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.

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. $\endgroup$ – Matthew Daws Jun 27 '18 at 10:03