$C^*$ algebra has a nonzero finite dimensional quotient

Question

I wonder what is the exact definition of quotient.If $C^*$ algebra $A$ has finite dimensional quotient,does this mean that there is a finite dimensional $C^*$ subalgebra $B$ such that the map from $A$ to $B$ is quotient map?

Answer 1

No, a quotient is not (necessarily) a subalgebra.

It means that there is a closed ideal $I$ inside $A$ such that $A/I$ (which has a naturally defined $C^*$-algebra structure) is a finite-dimensional $C^*$-algebra.

As an example take $A=\mathcal C(X)$ to be the $C^*$-algebra of (complex valued) continuous functions on a locally compact topological space $X$ and let $I_x$ be the ideal $I_x=\{f\in \mathcal A \, : f(x)=0\}$. Then $A/I_x\simeq \mathbb C$ is a $1$-dimensional $C^*$-algebra.

Answer 2

Let $X$ be a compact, Hausdorff and path-connected space. Let $x,y \in X$ be two different points and $I_{x,y} = \{ f \in C(X) : f(x) = f(y) = 0 \}$. Take $A = C(X)$ and $B = \mathbb{C} \oplus \mathbb{C} = A/I_{x,y}$. Clearly $\mathbb{C}^2$ is not a subalgebra of $C(X)$. Otherwise the image of $(0,1)$ will be a nontrivial projection in $C(X)$, i.e. a projection whose spectra contains both $0$ and $1$. But a continuous function in a connected space cannot take only the values $0$ and $1$, giving a contradiction.