Homotopy between idempotents of small difference Let $A$ be a unital $C^*$-algebra. It is known that if $p$ and $q$ are projections in $A$ with
$$\|p-q\|<1,$$
then $p$ and $q$ are homotopic through a path of projections.
Question: Does a similar statement hold for idempotents? More precisely, if $e$ and $f$ are idempotents in $A$, does there exist $\delta>0$ such that $e$ and $f$ are homotopic through idempotents whenever
$$\|e-f\|<\delta?$$
 A: Supposing that
$$
\Vert f-e\Vert < {1\over \Vert e \Vert +\Vert f \Vert},  \tag{*}
$$
let $u=ef+(1-e)(1-f)$.  Then
$$
  \Vert u-1\Vert =
  \Vert ef-e-f+ef\Vert=
  \Vert e(f-e) - (f-e)f \Vert \leq $$ $$ \leq
  \Vert e \Vert\Vert f-e\Vert + \Vert f-e \Vert\Vert f \Vert =
  (\Vert e \Vert +\Vert f \Vert)(\Vert f-e\Vert) <1.
  $$
This implies that $u$ is invertible.  Moreover we have that
$eu=uf$, whence $e=ufu^{-1}$.
Notice that the fact that $\Vert u-1\Vert <1$ implies not only that $u$ is invertible, but also that the power series defining the logarithm converges at $u$, so that $u=e^h$, for some $h$ in $A$. We then obtain a path of idempotents joining $f$ and $e$ by
$$u_t=e^{th}fe^{-th}.$$

Remarks:

*

*This does not provide a universal $\delta$, as required in the OP, but at least shows that being homotopically equivalent is invariant under small perturbations.


*My impression is that there is no universal $\delta$ and I'll report back should I be able to prove it.


*An elementary estimate shows that if $\Vert f-e\Vert <(4\Vert e\Vert )^{-1}$, then condition (*) above holds, so we deduce that $e$ is
homotopically equivalent to every idempotent element in a neighborhood of $e$.
A: By Lemma (11.2.7) in (Rørdam, M.; Larsen, F.; Laustsen, N., An introduction to (K)-theory for (C^*)-algebras, London Mathematical Society Student Texts. 49. Cambridge: Cambridge University Press. xii, 242 p. (2000). ZBL0967.19001.), for every idempotent $e$ in $A$, one has that
$$
  \rho (e):= ee^*(1 + (e - e^*)(e^*-e))^{-1}
  $$
is a projection (self-adjoint idempotent) and $e\sim_h\rho (e)$ (meaning that $e$ and $\rho (e)$ are  homotopic through a path
of idempotents).
Thus, given $e$ and $f$ satisfying $\Vert e-f\Vert <\delta $ (the precise value of $\delta $ to be filled in later), we have that
$e\sim_h\rho (e)$ and   $f\sim_h\rho (f)$, so if we can manage to prove that   $\rho (e)\sim_h\rho (f)$, we will get, by transitivity,
that $e\sim_h f$.
As noted in the OP, since $\rho (e)$ and $\rho (f)$ are projections, it would be enough to  prove that $\Vert \rho (e)-\rho (f)\Vert <1$.
It is not difficult to see that the range of an idempotent element $e$ coincides with the range of $\rho (e)$, so $\rho (e)$ is in fact the
orthogonal projection onto the range of $e$.
Given idempotents $e$ and $f$, let us henceforth write $E$ and $F$ for the ranges of $e$ and $f$, respectively, and by
$p$ and $q$ the orthogonal projections onto $E$ and $F$, which amounts to saying that $p=\rho (e)$ and $q=\rho (f)$.
Define
$$
  \alpha (E, F) = \sup\{\text{dist}(x,F): x\in E,\  \Vert x\Vert \leq 1\},
  $$
$$
  \beta (E, F) = \sup\{\text{dist}(x,E): x\in F,\  \Vert x\Vert \leq 1\}.
  $$
and finally put
$$
  d(E, F) = \max\{\alpha (E, F),\beta (E, F)\}.
  $$
Lemma 1. We have
$$
  d(E,F)\leq \Vert e-f\Vert .
  $$
If moreover  $e$ and $f$ are self-adjoint, then
$$
  \Vert e-f\Vert \leq 2d(E,F).
  $$
Proof.  For $x$ in $E$ with $\Vert x\Vert \leq 1$, we have
$$
  \text{dist}(x,F) \leq  \Vert x-f(x)\Vert  = \Vert e(x)-f(x)\Vert  \leq  \Vert e-f\Vert ,
  $$
so $\alpha (E, F)\leq \Vert e-f\Vert $, and it can be likewise proved that    $\beta (E, F)\leq \Vert e-f\Vert $, whence  $d(E, F)\leq \Vert e-f\Vert $.
Now assume that $e$ and $f$ are self-adjoint, so in particular  $\Vert e\Vert  \leq 1$ and  $\Vert f\Vert  \leq 1$.
For every $x$ in $H$ with $\Vert x\Vert \leq 1$, we have that   $e(x)\in E$ and $\Vert e(x)\Vert \leq 1$.  Moreover, the element in $F$ closest to $e(x)$
is $f(e(x))$, so
$$
  \Vert e(x)-f(e(x))\Vert  =  \text{dist}(e(x), F) \leq  \alpha (E,F)\leq d(E,F).
  $$
Taking the supremum for all $x$ in $H$ with $\Vert x\Vert \leq 1$, we  deduce that
$$
  \Vert e-fe\Vert \leq d(E,F),
  $$
and a symmetric  reasoning gives
$\Vert f-ef\Vert \leq d(E,F)$, so also
$$
  \Vert f-fe\Vert  =   \Vert (f-ef)^*\Vert  =   \Vert f-ef\Vert \leq d(E,F).
  $$
This said we obtain
$$
  \Vert e-f\Vert  =
  \Vert e-fe+fe-f\Vert  \leq
  \Vert e-fe\Vert +\Vert fe-f\Vert  \leq
  2d(E,F). 
  \tag*{$\blacksquare$}
  $$
Lemma 2.  We have
$$
  \Vert p-q\Vert \leq 2\Vert e-f\Vert .
  $$
Proof.  This follows from
$$
  \Vert p-q\Vert \leq 2d(E,F)\leq 2\Vert e-f\Vert .
  \tag*{$\blacksquare$}
  $$
Theorem.  If $\Vert e-f\Vert <1/2$, then $e\sim_hf$.
Proof.  By Lemma (2) we have
$$
  \Vert \rho(e)-\rho(f)\Vert =\Vert p-q\Vert \leq 2\Vert e-f\Vert <1,
  $$
so the conclusion follows as indicated above. $\qquad \blacksquare$
