The isomorphism between two $C^*$-algebras

Question

Let $A,B$ be two $C^*$-algebras. If there exist two injective $*$-homomorphisms $f:A\to B$ and $g:B\to A$, can we conclude that $A$ is $*$-isomorphic to $B$?

Answer 1

Recall that by the commutative Gelfand-Naimark theorem, the map $A \mapsto \text{Hom}(A, \mathbb{C})$ establishes a contravariant equivalence of categories between commutative unital $C^{\ast}$-algebras and compact Hausdorff spaces. This equivalence switches monomorphisms and epimorphisms, and with a little work you can show that the monomorphisms of $C^{\ast}$-algebras are precisely the injections and the epimorphisms of compact Hausdorff spaces are precisely the surjections. So, for commutative unital $C^{\ast}$-algebras, your question dualizes to:

If $X, Y$ are two compact Hausdorff spaces such that there exist continuous surjections $f : X \to Y, g : Y \to X$, then must $X$ and $Y$ be homeomorphic?

The answer is no. For example, take $X = S^1, Y = [-1, 1]$. There is a continuous surjection $X \to Y$ given by, say, the projection to the $x$-axis (thinking of $S^1$ as the unit circle in $\mathbb{R}^2$), and also a continuous surjection $Y \to X$ given by identifying the endpoints, or more explicitly given by $y \mapsto e^{\pi i y}$. But $X$ and $Y$ are not homeomorphic; for example, $Y$ can be disconnected by removing a point but $X$ cannot.

Answer 2

To complement Qiaochu's answer, the double embedding between non-isomorphic C$^*$-algebras is fairly common. Here are some examples.

• C$^{*\vphantom{j}}_r(\mathbb F_n)$, reduced C$^*$-algebras of free groups (this was mentioned by Aweygan in the comments). It was proven by Pimsner-Voiculescu that they are non-isomorphic, and the mutual embedding comes from the usual mutual group embedding between $\mathbb F_n$ and $\mathbb F_m$.

• Cuntz algebras $\mathcal O_n$. There is maybe a more ad-hoc argument, but it follows from Kirchberg-Phillips result that all exact C$^*$-algebras embed in $\mathcal O_2$ (and thus $\mathcal O_n$). The Cuntz algebras are nuclear and hence exact.

• In the non-separable case we can look at von Neumann algebras, and examples are also kind of common there, too. Looking at separable (in the von Neumann sense) factors, every type II$_\infty$ and type III factor contains a copy of $B(\ell^2(\mathbb N))$ (this, because II$_\infty$ factors are of the form $M\otimes B(\ell^2(\mathbb N))$ with $M$ of type II$_1$, and type III factors always satisfy $M\simeq M\otimes B(\ell^2(\mathbb N))$). So for any two properly infinite separable von Neumann factors $M_1$ and $M_2$ we have a mutual embedding.