Restricting and extending *-homomorphisms with norm-dense subalgebras

Question

Let $G$ and $H$ be discrete groups, and suppose that I have a surjective *-homomorphism: $$\pi:\mathrm{C}^*(G)\to C_{\text{r}}(H).$$

Is it true that the restriction to the group algebra $\pi_{|_{\mathbb{C}G}}$ maps onto $\mathbb{C}H$? Then I feel like I can extend $\pi_{|_{\mathbb{C}G}}$ to a surjective *-homomorphism $\tilde{\pi}:\mathrm{C}^*(G)\to\mathrm{C}^*(H)$, but the tension between $\pi$ and $\tilde{\pi}$ makes me feel like something has gone wrong.

Is the problem here with the restriction, or the extension (I don't think so), something else, or does this actually work?

The actual problem I have is different, but I feel an answer to the above should help. Speaking generally, what I have is a surjective *-homomorphism of unital $\mathrm{C}^*$-algebras $\pi:A_\text{u}\to B_{\text{r}}$. Both $A_{\text{u}}$ and $B_{\text{r}}$ have canonical dense *-algebras $\mathcal{A}$ and $\mathcal{B}$, with $A_{\text{u}}$ a maximal completion of $\mathcal{A}$ and $B_\text{r}$ in some sense a minimal completion of $\mathcal{B}$. I want to restrict $\pi$ to $\pi_{|_{\mathcal{A}}}$ to give a *-homomorphism $\mathcal{A}\to\mathcal{B}$ and then extend to $A_{\text{u}}\to B_{\text{u}}$.

(In fact $A_\text{u}$ is a certain universal $\mathrm{C}^*$-algebra and $B_{\text{r}}=C_{\text{r}}(S_N^+)$, and I am trying to show that either $A_{\text{u}}=C_\text{u}(S_N^+)$ or it doesn't admit a compact quantum group structure... because if it did admit a compact quantum group structure, say $A_{\text{u}}=C_{\text{u}}(\mathbb{G})$, then it would admit a dense Hopf*-algebra $\mathcal{O}(\mathbb{G})$, and we would end up with $\mathbb{G}=S_N^+$ after all if we could restrict and extend as above.)

Answer 1

From the context in the question and the comments, it appears that the OP is after the following result:

Proposition: Let $\mathbb{G}$ and $\mathbb{H}$ be two compact quantum groups and $\pi: C(\mathbb{G})\to C(\mathbb{H})$ a surjective unital, coproduct-preserving $*$-homomorphism. Then $\pi(\mathcal{O}(\mathbb{G}))= \mathcal{O}(\mathbb{H})$ with $\mathcal{O}(-)$ the dense Hopf $*$-subalgebras.

Proof: If $x\in \mathcal{O}(\mathbb{G})$, we find a finite-dimensional representation $U\in B(K)\odot \mathcal{O}(\mathbb{G})$ and $\omega \in B(K)_*$ such that $x= (\omega \odot \iota)(U)$. Then we have $$\pi(x) = \pi((\omega \odot \iota)(U))= (\omega\odot \iota)(\iota \odot \pi)(U)\in \mathcal{O}(\mathbb{H})$$ where we used that $(\iota \odot \pi)(U)$ is a representation of $\mathbb{H}$ since $$(\iota \odot\Delta_{\mathbb{H}})((\iota \odot \pi)(U)) = (\iota \odot \pi \odot \pi)(\iota \odot \Delta_{\mathbb{G}})(U) = (\iota \odot \pi\odot \pi)(U_{12}U_{13}) = (\iota \odot \pi)(U)_{12}(\iota \odot \pi)(U)_{13}.$$ This shows that $\pi(\mathcal{O}(\mathbb{G}))\subseteq \mathcal{O}(\mathbb{H})$.

Next, note that $A=\pi(\mathcal{O}(\mathbb{G}))$ is dense in $C(\mathbb{H})$ and it is a Hopf $^*$-subalgebra of $C(\mathbb{H})$ (meaning that $\Delta_{\mathbb{H}}(A)\subseteq A\odot A$ and with the restriction of the coproduct $A$ becomes a Hopf $^*$-algebra). Necessarily, it must follow that $A=\mathcal{O}(\mathbb{H})$, by Theorem 5.1 in this paper.

Answer 2

For completeness, let me give a counter example to the question stated in the question. Consider $G=H=\mathbb{Z}$ so $C^*(G), C^*_r(H)$ are both identified with $C(S^1)$, where $S^1$ denotes the unit circle $\{z\in\mathbb{C}: |z|=1\}$. Moreover, the group algebras $\mathbb{C}G$ and $\mathbb{C}H$ are both identified with the dense $*$-subalgebra of trigonometric polynomials, i.e. the subalgebra of functions of the form $z\mapsto\sum_{k=-N}^N\alpha_kz^k$, where $N\in\mathbb{N}$ and $\alpha_k\in\mathbb{C}$, i.e. if $\zeta$ denotes the identity map $\zeta(z):=z$, then $A$ is the linear span of $\{\zeta^k: k\in\mathbb{Z}\}$.

Now take even some automorphism of $C(S^1)$, which by Gelfand duality is necessarily of the form $h_*:C(S^1)\to C(S^1)$, where $h\in\mathrm{Homeo}(S^1)$, i.s. $h_*(f)=f\circ h$ for all $f\in C(S^1)$. Clearly, $h_*(\zeta)=h$, since $$h_*(\zeta)(z)=(\zeta\circ h)(z)=\zeta(h(z))=h(z)$$ so a counter-example is immediate if one takes $h\in\mathrm{Homeo}(S^1)$ be a map that is not a trigonometric polynomial. Now I can't write down any such $h$ but they surely must exist... (e.g. take any weird self-map of a small interval that preserves the endpoints, this will induce a homeomorphism of an arc of the circle that preserves the endpoints, then extend this by identity on the rest of the circle.. more here)