\section{*-Homomorphisms}
\label{section:c-star-homomorphism}

\begin{definition}[*-Homomorphism]
\label{definition:c-star-homomorphism}
    Let $A, B$ be involutive algebras over $\complex$ and $\phi: A \to B$, then $\phi$ is a \textbf{*-homomorphism} if:
    \begin{enumerate}[label=(SH\arabic*)]
        \item For each $x, y \in A$ and $\lambda \in \complex$, $\phi(\lambda x + y) = \lambda \phi(x) + \phi(y)$. 
        \item For each $x, y \in A$, $\phi(xy) = \phi(x)\phi(y)$. 
        \item For each $x \in A$, $\phi(x^*) = \phi(x)^*$.
    \end{enumerate}

    If $A$ and $B$ are unital, then $\phi$ is \textbf{unital} if:
    \begin{enumerate}
        \item[(U)] $\phi(1_A) = 1_B$. 
    \end{enumerate}
\end{definition}


\begin{proposition}
\label{proposition:star-homomorphism-contractive}
    Let $A, B$ be unital $C^*$-algebras and $\phi: A \to B$ be a unital *-homomorphism, then for each $x \in A$, 
    \begin{enumerate}
        \item $\sigma_B(\phi(x)) \subset \sigma_A(x)$. 
        \item $\norm{\phi(x)}_B \le \norm{x}_A$. 
    \end{enumerate}
\end{proposition}
\begin{proof}
    (1): Since $\phi$ is unital, $\phi(G(A)) \subset G(B)$, so $\sigma_B(\phi(x)) \subset \sigma_A(x)$. 
    
    (2): By (1) and \autoref{corollary:c-star-unique-norm}, 
    \begin{align*}
    \norm{\phi(x)}_B^2 &= \sup\bracsn{|\lambda|\ | \lambda \in \sigma_B(\phi(x^*x))} \\
    &\ge \sup\bracsn{|\lambda|\ | \lambda \in \sigma_A(x^*x)} = \norm{x}_A^2
    \end{align*}
    
\end{proof}

\begin{theorem}
\label{theorem:continuity-of-homomorphism-c-star}
    Let $A, B$ be unital $C^*$-algebras and $\Phi: A \to B$ be a unital *-homomorphism, then $\Phi(A)$ is closed. 
\end{theorem}
\begin{proof}[Proof, {{\cite[Theorem 11.1]{Zhu}}}. ]
    Let $y \in \ol{\Phi(A)} \cap B_{sa}$, then there exists $x \in A_{sa}$ such that $\norm{y - \Phi(x)}_B \le \norm{y}_B/2$. Let
    \[
    f: \complex \to \complex \quad z \mapsto \begin{cases}
        z & |z| \le 2\norm{y}_F \\
        2\norm{y}_F \cdot \sgn z = 2\norm{y}_F \cdot \frac{z}{|z|} & |z| \ge 2\norm{y}_F
    \end{cases}
    \]

    then $f \in C(\complex; \complex)$. Since $\norm{\Phi(x)}_B \le \norm{y}_B + \norm{y - \Phi(x)}_B \le 2\norm{y}_B$, $\sigma_B(\Phi(x)) \subset \ol{B_\complex(0, 2\norm{y}_B)}$, and $f|_{\sigma_B(\Phi(x))}$ is the identity. Thus by the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus}, $\Phi(x) = f(\Phi(x)) = \Phi(f(x))$. By the \hyperref[Spectral Mapping Theorem]{theorem:spectral-mapping-continuous}, $\sigma_A(f(x)) = f(\sigma_A(x))$. By \autoref{theorem:c-star-normal-spectral-radius}, $\norm{f(x)}_A = [f(x)]_{sp} \le \norm{f}_u = 2\norm{y}_F$. 

    The above setup implies that for every $y \in \ol{\Phi(A)} \cap B_{sa}$, there exists $z \in A_{sa}$ such that $\norm{y - \Phi(z)}_{B} \le \norm{y}_B/2$, and $\norm{z}_A \le 2\norm{y}_B$. By the \hyperref[method of successive approximations]{theorem:successive-approximation}, $\phi(A_{sa}) = \ol{\Phi(A)} \cap B_{sa}$. Therefore $\Phi(A) = \ol{\Phi(A)}$. 
\end{proof}

\begin{definition}[Representation of $C^*$-Algebra]
\label{definition:representation-cstar-algebra}
    Let $A$ be a $C^*$-algebra, then a \textbf{representation} of $A$ is a pair $(H, \pi)$, where $H$ is a Hilbert space, and $\pi: A \to B(H)$ is a *-homomorphism. 
\end{definition}

\begin{definition}[Unitary Equivalence]
\label{definition:representation-unitary-equivalent}
    Let $A$ be a $C^*$-algebra and $(H_1, \pi_1), (H_2, \pi_2)$ be representations of $A$, then $(H_1, \pi_1)$ and $(H_2, \pi_2)$ are \textbf{unitarily equivalent} if there exists an isometry $U \in L(H_1; H_2)$ such that the following diagram commutes
    \[
    \xymatrix{
    H_1 \ar@{->}[r]^{U} \ar@{->}[d]_{\pi_1(x)} & H_2 \ar@{->}[d]^{\pi_2(x)} \\
    H_1 & H_2 \ar@{->}[l]^{U^*}
    }
    \]

    for all $x \in A$. 
\end{definition}