**Lemma**. Suppose $A$ is a unital Banach algebra and $\gamma$ is a continuous path in the idempotents of $A$. Then there is a continuous path $u$ in invertible elements of $A$ such that $u(t)\gamma(t) u^{-1}(t)=\gamma(0)$.

*Proof*. For any $s\in [0,1]$, let $$u_s(t)=\gamma(s)\gamma(t)+(1-\gamma(s))(1-\gamma(t)),$$ then $u_s(t)\gamma(t)=\gamma(s)u_s(t)$ and $\|u_s(t)-1\|=\|(2\gamma(s)-1)(\gamma(t)-\gamma(s))\|\leq \|2\gamma(s)-1\|\|\gamma(s)-\gamma(t)\|.$
Let $r_s>0$ be such that $\|\gamma(s)-\gamma(t)\|<\frac{1}{\|2\gamma(s)-1\|}$ for all $t\in (s-r_s, s+r_s)$, then $u_s$ is invertible on the open set $(s-r_s, s+r_s)$. Since $[0,1]$ is compact, we can find finite points $s_1< s_2<\cdots<s_n$ such that $\{(s_i-r_{s_i}, s_i+r_{s_i}): i=1,2,\cdots,n\}$ is a cover of $[0,1]$, $(s_i-r_{s_i}, s_i+r_{s_i}) \cap(s_{i+1}-r_{s_{i+1}}, s_{i+1}+r_{s_{i+1}})\neq\emptyset$, $0\in (s_1-r_1, s_1+r_1)$ and $1\in (s_n-r_{n}, s_n+r_{s_n})$.

Pick $s^i\in (s_i-r_{s_i}, s_i+r_{s_i})\cap (s_{i+1}-r_{s_{i+1}}, s_{i+1}+r_{s_{i+1}})$ such that $s_i<s^i<s_{i+1}$, $i=1,\cdots,n-1$ and s^0=0, s^n=1. For each $i\in \{1,2,\cdots, n\}$, define $$v_i(t)=\begin{cases}u_{s_i}(s^{i-1})^{-1}u_{s_i}(t), t\in [s^{i-1}, s^i]\\ 1, t\in [0,s^{i-1}]\\ u_{s_i}(s^{i-1})^{-1}u_{s_i}(s^i), t\in [s^i,1]\end{cases}$$ and let $u=v_1v_2\cdots v_n$, then $u$ is a continuous path in invertible elements of $A$ and $u(t) \gamma(t) u^{-1}(t)=\gamma(0)$.