Lifting idempotents from a quotient of a Banach algebra In a quotient of a Banach algebra $A$, if an invertible element is connected to the identity by a continuous path of invertibles, then it can be lifted to an invertible element in $A$. Is there an analogous result for idempotents or at least something that comes close to the corresponding statement for idempotents, such as lifting idempotents to almost idempotent elements in $A$?
The specific situation I am in is the following. Let $e\in A$ be almost idempotent in the sense that $||e^2-e||<\varepsilon$ (where $\varepsilon$ is very small). Let $q:A\rightarrow A/J$ be the quotient homomorphism. Suppose that $q(e)$ is connected to $0\in A/J$ by a continuous path $(p_t)_{t\in[0,1]}$ where $||p_t^2-p_t||<\varepsilon$ for each $t$. An additional assumption may be that there is a contractive linear (but not necessarily multiplicative) map $s:A/J\rightarrow A$ such that $q\circ s=id_{A/J}$. Can $p_t$ be lifted to an almost idempotent element in $A$ with possibly larger $\varepsilon$? Possibly another way to look at this is whether an $\varepsilon$-idempotent in $C([0,1],A/J)$ can be lifted to an $N\varepsilon$-idempotent in $C([0,1],A)$ for some $N$.
 A: This may or may not be as strong a result as you would like or necessarily in the right direction, but hopefully it helps.
See Lemma 2.1 on page 411 of Larson's "Nest Algebras and Similarity Transformations" in the Annals of Mathematics from 1985. As Larson mentions, this is a generalization of a standard result.
Basically this result says that for a Banach algebra $\mathcal{A}$ with Jacobson radical $\mathcal{R}$ and any ideal $\mathcal{R}_0$ of $\mathcal{A}$ which is contained in the Jacobson radical, then if the image of $A \in \mathcal{A}$ in $\mathcal{A}/\mathcal{R}_0$ is idempotent, then there is an idempotent $P \in \mathcal{A}$ which commutes with $A$ and has the same image in the quotient algebra.
In particular, any idempotent in $\mathcal{A}/\mathcal{R}_0$ lifts to $\mathcal{A}$.
Edit
The result hoped for by the OP does not exist in general, even for $C^*$-algebras. Indeed, consider the commutative $C^*$-algebra $\mathcal{A} = C([0,1])$ of continuous functions on $[0,1]$, and the ideal $\mathcal{J} = C_0((0,1))$ of continuous functions vanishing at both zero and one.
Note that the only idempotents of $\mathcal{A}$ are the trivial ones, namely, the identity and zero (since an idempotent in this algebra takes only the values zero and one, it is a continuous function, and $[0,1]$ is connected). However, if $f$ is any continuous function on $[0,1]$ with $f(0)=0$ and $f(1)=1$, then the image of $f$ is idempotent in $\mathcal{A}/\mathcal{J}$ since $f^2-f \in \mathcal{J}$. Moreover, any element $g \in \mathcal{A}$ which maps to this idempotent satisfies $f-g \in \mathcal{J}$, and therefore $g(0) = f(0) = 0$ and $g(1) = f(1) = 1$, and so by the previous remarks, $g$ cannot be idempotent.
Perhaps a better way to comprehend the results of the previous paragraph is that $\mathcal{A}/\mathcal{J}$ is naturally isomorphic to $C(\{0,1\})$, an algebra with four idempotents. Since $\mathcal{A}$ only has two idempotents, not all of the idempotents in the quotient can lift.
Perhaps also of interest regarding this discussion and the one prior to the edit: $C^*$-algebras have trivial Jacobson radical (see the last bullet).
