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: 
Another proof that may be circular depending on how you proved the Rank-Nullity theorem is as follows:
Let $\pi:V\to V/W$ be the canonical projection. Notice that it is linear and surjective and the kernel by definition is $W$ hence:

$$\dim(V)=\dim(V/W)+\dim(W)$$

And your equality follows.
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.