Let $(b_{1},b_{2},b_{3})$ be the coordinates on $B$ and $\Delta\subseteq B$ as in Proposition 4.10. To avoid cumbersome notation let us assume $B=\mathbb{R}\times\mathbb{R}^{2}$. We may identify $\mathbb{R}^{2}$ with $\{0\}\times\mathbb{R}^{2}$. Then $\Delta\subset\mathbb{R}^{2}$. Let $\lambda_{1},\lambda_{2},\lambda_{3}$ be the periods of $\mathcal{F}$ as in (19). We want to show that the affine structure on $B-\Delta$ induced by $\mathcal{F}$ is isomorphic to the one given in Examples 3.10 or 3.11. To do this we will consider the locally defined map $A=(A_{1},A_{2},A_{3})$, where each $A_{j}$ is a suitable branch of a primitive of $\lambda_{j}$ such that $A_{j}(0)=0$. First we will show that –perhaps after replacing $B$ by a smaller neighborhood of $0$– the map $A$ extends to a homeomorphism $A:B\rightarrow A(B)\subseteq\mathbb{R}^{3}$. Let

	$\displaystyle R$	$\displaystyle=$	$\displaystyle\mathbb{R}\times\Delta$
	$\displaystyle R^{+}$	$\displaystyle=$	$\displaystyle\mathbb{R}_{\geq 0}\times\Delta$
	$\displaystyle R^{-}$	$\displaystyle=$	$\displaystyle\mathbb{R}_{\leq 0}\times\Delta$

and take the open cover $\{U_{1},U_{2}\}$ of $B-\Delta$ where

	$\displaystyle U_{1}$	$\displaystyle=$	$\displaystyle B-R^{+},$
	$\displaystyle U_{2}$	$\displaystyle=$	$\displaystyle B-R^{-}.$		(20)

On $U_{1}$ we can choose an affine coordinates map given by

$$A(b_{1},b_{2},b_{3})=(\psi_{1}(b_{1},b_{2},b_{3}),2\pi b_{2},2\pi b_{3}),$$

where $\psi_{1}$ is a primitive of $\lambda_{1}$. Clearly $A(R^{-})\subset R$. We now show that $A$ extends continuously to $B$. The key observation is that the symplectic form $\omega$ is exact in a neighborhood of the singular fibre over the vertex of $\Delta$. This is straightforward in the case of Example 4.4, where $\omega$ is the standard symplectic form on $\mathbb{C}^{3}$ but it is also true in general. So assume $\omega=d\eta$ for some 1-form $\eta$. Now let us fix a basis $e=(e_{1},e_{2},e_{3})$ of $H^{1}(f^{-1}(U_{1}),\mathbb{Z})$, corresponding to the periods $\lambda_{1},\lambda_{2}$ and $\lambda_{3}$ respectively. Recall that action coordinates can be computed by

$$A(b)=\left(-\int_{e_{1}(b)}\eta,\ -\int_{e_{2}(b)}\eta,\ -\int_{e_{3}(b)}\eta\right),$$

where $e_{j}(b)$ is a $1$-cycle, contained in $f^{-1}(b)$, representing $e_{j}$. We prove first that $A$, as a map, extends continuously to $B-\Delta$. Notice that $e_{2}$ and $e_{3}$ are monodromy invariant, so we may assume that $e_{2}(b)$ and $e_{3}(b)$ are well defined for all $b\in B-\Delta$ and that

$$-\int_{e_{j}(b)}\eta=2\pi b_{j},$$

(21)

for $j=2,3$. In particular, $A_{2}$ and $A_{3}$ are defined on $B$. Let us study

$$\psi_{1}(b)=-\int_{e_{1}(b)}\eta.$$

Suppose that $\psi_{1}(\bar{b})=0$ for a fixed point $\bar{b}\in U_{1}$. Given another point $b\in U_{1}$ let $\Gamma:[0,1]\rightarrow U_{1}$ be a path such that $\Gamma(0)=\bar{b}$ and $\Gamma(1)=b$. Consider the cylinder $S$ inside $f^{-1}(U_{1})$ spanned by the cycles $e_{1}(\Gamma(t))$. Then one can see that

$$\psi_{1}(b)=\int_{S}\omega.$$

(22)

We may use (22) to define $\psi_{1}(b)$ for $b\in R^{+}-\Delta$. Since $B-\Delta$ is not simply connected, this expression of $\psi_{1}$ is well defined provided that it is independent of the chosen path $\Gamma$. Suppose that $\Gamma_{1}$ and $\Gamma_{2}$ are two different paths from $\bar{b}$ to $b$ such that $\Gamma_{1}-\Gamma_{2}$ is not homotopically trivial in $B-\Delta$, then we have to show that if $S_{1}$ and $S_{2}$ are the corresponding cylinders, then

$$\int_{S_{1}-S_{2}}\omega=0.$$

Denote by $e_{1}^{+}(b)$ and $e_{1}^{-}(b)$ those boundary components of $S_{1}$ and $S_{2}$ respectively, which lie on top of $b$ (the endpoint of both $\Gamma_{1}$ and $\Gamma_{2}$). Then

