Let $A$ be a $C^*$-algebra and $$A = I_1 \supset I_2 \supset I_3 \supset\ldots$$ be a sequence of embedded ideals in $A$ such that $\bigcap_{i=1}^\infty I_i = 0$. Is it true, that the projective limit of the projective system $$A/I_1 \leftarrow A/I_2 \leftarrow A/I_3 \leftarrow\ldots$$ is isomorphic to $A$?


This is not true. Let $A=C[0,1]$. Consider the increasing sequence of closed sets closed sets $F_0=\varnothing$, $F_i=[0,2^{-i}]$, and set $I_i=\left\{f\in A:f|_{F_i}=0\right\}$. Then we are in the setting you described.

We can identify $A/I_i$ with $C(F_i)$, via the map $A/I_i\to C(F_i)$, $f+I_i\mapsto f|_{F_i}$. With this $A/I_{i+1}\to A/I_i$ is identified with the restriction $C(F_{i+1})\to C(F_i)$, $f\mapsto f|_{F_i}$.

Now let's show that the inverse limit of $A/I_i=C(F_i)$ is $C_b([0,1))$, the space of bounded continuous functions on $[0,1)$. We will use the universal property. Define $\pi_i:C_b[0,1)\to C(F_i)$ as restriction: $\pi_i(f)=f|_{F_i}$. Let's prove universality.

Given any $\text{C}^*$-algebra $Y$ with morphisms $\psi_i:Y\to C(F_i)$ which respect the restriction maps $C(F_{i+1})\to C(F_i)$, define $\psi:Y\to C_b([0,1))$ by $\psi(y)(x)=\psi_i(y)(x)$ for all $y\in Y$ and $x\in[0,1)$, where we choose $i$ in such a way that $x\in F_i$. The function $\psi(y)$ is clearly continuous. To see that $\psi(y)$ is bounded (and so $\psi$ is well-defined), we know that $*$-homomorphisms of $\text{C}^*$-algebras are norm-decreasing (contractive), and so $$\sup_{x\in[0,1)}|\psi(y)(x)|=\sup_i\sup_{x\in F_i}|\psi_i(y)(x)|=\sup_i\Vert\psi_i(y)\Vert\leq\Vert y\Vert<\infty$$ so $\psi$ is well-defined. The commutativity of the relevant diagrams is easy enough.

Therefore, $\underset{\leftarrow}{\lim}A/I_i=C_b[0,1)$. This is different from $A=C[0,1]$, because the spectrum of $C_b[0,1)$ is the Stone–Čech compactification $\beta[0,1)$, which is different from $[0,1]$, which is the spectrum of $C[0,1]$.


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