Let $B$ be a unital Banach algebra and $E$ a Banach right $B$-module. In this paper in lemma 2.2. (p. 6), Dadarlat proves that if $e_0,f_0,e_1,f_1\in L_B(E)$ are idempotents such that $e_i-f_i\in K_B(E)$, and if $\|e_0-e_1\|,\|f_0-f_1\|$ are small enough (depending on $\|e_i\|,\|f_i\|$) then there are homotopies $e_t,f_t$ through idempotents such that $e_t-f_t\in K_B(E)$ for all $t$.
As far as I can understand it, the argument goes so: Take linear homotopies $x_t,y_t$. They will not be point-wise idempotents, but $x_t^2-x_t$ and $y_t^2-y_t$ are small enough so that the spectra of $x_t$ and $y_t$ are both contained in $\Omega=\mathbb C-(\frac 12+i\mathbb R)$. Let $\chi\colon\Omega\to\{0,1\}\subset\mathbb C$ be the locally constant (hence holomorphic) function which maps the component of $0$ onto $0$ and the component of $1$ onto $1$. Then $\chi(x_t),\chi(y_t)$ are defined via holomorphic functional calculus, and $\chi(x_t)-\chi(y_t)$ is compact by naturality of the holomorphic functional calculus.
Now my question is: Why is this a homotopy between the $e_i$ or the $f_i$? I would think that this follows again from the naturality of the functional calculus, applied to the evaluation maps at time $0$ and $1$, but then one would need that $\chi(p)=p$ whenever $p$ is actually an idempotent. Of course, the Spectral Mapping Theorem gives that the spectrum of $\chi(p)-p=(\chi-\mathrm{id})(p)$ contains only $0$. But I cannot see that actually $\chi(p)-p=0$.
Best Answer
Let $e$ be an idempotent, and for $j=0,1$, let $\gamma_j$ be the positively oriented circle centered at $j$ with radius $\frac14$, and let $\Gamma$ be the union of these contours. For $z\in\mathbb C\setminus\{0,1\}$, we have $$(z-e)^{-1}=(z-1)^{-1}e+z^{-1}(1-e),$$ so that \begin{align*} \chi(e)&=\frac{1}{2\pi i}\int_\Gamma\chi(z)(z-e)^{-1}\ dz\\ &=\frac{1}{2\pi i}\int_{\gamma_0}\chi(z)(z-e)^{-1}\ dz+\frac{1}{2\pi i}\int_{\gamma_1}\chi(z)(z-e)^{-1}\ dz\\ &=0+\operatorname{Ind}_{\gamma_1}(1)e+\operatorname{Ind}_{\gamma_1}(0)(1-e)\\ &=e. \end{align*}
That this is a homotopy follows from the fact that $\cup_{t\in[0,1]}\sigma(x_t)$ lies in a compact subset of $\mathbb C\setminus\{\frac12+it:t\in\mathbb R\}$ (and similarly for the $y_t$). So you can take a single contour to define $\chi(x_t)$ (resp. $\chi(y_t)$) for all $t$. Then basic norm estimates show that $t\mapsto e_t$ (resp. $t\mapsto f_t$) is continuous.