Isomorphism theorem for subfactors?

Question

It's about the existence of a generalization of the first isomorphism theorem for groups, for subfactors :

Let $(N \subset M)$ and $(N' \subset M')$ be irreducible inclusions of hyperfinite $II_1$ factors.
Let $\phi: M \to M'$ be a $W^*$-morphism with $\phi(N) = N'$ and $\phi|_N : N \to N'$ an isomorphism.

Question: Is there an isomorphism of subfactors: $(\phi^{-1}(N') \subset M ) \simeq (N' \subset \phi(M))$ ?
(in other words, is there $\psi: M \to \phi(M)$ a $W^*$-isomorphism, with $\psi(\phi^{-1}(N')) = N' $ ?)

Example: let $N=N'=R$, $M=R \rtimes G $, $M'=R \rtimes G'$ with $G$ and $G'$ finite groups.
Let $f: G \to G'$ be a group-morphism.
First isomorphism theorem for groups : $G /ker(f) \simeq im(f)$.
Let $\phi: R \rtimes G \to R \rtimes G'$ be the canonical $W^*$-morphism coming from $f$.
Then, $\phi^{-1}(N') = R \rtimes ker(f)$ and $\phi(M)=R \rtimes im(f)$, so:

• $(\phi^{-1}(N') \subset M ) = (R \rtimes ker(f) \subset R \rtimes G) \simeq (R \subset R \rtimes G /ker(f))$
• $ (N' \subset \phi(M)) \hspace{0.5cm} = \hspace{0.5cm} (R \subset R \rtimes im(f)) \hspace{0.6cm} \simeq (R \subset R \rtimes G /ker(f)) $

Edit (after Dave Penneys's answer): The $W^*$-morphisms are too strong, we need to find weaker maps for this generalization be relevant and this example, correct.

Answer 1

A II$_1$-factor is algebraically simple, so each morphism of II$_1$-factors is either injective or zero. Thus every non-zero morphism is an isomorphism onto its image. So $\phi: M \to \phi(M)$ is an isomorphism that takes $\phi^{-1}(N')$ to $N'$.

I don't think the canonical surjection $G\to G'=G/\ker(f)$ actually gives you a map of factors $M\rtimes G\to M\rtimes G'$. In particular, if we denote the implementing unitaries as $u_g$ for $g\in G$, the map $u_g\mapsto u_{g\ker(f)}$ does not extend to a non-zero map of II$_1$-factors if $\ker(f)$ is non-trivial. The element $u_g-u_{g'}$ would map to zero if $g,g'\in \ker(f)$, and a non-trivial map of II$_1$-factors must be injective.

Answer 2

Here is a generalization of the answer of Dave to all the bimodule-morphisms :

Let $R$ be the hyperfinite II$_1$ factor, and let $G$, $G'$ be finite groups.
Then $R \rtimes G$ and $R \rtimes G'$ are (algebraic) $R$-$R$ bimodules.

By definition of the cross-product: $\forall x \in R$ and $\forall g \in G$, then $u_g . x = \sigma_g(x) . u_g$

The bimodule-morphisms are the $R$-$R$ bilinear maps.
If $\forall f: G \to G'$ (group-morphism), $\exists \phi_f: R \rtimes G \to R \rtimes G'$ (bimodule-morphism) with :

• $\phi_f(x)=x$ $\forall x \in R$
• $\phi_f(u_g)=u_{f(g)}$ $\forall g \in G$

Then $\phi_f(u_g . x) = \phi_f(u_g).x $ and $ \phi_f(\sigma_g(x) . u_g) = \sigma_g(x) . \phi_f(u_g) $ by $R$-$R$ bilinearity.
So, $ \sigma_g(x) . u_{f(g)}= u_{f(g)}.x$.

Now if we take $f$ to be the trivial group-morphism, then: $\sigma_g(x)=x$, contradiction.

Conclusion : The group-morphisms are not "compatible" with the bimodules-morphisms.This shows that the category of finite groups is not a (natural) subcategory of the category of $R$-$R$ bimodules.