The functoriality of group C* algebra structure Let $G$ and $H$ be discrete groups and $f:G \rightarrow H$ be any homomorphism of these groups. I have three questions about it:
1) How to prove the functoriality of the construction of universal $C^*$-algebra of discrete
group (the existence of induced homomorphism $C^*(G) \to C^*(H)$)?
2) How to prove that the construction of reduced $C^*$-algebra of discrete group is not functorial (I am especially interested in counterexamples) and in which case (I mean conditions for group homomorphism) it'll be functorial? 
3) Let us consider the case when $G = \mathbb{Z}$ (integers) and $H = \mathbb{Z}/ n\mathbb{Z}$. How to describe the kernel and the image of the induced homomorphism of group $C^*$-algebras?
 A: Regarding Q2: the reduced $C^\star$-algebra is functorial wrt homomorphisms with amenable kernels. Indeed, let $N$ be a normal, amenable subgroup of $G$; since the trivial representation of $N$ is weakly contained in the regular representation of $N$ (by amenability), by continuity of induction the regular representation of $G/N$ is weakly contained in the regular representation of $G$, which means that the reduced $C^\star$-algebra of $G$ maps onto the one of $G/N$.
A: Actually, I'm not quite sure where this is explained completely basically in the literature.  So maybe here's a hint, at least for Q1.  The definition of $C^*(G)$ is that it's the completion of L^1(G) with respect to the biggest C*-norm.  So if $\phi:G\rightarrow H$ is a continuous group homo, we immediately get a *-homomorphism $\ell^1(G) \rightarrow \ell^1(H)$, and so by inclusion, a *-homo $\ell^1(G) \rightarrow C^*(H)$.  But this defines some C*-norm on $\ell^1(G)$, so the norm on $C^*(G)$ must dominate this, and hence we get a continuous extension to $C^*(G) \rightarrow C^*(H)$.
For Q2, find a proof in the literature (I think this goes back to Godemont?) that $C^*_r(G) = C^*(G)$ if and only if the left-regular representation contains the trivial representation, if and only if G is amenable.  Put another way, if G is not amenable, then the trivial homomorphism $G\rightarrow\{1\}$ doesn't induce a map $C^*_r(G)\rightarrow C^*(\{1\}) = \mathbb C$.