$$\partial(S_{1}-S_{2})=e_{1}^{+}(b)-e_{1}^{-}(b),$$

and

$$\int_{S_{1}-S_{2}}\omega=\int_{e_{1}^{+}(b)-e_{1}^{-}(b)}\eta.$$

Because of monodromy, $e_{1}^{+}(b)$ and $e_{1}^{-}(b)$ may not coincide and it is not obvious that the above integral vanishes. Nevertheless, we know that $b\in R^{+}$ and there are three cases: if $b=(b_{1},b_{2},b_{3})$ then either $b_{2}=0$, $b_{3}=0$ or $b_{2}=b_{3}$. Let us look at that the latter case. With respect to the basis $e=(e_{1},e_{2},e_{3})$ as above, the monodromy matrices $T_{1}$, $T_{2}$ and $T_{3}$ corresponding respectively to generators $g_{1}$, $g_{2}$ and $g_{3}$ of $\pi_{1}(B-\Delta)$ as depicted in Figure 3 are those given in Proposition 4.10.

Let $\bar{b}$, $b$, $\Gamma_{1}$ and $\Gamma_{2}$ be given as in Figure 8, then one can see that $\Gamma_{1}-\Gamma_{2}=g_{1}^{-1}g_{2}^{-1}$. This implies that

$$e_{1}^{+}(b)=e_{1}^{-}(b)-e_{2}(b)+e_{3}(b)$$

and therefore that

$$\int_{e_{1}^{+}(b)-e_{1}^{-}(b)}\eta=\int_{-e_{2}(b)+e_{3}(b)}\eta=2\pi(b_{2}-b_{3})=0,$$

where in the second equality we have used (21). Similarly one treats the cases $b_{2}=0$ or $b_{3}=0$ using monodromy matrices $T_{1}$ and $T_{2}$ respectively. This shows that $\psi_{1}$ extends continuously to $B-\Delta$. It can be easily seen that it also extends continuously to points in $\Delta$. In fact one can use (22) as a definition of $\psi_{1}(b)$ when $b\in\Delta$. This makes sense since the cycles $e_{1}(b)$ spanning $S$ can be extended as cycles on singular fibres when $b\in\Delta$, e.g. when $b=0$, $e_{1}(0)$ is a homologically non trivial closed curve passing through the singularity of $f^{-1}(0)$, in particular $e_{1}(0)$ is the generator of $H_{1}(f^{-1}(0),\mathbb{Z})=\mathbb{Z}$.

We argue that $A$ is injective onto its image, at least when restricted to a smaller neighborhood of $b=0$. This would imply that $A$ is a homeomorphism. Clearly, $A$ is injective if and only if for fixed values of $b_{2}$ and $b_{3}$, the function $\psi_{1}(\,\cdot\,,b_{2},b_{3})$ is injective in a neighborhood of $b=0$. Since $d\psi_{1}=\lambda_{1}$, this holds if the coefficient of $db_{1}$ in $\lambda_{1}$ is never zero in a neighborhood of $b=0$. In fact, it was shown in §4 of [1] that this coefficient blows up to infinity as $b\rightarrow 0$, in particular it never vanishes.

One can easily check that $A$ defines an isomorphism between the affine structure with singularities induced on $B$ by the fibration $\mathcal{F}$ and the one described in Example 3.11, where $\tau:\Delta\rightarrow\mathbb{R}$ is given by $\tau=\psi_{1}|_{\Delta}$. We only need to verify that $\tau$ is smooth. In fact, it turns out that $\tau=H|_{\Delta}$ where $H$ is the smooth function in (19); this follows from the computation of $\lambda_{0}$ given in [1]§4. Consider the fibration $F:\mathbb{C}^{3}\rightarrow\mathbb{R}^{3}$ of Example 4.3. This is the local model for the singularity of a positive fibration. Consider two sections $\sigma_{-}$ and $\sigma_{+}$ of $F$, disjoint from $\Crit(F)$ and such that for every $b\in\Delta$, $\sigma_{-}(b)$ and $\sigma_{+}(b)$ lie on distinct connected components of the smooth part of the fibre over $b$. For every $b\in\mathbb{R}^{3}$ consider a curve $\gamma(b)$ contained $F^{-1}(b)$ joining $\sigma_{-}(b)$ to $\sigma_{+}(b)$ and define the function

$$a_{0}(b)=-\int_{\gamma(b)}\eta.$$

Then $\lambda_{0}=da_{0}$. Clearly $a_{0}$ can be continuously defined on $\mathbb{R}^{3}$. Using the fact that $F$ satisfies $F(-z_{1},z_{2},z_{3})=(-b_{1},b_{2},b_{3})$, where $F(z_{1},z_{2},z_{3})=(b_{1},b_{2},b_{3})$, one can show that $a_{0}$ satisfies $a_{0}(-b_{1},b_{2},b_{3})=-a_{0}(b_{1},b_{2},b_{3})$ and therefore that $a_{0}|_{\Delta}=0$. This proves that $\tau=H|_{\Delta}$. ∎