Equivalence of two elements in a quotient $C^*$-algebra

Question

Let $A$ be a unital $C^*$-algebra, and $I_1$ and $I_2$ closed (two-sided) $C^*$-ideals of $A$ such that $A=I_1+I_2$. Suppose we have two positive elements $a_1\in I_1$ and $a_2\in I_2$ such that $$a_1^2+a_2^2=1.$$

Question: Then is it true that $a_1$ and $1$ define the same class in the quotient $A/I_2$?

Thoughts: I want to say that $a_1 = \sqrt{1-a_2^2}$, and then expand this as a power series and cancel the leading $1$, whence the remaining terms all lie in $I_2$. But I am not sure if this series converges in $A$. Perhaps there is some basic fact that I am missing here about power series in $C^*$-algebras.

Answer 1

You don't need the assumption $A=I_1+I_2$.

You have $\overline{a_1}^2=1$ in the quotient $A/I_2$. But the quotient is again a unital $C^*$-algebra and $\sigma_{A/I_2}(\overline{a_1}) \subseteq \sigma_A(a_1)\subseteq [0,\infty[$ so also $\overline{a_1}$ is positive in $A/I_2$. The assumption $a_1^2+a_2^2=1$ also implies that $1$ is a positive element. Thus the same reasoning gives that $1$ is positive in $A/I_2$. Since every positive element in a unital $C^*$-algebra has a unique positive square root (Theorem 2.2.1 in Murphy's "$C^*$-algebras and operator theory"), we conclude $1=\overline{a_1}$.

Answer 2

Here is a slightly different way to look at it: Functional calculus commutes with unital $\ast$-homomorphisms, i.e., $f(\pi(a))=\pi(f(a))$. For polynomials this is immediately clear from the definition of $\ast$-homomorphisms, and for arbitrary continuous functions, one can use approximation by polynomials.

In particular, if $\pi\colon A\to A/I_2$ is the quotient map, then $\pi(a_1)=\sqrt{\pi(1)-\pi(a_2)^2}=\pi(1)$.

This is close to the approach suggested in the OP, you just don't need to worry about convergence of power series, the polynomial approximation guaranteed by the Stone-Weierstrass theorem is enough.