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.
