Every separable Banach space is a linear quotient of $\ell\_1$, however not every separable Banach algebra is a Banach-algebra quotient of $\ell\_1(G)$ for some group $G$ (these are the so called unitary Banach algebras). 


Is every separable Banach algebra a quotient of $\ell\_1(S)$ for some countable semigroup?


I'd be interested in knowing what could be the Banach-algebra quotients of $\ell\_1(\mathbf{N}\_0)$ and $\ell\_1(\mathbf{Z})$. 


Does either of them quotient onto $C(X)$ for some infinite compact metric space $X$?



What algebras are quotients of $\ell_1(\mathbf{N})$?
(What follows is largely the result of digging around online, based on knowing a few more magic words than the OP.)


**Answer to the first question (I think).**


Let $V$ be a separable Banach space. The standard proof that $V$ is isometrically isomorphic to a linear quotient of $\ell\_1$ works by choosing a countable set $X$ that is a dense subset of the unit sphere of $V$, defining $f:\ell\_1(X) \to V$ in the obvious way, and then verifying two things:


(i) $f$ is surjective;


(ii) the natural map $f\_1: \ell\_1(X)/\ker(f) \to V$ is an isometry.


Suppose $V$ is actually a unital Banach algebra. Let $S$ be the multiplicative subsemigroup of $V$ generated by $X$; of course $S$ is countable. Although $S$ might not lie in the unit sphere of $V$, it does lie in the unit ball, and so we do get a norm-decreasing homomorphism $h: \ell\_1(S) \to V$.


We may factorize $f$ as $h\circ \iota$ where $\iota: \ell\_1(X)\to \ell\_1(S)$ is the natural isometric embedding induced from $X\subset S$. Since $f$ is surjective, so is $h$. Moreover, since $f\_1$ is an isometry: given $b$ in the unit ball of $V$ and $K>1$ we can find $\xi\in \ell\_1(X)$ with $f(\xi)=b$ and $\Vert\xi\Vert\_1 < K$. So $b=h(\iota(\xi))$ where $\Vert\iota(\xi)\Vert\_1 < K$. This shows that $h$ is actually a quotient map of Banach spaces, as required.


**Comment on the intermediate question.** I am not sure right now what one can say about the Banach algebra quotients of these algebras in general.


**Partial answer to the final question.** Via the Gelfand transform we can regard $\ell\_1({\bf Z}\_+)$ as an algebra of continuous functions on the closed unit disc, and hence for any closed subset $E$ of the unit disc, there is a norm-decreasing homomorphism $r\_E: \ell\_1({\bf Z}\_+)\to C(E)$ given by restriction of these functions. It is known that for some infinite closed $E\subset {\bf T}$ this restriction homomorphism $r\_E$ can be surjective: according to remarks in Volume 2 of Hewitt&Ross, these were originally called Carleson sets, but Wik showed that this coincides with the more widely known notion of a Helson subset of ${\bf T}$, which corresponds to $r\_E: \ell\_1({\bf Z}) \to C(E)$.



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> [MR0125404](http://www.ams.org/mathscinet-getitem?mr=125404) (23 #A2707) I. Wik, *On linear dependence in closed sets.* Ark. Mat. 4 1961 209–218. 
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The condition that $r\_E:\ell\_1({\bf Z})\to C(E)$ be a quotient map of Banach spaces corresponds to the classical notion of $E$ being a subset of ${\bf T}$ with Helson constant one. Such sets are known to exist, but I confess I don't know anything about the proofs/constructions. Wik's result shows that Helson sets are Carleson sets, but it is not clear to me from his proof if "1-Helson sets" are automatically "1-Carleson".


So I can't answer to your question in the $\ell\_1({\bf Z}\_+)$ case, but the answer is positive in the $\ell\_1({\bf Z})$-case.


