diff --git a/src/fa/norm/normed.tex b/src/fa/norm/normed.tex
index a3a5c1e..3238188 100644
--- a/src/fa/norm/normed.tex
+++ b/src/fa/norm/normed.tex
@@ -68,7 +68,7 @@
 \end{proof}
 
 
-\begin{theorem}[Successive Approximation]
+\begin{theorem}[Successive Approximations]
 \label{theorem:successive-approximation}
     Let $E, F$ be normed vector spaces, $T \in L(E; F)$, $C \ge 0$, and $\gamma \in (0, 1)$. If for all $y \in F$, there exists $x \in E$ such that:
     \begin{enumerate}
@@ -84,7 +84,7 @@
 
     In particular, if $E$ is a Banach space, then for every $y \in F$, there exists $x \in E$ such that $\norm{x}_E \le C\norm{y}_F/(1 - \gamma)$ and $Tx = y$.
 \end{theorem}
-\begin{proof}
+\begin{proof}[Proof, learned from Anson Li (https://ansonli0.com/). ]
     Let $y_1 = y \in F$. Let $n \in \natp$ and suppose inductively that $\bracs{x_k| 0 \le k < n}$ and $\bracs{y_k| 0 \le k \le n}$ has been constructed. By (a) and (b), there exists $x_n \in E$ such that $\norm{x_k}_E \le C\norm{y_k}_F$ and $\norm{y_n - Tx_n}_F \le \gamma \norm{y_{n}}_F$. Let $y_{n+1} = y_n - Tx_n$, then $\norm{y_{n+1}}_F \le \gamma \norm{y_n}_F$.
 
     For each $n \in \nat$,
diff --git a/src/op/c-star/homomorphism.tex b/src/op/c-star/homomorphism.tex
index 429062f..f35d588 100644
--- a/src/op/c-star/homomorphism.tex
+++ b/src/op/c-star/homomorphism.tex
@@ -41,6 +41,17 @@
     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}