Question: Let $A$ be a C$^\*$-algebra, $f$ a representation of $A$, $F$ the universal representation of $A$, and $g=f\circ F^{-1}$. For an ultraweakly continuous linear functional $w$ on $f(A)$, $w\circ g$ is bounded and hence according to a well-known theorem, is ultraweakly continuous linear functional on $F(A)$.
My question: it results that g is ultraweakly continuous. How is its proof?
Thank you for consideration.

Answer: 
Suppose $\{ w\_1, w\_2, \dots, w\_r \}$ is a basis of $W$. You can complete it to get a basis $\{ w\_1, \dots, w\_r, v\_{r+1}, \dots, v\_n \}$ of the whole space. The clases $\{ v\_{r+1} + W, \dots, v\_n + W \}$ are a basis of the quotient space (Why?) A proof of the dimension now follows easily.

Since you ask for another proof. If you have studied First Isomorphism Theorem you can apply it to the map $\phi: V \to V\W$ where $\phi(x) = x + W$.
Let $\{w\_i\}$ be a basis for $W$ and extend by $\{v\_j\}$ to get a basis for $V$. Then you can show $\{v\_j + W\}$ is a basis for the quotient.