Proposition 43.4.9.label Let $M$ be a von Neumann algebra, $H$ be a complex Hilbert space, and $\pi: M \to B(H)$ be a normal unital *-homomorphism, then $\pi(M)$ is a von Neumann algebra.
Proof. Let $N = \pi(M)''$ be the von Neumann algebra generated by $\pi(M)$. By the Kaplansky Density Theorem, $\ol{B_N(0, 1)}^{B(H)}= \ol{B_{\pi(M)}(0, 1)}^{\text{\small WOT}}$. By Proposition 5.16.3 and Proposition 43.1.3, $\ol{B_M(0, 1)}^{M}$ is weak-operator compact, so $\pi(\ol{B_M(0, 1)}^{M})$ is weak-operator compact and contains $\pi(B_{M}(0, 1))$. Thus
\[\ol{B_{\pi(M)}(0, 1)}^{B(H)}\supset \pi(\ol{B_M(0, 1)}^{M}) \supset \ol{B_{\pi(M)}(0, 1)}^{\text{\small WOT}}\supset \ol{B_N(0, 1)}^{B(H)}\]
and $\ol{B_{\pi(M)}(0, 1)}^{B(H)}= \ol{B_{N}(0, 1)}^{B(H)}$, so $\pi(M) = \ol{\pi(M)}^{B(H)}= N$ by Theorem 39.13.3.$\square$
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