Since the problem is local, we may assume for all $r\leq 1$ that the above map $\phi_{r}$ exists and $\epsilon(1)$ is as small as we like. For simplicity we denote $\phi_{k}=\phi_{10^{-k}}$, and $\epsilon_{k}=\epsilon(10^{-k})$. Now we first define $F_{0}(x)=\phi_{0}(x)$ on $\hat{A}(1,100)$. Inductively suppose $F_{k}$ is defined on $\hat{A}(10^{-k},10^{-k+2})$ satisfying $F_{k}(x)=\phi_{k}\circ R_{k}^{-1}(10^{k}x)$ on $\hat{A}(10^{-k},20\cdot 10^{-k})$ for some rotation $R_{k}\in O(4)$, then we apply Lemma 5.9 to the two maps $\phi_{k}\circ R_{k}^{-1}$ and $\phi_{k+1}$ with $r=10^{-k}$ and $\epsilon=\max(\epsilon_{k-1},\epsilon_{k})$, and obtain a map $f_{k+1}$ defined on $\hat{A}(1/10,100)$ satisfying (2). Then we define $F_{k+1}(x)$ to be $f_{k+1}(10^{k}x)$ on $\hat{A}(10^{-k-1},10^{-k+1})$. By Lemma 5.9 we see that all the $F_{k}$’s match together to a map $F$ from $\hat{B}^{*}$ to $B(q,200)$, and we can modify $F$ slightly near $\partial\hat{B}$ so that the image is exactly $B^{*}$. It is easy to see that $|F^{*}g-g_{0}|_{L^{\infty}(\hat{A}(10^{-k},10^{-k+1}))}=|10^{2k}f_{k+1}^{*}g-g_{0}|_{L^{\infty}(\hat{A}(10,100))}\leq G(\max(\epsilon_{k-1},\epsilon_{k}))$, and $F^{*}g$ extends to a continuous metric tensor over $B$. ∎