I am going through the proof of the Sherman-Takeda theorem and Fillmore's book "A User's Guide on Operator Algebras" seems to have a nice approach, but something seems off to me:

We need to prove that if $A$ is a $C^*$-algebra then $A^{**}$ is isometrically isomorphic to $\pi_u(A)''$, where $(H_u,\pi_u)$ is the universal representation.

The idea is finding an isometric isomorphism $e:A^*\to(\pi_u(A)'')_*$ taking the adjoint map and employing the uniqueness of the predual in von Neumann algebras. To that end, Fillmore takes a state $\rho\in S(A)$ and says that $\rho$ extends uniquely to a vector state on $\pi_u(A)''$. I agree with this and I can see why this is true. The map $e:A^*\to(\pi_u(A)'')_*$ is then constructed as follows: an arbitrary functional of $A^*$ is written as a linear combination of (four) states, so the above observation allows us to extend any functional of $A^*$ to a linear combination of (four) vector states on $\pi_u(A)''$, so $e$ takes our functional to that extension. I understand why this map $e:A^*\to(\pi_u(A)'')_*$ is a linear isometry but I find it weird that it is onto. Fillmore's proof says that, if $\tau\in(\pi_u(A)'')_*$ is a normal linear functional, then the restriction $\tau\vert_{\pi_u(A)}$ is a functional of $\pi_u(A)^*\cong A^*$ so $e(\tau\vert_{\pi_u(A)})=\tau$. Even though this makes sense to me, there's this awkward part:

Doesn't this imply that all normal functionals on $\pi_u(A)''$ are linear combinations of vector states, hence (SOT) continuous? Is this true or is it a hint that something has gone wrong?

I feel awkward about this because I have the impression that it is very rare for SOT continuous functionals to be the same as the ultraweakly continuous ones.