In Takesaki, Theory of Operator Algebras I, we find the following theorem and proof (page 184):
I do not understand the last step (starting from "hence"). I manage to see that $\pi(\mathcal S)'' \subseteq \pi(\mathcal S'')$, but I do not see the argument why we would have $\pi(\mathcal S)'' \supseteq \pi(\mathcal S'')$. Can someone enlighten me?
For completeness, Proposition 1.4 referenced in the proof says:
And $U_i$ refers to the isometry $\xi\mapsto\xi\otimes\epsilon_i$ for an orthonormal basis $\{\epsilon_i\}$.
I found the following proof. I am not sure that's what Takesaki had in mind.
Claim 1: $x$ and $y$ commute for $x\in\pi(S'')$ and $y\in\pi(S)'$.
To show this, it is sufficient to show $U_i^* xy U_j = U_i^*yxU_j$ for all $i,j$.
Define $y_0 := U_i^* y U_j$.
We now show that $y_0\in S'$: For any $z\in S$, we have:
$$y_0 z = U_i^* y U_j z = U_i^* y U_j U_j^* (z \otimes 1) U_j = U_i^* y (z \otimes 1) U_j U_j^* U_j$$ (For the last equality, we use that $U_j U_j^*=1\otimes \epsilon_j\epsilon_j^*$ and thus commutes with $z\otimes 1$.)
$$\dots = U_i^* y (z \otimes 1) U_j = U_i^* (z \otimes 1) y U_j$$ (For the last equality, use that $z\otimes 1\in \pi(S)$ and $y\in\pi(S)'$, so they commute.)
$$\dots = z y_0.$$ (Analogous to the beginning of the computation.)
So $y_0$ commutes with $z$. Since this holds for all $z\in S$, we have $y_0\in S'$.
We proceed with the proof of the claim.
Since $x\in\pi(S'')$, we have $x=\pi(w)$ for some $w\in S''$.
$$U_i^* xy U_j = U_i^*U_iU_i^* (w\otimes 1) yU_j = U_i^* (w\otimes 1) U_iU_i^* yU_j$$ (The last equality holds because $U_i U_i^*=1\otimes \epsilon_i\epsilon_i^*$ commutes with $w\otimes 1$.)
$$\dots = wy_0 = y_0w$$ (The last equality because $w\in S''$ and $y_0\in S'$.)
$$\dots = U_i^* yx U_j.$$ (Analogous to the beginning of the computation.)
So we have shown $U_i^* xy U_j = U_i^* yx U_j$. Since this holds for all $i,j$, we have $xy=yx$. This proves Claim 1.
From Claim 1, we immediately get $\pi(S'')\subseteq \pi(S)''$.