Why is a contractive algebra homomorphism between C*-algebras necessarily a C*-homomorphism? Suppose Φ is an algebra homomorphism from a unital C*-algebra into another unital C*-algebra. How to prove that if $\| \Phi(x)\| \le \|x\|$ for all $x$, then $\Phi$ is a C*-homomophism? In other words: How to prove that $\Phi$ preserves involutions? This is an exercise from Kehe Zhu's "An introduction to operator algebras".   
 A: Because $\Phi(1)=1$ it follows that $\Phi(x^{-1})=\Phi(x)^{-1}$ for each invertible $x$ in the domain.  In particular, if $u$ is unitary, then $\Phi(u)$ and $\Phi(u^*)$ are inverses of one another with norm at most $1$.  If you will take for granted that $C^*$-algebras are $*$-isomorphic to $C^*$-algebras of operators on Hilbert space, this shows that $\Phi(u)$ is unitary and $\Phi(u^*)=\Phi(u)^{*}$.  (If $T\in B(H)$ is invertible and $\|Tx\|\neq \|x\|$ for some $x$, then either $T$ or $T^{-1}$ has norm greater than $1$.)
Because $\Phi$ preserves involutions on unitary elements and $C^*$-algebras are spanned by their unitary elements, $\Phi$ preserves involutions.

Added: Without appealing to representation on Hilbert space, but (implicitly) using Gelfand theory or other means of knowledge about positive elements on $C^*$-algebras, here's proof that if $\|x\|=\|x^{-1}\|=1$ then $x$ is unitary.  
Note that $\sigma(x^*x)^{-1}=\sigma((x^*x)^{-1})=\sigma(x^{-1}(x^{-1})^*)$.  Because $x^*x$ and $x^{-1}(x^{-1})^*$ are positive with norm $1$, they have spectrum contained in $(0,1]$.  Hence $\sigma(x^*x)=\{1\}$ which implies $x^*x=1$.
