Kernel inclusion implies the norm inequality

The map $T$ on $\rho (C^{\ast }(G))$ defined by $T(\rho (a)):=\pi (a)$ is a well-defined algebraic star-homomorphism: if $\rho (a)=0$ then $a\in \ker \rho \subseteq \ker \pi$, so $\pi (a)=0$; linearity, multiplicativity and star preservation follow from the corresponding properties of the extended representations, and the definition is compatible with sums and products because $\rho$ and $\pi$ are star-homomorphisms. If $\rho =0$, kernel inclusion forces $\pi =0$ and $T=0$ is bounded. Otherwise factor $\pi$ through $C^{\ast }(G)\mathrm{/}\ker \rho$ using the quotient factorization in [F1]; the induced bounded map $\dot{\pi }$ satisfies $\mathrm{\parallel }\dot{\pi }\mathrm{\parallel }\le \mathrm{\parallel }\pi \mathrm{\parallel }$ by taking the infimum over coset representatives. Since $\dot{\rho }$ is an isometry onto its image, $T=\dot{\pi }\circ {\dot{\rho }}^{-1}$ is bounded. It is therefore a star-homomorphism in the library's bounded sense.

The homomorphism $T$ is contractive by [F2], so for every $a\in C^{\ast }(G)$ one has $\mathrm{\parallel }\pi (a)\mathrm{\parallel }=\mathrm{\parallel }T(\rho (a))\mathrm{\parallel }\le \mathrm{\parallel }\rho (a)\mathrm{\parallel }$.

The Axiom of Choice is inherited from the quotient and representation-correspondence suppliers; the factorization uses no further choice (The Axiom of Choice).