Schröder–Bernstein for representations of operator algebras

This is claimed in a Wikipedia Article:

If two representations (of a $C^*$-algebra $A$) $\rho$ and $\sigma$, on Hilbert spaces $H$ and $G$ respectively, are each unitarily equivalent to a subrepresentation of the other, then they are unitarily equivalent.

I don't really follow the proof given. Indeed, what bothers me is the following: if we let $I,J$ be sets and $f:I\rightarrow J, g:J\rightarrow I$ be injections inducing isometries $\ell^2(I)\rightarrow\ell^2(J)$ and $\ell^2(J)\rightarrow\ell^2(I)$ and we just let $A$ be the complex numbers, then the claim is that there is a unitary $U:\ell^2(I)\rightarrow\ell^2(J)$. But furthermore, the proof indicates that $U$ is constructed from the original isometries, so $U$ maps basis vectors to basis vectors and hence immediately induces a bijection $I\rightarrow J$. That is, we seem to have reproved Schröder–Bernstein using nothing but induction...

Indeed, I am fairly sure I can prove this result using Schröder–Bernstein for projections, working with $\lambda=\pi\oplus\rho$ and using comparison of projections in $\lambda(A)'$. Notice that in the projection case, using the notation of the Wikipedia article, you need to consider $R$.

Am I correct to be a little worried?

Furthermore, Schröder–Bernstein for projections is very standard, and in lots of books. Perhaps I have just not looked hard enough, but I cannot find a textbook proof of the result for the representations.

Is there a reference to this result in a textbook or paper?

Edit: Let me re-write the Wikipedia proof, in the special setting I outlined about. It boils down to observing that we can partition $I = I' \sqcup g(J)$ and $J = J' \sqcup f(I)$ and then repeatedly applying $f$ and $g$, so $$J = J' \sqcup f(I) = J' \sqcup f(I') \sqcup fg(J) = J' \sqcup f(I') \sqcup fg(J') \sqcup fgf(I) = \cdots$$ and $$f(I) = f(I') \sqcup fg(J) = f(I') \sqcup fg(J') \sqcup fgf(I) = \cdots$$ But now we come to Ruy's objection: you cannot conclude that e.g. $$J = \bigsqcup_{n\geq0} (fg)^n(J') \sqcup \bigsqcup_{n\geq 0} (fg)^nf(I')$$ The argument (for example) completely missing the possibility of "cycles", in Andreas's language.

So, I think I've answer my first question. I know how to prove the result. I am still interested in an actual reference in e.g. a textbook.

• In my opinion the argument in the Wikipedia article is fallacious. I do not see how the finite decomposition of each representation given there leads to the claimed infinite decomposition. Induction gives the result for arbitrarily large natural numbers, but not necessarily for infinite! – Ruy Jun 23 '18 at 20:51
• I do not think that there is a problem with the argument. The Schröder-Bernstein theorem is almost a triviality once you get the right picture. Just draw a bipartite graph with edges coming from $f$ and $g$ and consider the connected components -- either its a cycle, a two-sided line or a one-sided line. Hence, for each component, there is a trivial bijection (supported on the bipartite graph). All other proofs of generalizations follow this simple recipe, maybe involving some partition process to get started. – Andreas Thom Jun 26 '18 at 21:46
• @AndreasThom I agree that this is how one should (or could) prove Schröder-Bernstein. My complaint is that the proof offered by Wikipedia does not follow this logic. I've edited the question to make this clearer, I hope. – Matthew Daws Jun 27 '18 at 10:03
• Ok, I do not understand the proof either (unless it is the argument from above). The confusion might arise from the fact that many representations are "isomorphic", but this is not distinguished from being "identical" in the Wikipedia article. – Andreas Thom Jun 27 '18 at 13:13
• I have never interfered with Wikipedia articles, so it would take me some time to figure out how to do it. However it might be interesting to either edit the article mentioned by Mathew, or at least to start a discussion in its talk page (currently empty) pointing out these problems. Any volunteers? – Ruy Jun 27 '18 at 23:15