (m_r, \beta_r)$ and stabilizers $\mathbb Z/m_i$, of the principal $S^1$-bundle, $W \to B$, a section
$$\sigma : \overline{B - \cup \mathcal D_i} \to \overline{W - \cup \mathcal V_i} $$ where $\mathcal D_i$ are discs around $r$ points corresponding to the exceptional orbits cut from the quotient surface $B$, also removes $\mathcal D_0$ around a point corresponding to any principal orbit and $\mathcal V_i$ are the tubular neighborhoods in $W$ all of which are solid torus. On each neighborhood, we get a meridian $a_i$ which is closed curve as the boundary of a disc section of the torus, and an $S^1$-section $b_i$ on $\partial \mathcal V_i$, which can be written as
$$a_i = m_i\partial_i \sigma_\varphi + (\beta_i - d_i m_i)b_i $$
where $\varphi : B - \cup \mathcal D_i \to S^1$ is a modifying map used to change $\sigma$ by multiplication
$$\sigma_\varphi (x) = \varphi (x) \cdot \sigma (x)$$ and each $d_i$ is the degree of $\varphi|_{\mathcal D_i}$. She defines the Euler class of the bundle as the number of times the curve $\partial \sigma_\varphi$ winds around $b_i$ with oppositely oriented choice, which in this case it should be (I guess) $\frac{\beta_i}{m_i} - d_i$ for each connected component of $\partial \sigma$ summing this up from $1$ to $r$ gives us
$$\sum_{i=1}^r \frac{\beta_i}{m_i} + d_0$$
where $d_0 = - (d_1 + \ldots + d_r)$ from Lemma I.3.6. I don't know how to get to the conclusion in (1).
Could someone, please, help?
I'm not sure if I'm correct but I think there is something wrong in the book. On the page 29 I think we should have $a_0 = \partial_0 \sigma - \beta b_0$, because in the case where $W$ is principal we should have $\beta = e$, where we consider $\partial \sigma$ with the opposite orientation of the boundary orientation.
It is proved in the book that $$a_i = m_i\partial_i \sigma + \beta_i b_i,\quad \textrm{for}\quad i=1,\ldots,r,$$ for some section $\sigma$ and for this section we must have $a_0 = \partial_i \sigma - \beta b_0$.
Therefore we have in $H_1(W,\mathbb Z)$ the following identities: $$[\partial_i \sigma] = -\frac{\beta_i}{m_i}[b_i]\quad \textrm{for}\quad i=1,\ldots,r,$$ $$[\partial_0 \sigma] = \beta[b_0],$$ because $a_i= \partial V_i$ and $V_i$ is a disk. Notice that $[b_0]= \cdots = [b_r]$ and lets define $b=b_0$.
On the other hand, $[\partial \sigma] = e [b]$.
Since $[\partial \sigma] = \sum_{i=0}^r[\partial_i \sigma],$ we have $$[\partial \sigma] = \left(\beta - \sum_{i=1}^r \frac{\beta_i}{m_i}\right)[b]$$ and, therefore, $$ e=\beta - \sum_{i=1}^r \frac{\beta_i}{m_i}. $$