Automorphism of algebras with certain initial conditions on given idempotents

Question

The First question Let $A$ be a Banach or a $C^*$ algebra. Assume that $e,f$ are two idempotents or prjections in $A$ which satisfy $ef=fe=0$. Assume that there are two automorphisms $\phi, \psi: A \to A$ with $\phi(e)\psi(f)=\psi(f)\phi(e)=0$.

1) Is there an automorphism $\rho:A\to A$ with $\rho(e)=\phi(e),\;\rho(f)=\psi(f)$?

2) A topological version: Let $X$ be a compact Hausdorff space and $B, C$ be two disjoint Clopen subsets of $X$. Assume that $F, G$ are two homeomorphisms of $X$ with $F(B)\cap G(C)= \varnothing$.Is there a homeomorphism $H$ of $X$ with $H(B)=F(B),\;H(C)=G(C)$?

The second question: Let $A$ be a (Banach or a $C^*$ ) algebra or a ring. Assume that $e,f$ are two idempotents or projections in $A$. Assume that $\phi,\psi A:\to A$ are two automorphisms. Is there an automorphism $\alpha$ of $M_2(A)$ with $ \alpha (e\oplus f)=\phi(e)\oplus \psi(f)$?

Note: A positive answer to these questions enable us to stablish a group structure on a possible new kind of K-theory of rings or algebras

Answer 1

Edit: The answer to both of your questions is "no".

The counterexamples below all carry a similar flavor.

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Counterexample to the topological version of the first question. (Taking the $C^*$-algebra of continuous $\mathbb{C}$-valued functions on the counterexample gives a counterexample to the $C^*$-algebra version.)

Let $X$ be the following subset of $\mathbb{R}$ with the subspace topology: $$ X=\{-2,0,2\}\cup\{2-\frac{1}{n}\,|\,n\in\mathbb{N}\}\cup\{-2+\frac{1}{n}\,|\,n\in\mathbb{N}\}. $$ There is a homeomorphism $F:X\to X$ fixing $-2$ and $2$ and taking each $x\in X-\{\pm 2\}$ to $\min (X\cap (x,2])$ (i.e. the closest point on the right).

Let $B=\{2-\frac{1}{n}\,|\,n=1,2,\dots\}\cup\{2\}$ and $C=\{-2+\frac{1}{n}\,|\,n=1,2,\dots\}\cup\{-2\}$ and write $G=F^{-1}$.

Then $F(B)=\{2-\frac{1}{n}\,|\,n=2,3,\dots\}\cup\{2\}$ and $G(C)=\{-2+\frac{1}{n}\,|\,n=2,3,\dots\}\cup\{-2\}$ are disjoint, but there is no homeomorphism $H:X\to X$ taking $B$ to $F(B)$ and $C$ to $G(C)$. Indeed, if $H$ existed, then it would restrict to a bijection between $X-B-C=\{0\}$ and $X-F(B)-G(C)=\{-1,0,1\}$, which is absurd.

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Counterexample to the the second question and the $C^*$-algebra version of the first question.

Take $A$ to be the abstract ring $\mathbb{C}^{\mathbb{Z}}$. Alternatively, take the subring of $\mathbb{C}^{\mathbb{Z}}$ consisting of sequences $(a_n)_{n\in \mathbb{Z}}$ so that both $\lim_{n\to \infty }a_n$ and $\lim_{n\to -\infty}a_n$ exist --- this is a $C^*$-algebra which is isomorphic to $C(X)$ with $X$ as above.

Let $e_s$ denote the sequence $(a_n)_{n\in \mathbb{Z}}$ for which $a_n=0$ if $n<s$ and $a_n=1$ if $n\geq s$, and let $f_s=1_A-e_s$.

It is easy to see that there are automorphisms of $A$ carrying $e_0$ to $e_1$ and $f_0$ to $f_{-1}$. However, there is not automorphism of $A$ carrying $e_0$ to $e_1$ and $f_0$ to $f_{-1}$ because $e_0+f_0=1_A$ while $e_1+f_{-1}\neq 1_A$. Furthermore, there is no automorphism of $\mathrm{M}_2(A)$ taking $e_0\oplus f_0$ to $e_1\oplus f_{-1}$, because $\mathrm{Ann}_{\mathrm{Cent}(\mathrm{M}_2(A))}(e_0\oplus f_0)=0$ while $\mathrm{Ann}_{\mathrm{Cent}(\mathrm{M}_2(A))}(e_1\oplus f_{-1})$ is nonzero.