It's about the existence of a generalization of the [first isomorphism theorem for groups](http://en.wikipedia.org/wiki/Isomorphism_theorem#Groups), for [subfactors](http://en.wikipedia.org/wiki/Subfactor) :


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](http://en.wikipedia.org/wiki/Isomorphism_theorem#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.



Isomorphism theorem for subfactors?
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. 


