# Double commutant of an injective subalgebra still injective?

Consider a von Neumann algebra $$A_0$$ and an injective *-isomorphism $$\pi : A_0 \to B(H)$$.

Then we have a *-subalgebra $$\pi(A) \subset B(H)$$, which is abstractly *-isomorphic to the von Neumann algebra $$A_0$$, but which might not be a "von Neumann subalgebra" of $$B(H)$$, i.e. we are not guaranteed $$\pi(A)''=\pi(A)$$. (E.g. this answer.)

If $$A_0$$ is injective, can we conclude $$\pi(A)''$$ injective? If it helps, I'm interested in the case $$A_0$$ is the hyperfinite $$II_1$$ factor.

No, it isn't. An irreducible representation of a II$$_1$$ factor represents faithfully as a dense sot-subalgebra of a non-separable $$B(H)$$, which then cannot be injective.