Let $M$ be a von Neumann algebra and let $N$ be a von Neumann subalgebra and let $E$ be a conditional expectation $M\to N$. Let $i$ be the canonical inclusion of $M$ into $M^{**}$.
Claim: $E$ is normal.
"Proof": (1) Let $E^{**}$ be the induced conditional expectation $M^{**}\to N^{**}$. It is normal and has the property $E^{**}(i(x))=i(E(x))$ for all $x\in M$. (2) The map $i$ is also normal, as it can be thought of as the composition of the universal normal representation $\nu$ (which is normal) and the inclusion of $\nu(M)$ as a direct summand of $M^{**}$ (which is also normal). (3) There is a normal *-homomorphism $\Phi:N^{**}\to N$ which extends the identity map. Conclusion: $E=\Phi\circ E^{**}\circ i$, so $E$ is normal.
Of course, the claim is false: non-normal conditional expectations exist. (1) and (3) are true; references for these abound. (2) also seems true, though I have no reference for the claim it is normal. Also, there is nothing special about conditional expectations here: the above "proves" many interesting linear maps between von Neumann algebras are normal.
Question: what is wrong with the "proof" above?
Any references on the second dual of a von Neumann algebra are appreciated. There are plenty of sources on the enveloping von Neumann algebra of a C*-algebra, but I can't find any that tell me much when that C*-algebra is already a von Neumann algebra (except for a short blurb in Blackadar).
I don't think the map $i $ can be normal when $M $ is not finite-dimensional. Let $f $ be a non-normal state of $M $; then there exists an increasing net $x_j $ of selfadjoints with $x_j\nearrow x $ and $c=\sup f (x_j)<f (x) $. Now, if $(\pi_f,H_f,\eta_f) $ denotes the gns triple for $f $, $$ \langle i (x_j)\tilde\eta_f,\tilde\eta_f\rangle=\langle \pi_f (x_j)\eta_f,\eta_f\rangle=f (x_j)<c <f (x)=\langle \pi_f (x)\eta_f,\eta_f\rangle=\langle i (x)\tilde\eta_f,\tilde\eta_f\rangle, $$ where $\tilde\eta_f\in\bigoplus_{g \text { state}}H_g $ is the vector with $\eta_f $ in the $f $-coordinate and zeroes elsewhere (note that $i=\bigoplus _{g \text { state}}\pi_g $). This shows that $i (x) $ is not the supremum of $\{i (x_j)\} $.