text: Q:
  
  $W^*$-algebras: Canonical choice of an isometric isomorphism $d: M \to ((M^*)_\sigma)^*$?
  
  Let $M$ be a $W^*$-algebra, i.e. a $C^*$-algebra which is isometrically isomorphic to the dual of a Banach space $M_*$. One can then show that $M_*$ is isometrically isomorphic to $(M^*)_\sigma$, the weak* continuous linear functionals on $M$. This allows a canonical choice of the predual, as $(M^*)_\sigma$ is determined entirely by the algebraic structure of $M$. However, I wondered if it is also possible to give the accompanying isometric isomorphism $d: M \to ((M^*)_\sigma)^*$ in a canonical way. The former arguments show that such an isomorphism must exist, however, it is not clear what $d$ exactly looks like. My guess was that one can choose $d$ as the dual pairing $A \mapsto \langle \cdot , A \rangle$, but I don't know how to show/disprove whether this $d$ is actually surjective. Does this work? If not, are there any other ways to give an isometric isomorphism $d$ in a canonical/explixit way?
  Edit: This article (last paragraph of section 4) states my exact claim, sadly, there is no proof/justification that goes along with it.
  
  A:
  
  I got the solution: Let $M_*$ be an abritrary predual and $\varphi: M_* \to (M_*)^*$ an arbitrary isometric isomorphism. Then one easily checks that $(\varphi^* \circ j)^* \circ d = \varphi$, where $*$ denotes the dual operator, $d$ is the above evaluation map/dual pairing, and $j: M_* \to (M_*)^{**}$ is the canonical embedding which is an isometric isomorphism onto $((M_*)^{**})_\sigma$. Hence $d = ((\varphi^* \circ j)^*)^{-1} \circ \varphi$, but the latter is a composition of isometric isomorphisms and thus an isometric isomorphism itself.
  
  