TITLE: Subspaces of a Subfactor
QUESTION [5 upvotes]: Is the following true?

Let $\mathcal N \subset \mathcal M$ be a subfactor. There is a bijective correspondence between the ultraweakly closed subspaces of $\mathcal M$ that are bimodules over $\mathcal N'\cap \mathcal M$, and the ultraweakly closed subspaces of $\mathcal N$.

If the statement is false, is there a simple way to modify it to make it true? I am particularly interested in the case that $\mathcal M$ is of type $\mathrm I$.
If $\mathcal V \subseteq \mathcal N$ is ultraweakly closed, then $\mathcal V (\mathcal N' \cap \mathcal M)$ is a bimodule over $\mathcal N'\cap \mathcal M$. If the subfactor admits a conditional expectation, then this function is injective. (Edit. Jesse points out that the conditional expectation should be ultrweakly continuous.)
Edit. Steven points out that the statement as written is trivially true by a counting argument. Of course, I'm asking about the function $\mathcal V \mapsto \mathcal V \mathcal (\mathcal N'\cap \mathcal M)$, or something similarly natural. He also notes that irreducible subfactors are a counterexample to the bijectivity of $\mathcal V \mapsto \mathcal V(\mathcal N' \cap\mathcal M )$ in general. This leaves a single concrete question:

Let $\mathcal N \subseteq \mathcal B (\mathcal H)$ be a factor. Is the function $\mathcal V \mapsto \mathcal V \mathcal N'$ a bijection between the ultraweakly closed subspaces of $\mathcal N$ and the ultraweakly closed subspaces of $\mathcal B(\mathcal H)$ that are $\\mathcal N'$ bimodules?

REPLY [2 votes]: I suppose the statement is true because the set of ultraweakly-closed subspaces of $N$ has the same cardinality as the set of $N^\prime\cap M$-bimodular subsets of $M$. You probably want an explicit description of the correspondence, preferably such that the $N^\prime\cap M$-bimodule that corresponds to $V\subset N$ is exactely $V(N^\prime\cap M)$.
This is certainly not true in general. For an irreducible subfactor $N\subset M$, i.e. $N^\prime\cap M=\C$, the condition that $V\subset M$ is an $N^\prime\cap M$-bimodule is empty, while there are strictly more subsets of $M$.
Of course, the bicommutant theorem shows that there are no irreducible subfactors of type I factors. I do not immediately see a counterexample in the type I case.