Quotients of C*-algebras It is known that every unital separable C*-algebra is a quotient of the full group C*-algebra $C^*(F_I)$, where $F_I$ is the free group generated by some index set $I$. 

Can we drop the separability assumption here? 

Any references would be appreciated.
 A: I think the idea in my comment works in detail as follows. Probably one can simplify the argument by applying the appropriate universal properties of the group $C^\ast$-algebra on the free group $F(U)$ on the unitaries $U$ of $A$, but I'm not very familiar with them, so here's a pedestrian's description:
By the Gelfand-Naimark theorem we can assume that $A$ is isometrically contained as a unital $C^\ast$-subalgebra of $B(H)$ for some Hilbert space $H$. The unitaries $U$ of $A$ act as unitaries on $H$, so we have a unitary representation of $U$ on $H$. This unitary representation extends to a *-representation of the complex group algebra $\mathbb{C}[U]$. Notice that the image of $\mathbb{C}[U]$ under this representation is contained in $A$. By the definition of the group $C^\ast$-algebra $C^\ast(U)$ this representation extends uniquely to a *-homomorphism $\rho \colon C^\ast(U) \to B(H)$.
I claim that $\rho$ is a homomorphism of $C^\ast(U)$ onto $A$. Since $\rho$ is continuous and $\rho(\mathbb{C}[U]) \subseteq A$ and $\mathbb C[U]$ is dense in $C^\ast(U)$, we have $\rho(C^\ast(U)) \subseteq A$. On the other hand, the elements of $U$ are unitary in $C^\ast(U)$, so they have norm one. Since $\rho$ is linear and norm-decreasing, the image of the unit ball of $C^\ast(U)$ contains the convex hull of the unitaries of $A$. But this convex hull is dense in the unit ball of $A$ by the Russo-Dye-Gardner theorem and by the open mapping theorem it follows that $\rho$ is onto $A$, as claimed.
Next, recall that the group $C^\ast$-algebra is functorial with respect to quotient homomorphisms, and $U$ is a quotient of the free group $F(U)$ generated by the set underlying $U$, so we have a surjective *-homomorphism $C^\ast(F(U)) \to C^\ast(U)$ and composing this surjection with $\rho$ we get the desired surjection $C^\ast(F(U)) \to A$ from the group $C^\ast$-algebra of some free group. 
If $A$ is separable, it suffices to take a countable dense subset of the unitaries on $A$, and the above argument shows that every separable $C^\ast$-algebra is a quotient of $C^\ast(F_\infty)$ where $F_\infty$ is the free group on countably many generators, so we see that $C^\ast(F_\infty)$ is universal for separable $C^\ast$-algebras in the same sense that $\ell_1$ is universal for separable Banach spaces.
