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Let $C^*(\langle a,b \rangle)$ be the full group $C^*$-algebra of free group on two letters $\langle a, b\rangle$. Is there an algebra morphism $C(S^1)\to C^*\left(\langle a,b \rangle\right)$ which extends $id_{S^1}\mapsto a$ ?

Note that it is clear what to do on polynomials, namely $f(z)=\sum_k t_kz^k$ is mapped to $\sum_k t_k a^k$.

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    $\begingroup$ Yes. $C(S^1)$ is the universal $C^{\ast}$-algebra generated by a unitary, so given any unital $C^{\ast}$-algebra $A$ and unitary $u\in A$, there is a homomorphism $\rho : C(S^1) \to A$ such that $\rho(z) = a$. $\endgroup$ Jun 3, 2017 at 13:24
  • $\begingroup$ Is the morphism $\rho$ uniquely determined by $\rho(z)=a$? @PrahladVaidyanathan $\endgroup$
    – user448378
    Jun 3, 2017 at 16:43
  • $\begingroup$ Yes, because $C(S^1)$ is generated by $z$ as a $C^{\ast}$-algebra. $\endgroup$ Jun 3, 2017 at 17:14

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Your idea works in general. In a C$^*$-algebra you have continuous functional calculus on normal operators (and $a$ is a unitary), so you can define $\gamma:C(S^1)\to C^*(\langle a,b\rangle)$ by $$ \gamma(f)=f(a). $$ This works because, with $a$ a unitary, you have $\sigma(a)\subset S^1$ (it is actually an equality, but we don't need it here).

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