# Image of a normal $*$-homomorphism

Let $$\mathcal M$$ be a von Neumann algebra. Let $$\pi:\mathcal M\to\mathcal M$$ be a normal $$*$$-homomorphism Is $$\pi(\mathcal M)$$ again a von Neuman algebra? By [J. Dixmier, Les algebres d’operateurs dans l’Espace Hilbertien, 2nd ed., Gauthier-Vallars, Paris, 1969., Part I, Chapter 4.3, Corollary 2] $$\pi(\mathcal M)$$ is a weakly closed $$*$$-subalgebra of $$\mathcal M.$$ Clearly, $$\pi(\mathcal M)$$ has a unit $$\pi(1).$$ So is it not enough to say that $$\pi(\mathcal M)$$ is again a von Neumann algebra? But many places I have seen such as Sunder, V (An Invitation to von Neumann algebra) where the author has emphasized on the injectiveness of $$\pi$$!!

The image $$\pi(\mathcal{M})$$ of a normal $$*$$-homomorphism $$\pi\colon \mathcal{M}\to\mathcal{N}$$ between von Neumann algebras $$\mathcal{M}$$ and $$\mathcal{N}$$ is indeed weakly closed in $$\mathcal{N}$$ (and thus a von Neumann algebra), also when $$\pi$$ is not injective.
In fact, one way to prove the general statement (in which $$\pi$$ need not be injective) is to reduce it to the 'injective case', as follows.
Note that the kernel $$\mathop{Ker}(\pi)$$ of $$\pi$$ is a weakly closed $$*$$-subalgebra of $$\mathcal{M}$$, and thus a von Neumann algebra. In particular, $$\mathop{Ker}(\pi)$$ has a greatest projection, $$c$$, (the unit of the von Neumann algebra $$\mathop{Ker}(\pi)$$.) Using the fact that $$\mathop{Ker}(\pi)$$ is in addition a two-sided ideal of $$\mathcal{M}$$ one can show that $$c$$ is central in $$\mathcal{M}$$ (see Theorem 6.8.8 of Kadison & Ringrose Vol. II, or perhaps 69{II,IV} of my thesis.)
Now the trick is to consider the von Neumann algebra $$(1-c)\mathcal{M}$$. Since $$\pi(c)=0$$, we have $$\pi(a)\ =\ \pi(ca+(1-c)a)\ =\ \pi((1-c)a)$$ for all $$a\in\mathcal{M}$$, and so $$\pi$$ can be written as the composition $$\mathcal{M} \stackrel{a\mapsto (1-c)a}\longrightarrow (1-c)\mathcal{M}\stackrel{\varrho}\longrightarrow \mathcal{N},$$ where $$\varrho$$ is simply the restriction of $$\pi$$ to $$(1-c)\mathcal{M}$$. Note that, $$\varrho$$ is a normal $$*$$-homomorphism that is in addition injective. Since moreover the image $$\varrho(\,(1-c)\mathcal{M}\,)$$ of $$\varrho$$ coincides conveniently with the image of $$\pi\equiv \varrho\circ ((1-c)(\,\cdot\,))$$, we see that the image $$\pi(\mathcal{M})$$ of the (not necessarily injective) normal $$*$$-homomorphism $$\pi$$ is a von Neumann algebra when the image of the injective normal $$*$$-homomorphism $$\varrho$$ is a von Neumann algebra.