By Lemma 4.5, there is a unique $C^{1}$-map

$$\sigma_{k}:B_{h_{E}}(0,\frac{3r}{2})\longrightarrow C^{1,\alpha}(d\Omega^{0}(T^{n})\oplus d^{*_{h}}\Omega^{2}(T^{n})),\ \ {\rm by}\ \ y\mapsto\sigma_{k}(y),$$

which satisfies

$$\mathfrak{F}_{k}(y,\sigma_{k}(y))=0,\ \ \ \|\sigma_{k}(y)\|_{C^{1,\alpha}(L,h)}\leq\delta\ll 1,$$

$${\rm and}\ \ \ \|D\sigma_{k}(y)\|\leq 2n\delta\overline{C}C\|(d\alpha_{k},d\beta_{k})\|_{C^{1,\alpha}(Y_{2r},g)}.$$

This implies

$$|\frac{\partial\sigma_{k,j}(y)}{\partial y_{i}}|\leq 2n\delta\overline{C}C\|(d\alpha_{k},d\beta_{k})\|_{C^{1,\alpha}(Y_{2r},g)}\ll 1,$$

for $k\gg k_{1}>1$.

Define a map $\Psi_{k}:Y_{\frac{3}{2}r}\longrightarrow Y_{2r}$ by

$$\Psi_{k}:(x,y)\mapsto(x,y_{1}+\sigma_{k,1}(y),\cdots,y_{n}+\sigma_{k,n}(y))=(x,y+\sigma_{k}(y)).$$

Note that the frame field $dx_{1},\cdots,dx_{n}$ induces local coordinates $x_{1},\cdots,x_{n}$ around any point on $L$, and the differential can be expressed as

$$d\Psi_{k}:(\dot{x},\dot{y})\mapsto(\dot{x}_{j}+\sum\frac{\partial\sigma_{k,j}(y)}{\partial x_{i}}\dot{x}_{i},\dot{y}_{j}+\sum\frac{\partial\sigma_{k,j}(y)}{\partial y_{i}}\dot{y}_{i})$$

under such local coordinates. Thus $d\Psi_{k}$ is an isomorphism when $k\gg 1$, which implies that $\Psi_{k}$ is an immersion. Furthermore, for $y_{1}\neq y_{2}\in\mathbb{R}^{n}$,

	$\displaystyle\Psi_{k}(x,y_{2})-\Psi_{k}(x,y_{1})$	$\displaystyle=$	$\displaystyle(x,\cdots,\int_{0}^{1}(1+\frac{\partial\sigma_{k,j}((1-t)y_{2}+ty_{1})}{\partial y_{j}}dt)(y_{2,j}-y_{1,j}),\cdots)$
		$\displaystyle\neq$	$\displaystyle 0.$

Hence $\Psi_{k}$ is an embedding.

Note that the $\Gamma$-action on $Y_{2r}=T^{n}\times B_{h_{E}}(0,2r)$ preserves $\omega_{k},g_{k},\Omega_{k},\omega,g,\Omega$, and is a product action on $T^{n}\times B_{h_{E}}(0,2r)$, i.e. there are $\Gamma$-actions on $T^{n}$ and $B_{h_{E}}(0,2r)$ such that $\gamma\cdot(x,y)=(\gamma\cdot x,\gamma\cdot y)$ for any $\gamma\in\Gamma$, $x\in T^{n}$, and $y\in B_{h_{E}}(0,2r)$. Under the identification map (1),

	$\displaystyle(\gamma\cdot x,\gamma\cdot y)=(\gamma\cdot x,\iota(\gamma_{*}\sum_{j}y_{j}\frac{\partial}{\partial y_{j}})\omega)$	$\displaystyle=$	$\displaystyle(\gamma\cdot x,\gamma^{*}\omega(\sum_{j}y_{j}\frac{\partial}{\partial y_{j}},\gamma^{-1}_{*}\cdot))$
		$\displaystyle=$	$\displaystyle(\gamma\cdot x,\gamma^{-1,*}\sum_{j}y_{j}dx_{j}).$

Thus

	$\displaystyle\gamma\cdot L(y,\sigma_{k}(y))$	$\displaystyle=$	$\displaystyle\{(\gamma\cdot x,\gamma\cdot(y_{1}+\sigma_{k,1}(y)(x),\cdots,y_{n}+\sigma_{k,n}(y)(x)))|x\in T^{n}\}$
		$\displaystyle=$	$\displaystyle\{(\gamma\cdot x,\gamma^{-1,*}\sum_{j}(y_{j}+\sigma_{k,j}(y)(x))dx_{j}|x\in T^{n}\}$
		$\displaystyle=$	$\displaystyle L(\gamma\cdot y,\gamma^{-1,*}\sigma_{k}(y)),$

for any $\gamma\in\Gamma$ and $y\in B_{h_{E}}(0,2r)$. Since the $\Gamma$-action preserves $\omega_{k}$ and $\Omega_{k}$, $L(\gamma\cdot y,\gamma^{-1,*}\sigma_{k}(y))$ are special lagrangian submanifolds. By the uniqueness of $\sigma_{k}(y)$, $\gamma^{-1,*}\sigma_{k}(y)=\sigma_{k}(\gamma\cdot y)\in C^{1,\alpha}(d\Omega^{0}(T^{n})\oplus d^{*_{h}}\Omega^{2}(T^{n}))$. Hence

	$\displaystyle\Psi_{k}(\gamma\cdot x,\gamma\cdot y)$	$\displaystyle=$	$\displaystyle(\gamma\cdot x,\gamma\cdot y+\sigma_{k}(\gamma\cdot y))$
		$\displaystyle=$	$\displaystyle(\gamma\cdot x,\gamma^{-1,*}\sum_{j}(y_{j}+\sigma_{k,j}(y))dx_{j})$
		$\displaystyle=$	$\displaystyle(\gamma\cdot x,\gamma\cdot(y_{1}+\sigma_{k,1}(y),\cdots,y_{n}+\sigma_{k,n}(y)))$
		$\displaystyle=$	$\displaystyle\gamma\cdot\Psi_{k}(x,y),$

i.e. $\Psi_{k}$ is a $\Gamma$-equivariant map.

We denote $\mathcal{P}:Y_{2r}\longrightarrow B_{h_{E}}(0,2r)$ the natural projection, $B_{k}=B_{h_{E}}(0,\frac{3}{2}r)$ and $W_{k}=\Psi_{k}(Y_{\frac{3}{2}r})$. Since the $\Gamma$-action on $B_{h_{E}}(0,2r)$ preserves the metric $h_{E}$ and $0$, $B_{k}=B_{h_{E}}(0,\frac{3}{2}r)$ is invariant. By $\delta\ll 1\ll r$, $W_{k}\supset Y_{r}$. Then $f_{k}=\mathcal{P}\circ\Psi_{k}^{-1}:W_{k}\longrightarrow B_{k}$ is a $\Gamma$-equivariant special lagrangian fibration of $(W_{k},\omega_{k},\Omega_{k})$ of phase $\theta_{k}$. We obtain the conclusion. ∎