Adjusted prose in example facts.

.vscode/project.code-snippets

@@ -171,5 +171,10 @@
   "scope": "latex",
     "prefix": "rank",
     "body": ["\\text{rk}"]
+  },
+  "Cite Proof": {
+  "scope": "latex",
+    "prefix": "cproof",
+    "body": ["{Proof, {{\\cite[$1]{$2}}}. }$0"]
   }
 }

src/op/banach/fc.tex

@@ -23,12 +23,18 @@
         \]
 
         for all $f \in H(U; \complex)$. 
+        \item For each $U \subset \complex$ open and $f \in H(U; \complex)$, the mapping
+        \[
+        \bracs{x \in A| \sigma_A(x) \subset U} \to A \quad x \mapsto f(x)
+        \]
+
+        is continuous. 
     \end{enumerate}
 
     The mapping $f \mapsto f(x)$ is the \textbf{holomorphic functional calculus} of $x$. 
 \end{definition}
 \begin{proof}[Proof, {{\cite[Proposition I.2.7]{Takesaki1}}}. ]
-    (Definition): Let $U, V \in \cn_\complex(\sigma_A(x))$ such that $\ol V$ is a compact subset of $U$, then by \autoref{proposition:existence-curves}, there exists closed rectifiable curves $\seqf{\gamma_j}$ on $V \setminus \sigma_A(x)$ such that 
+    (Definition), (3): Let $U, V \in \cn_\complex(\sigma_A(x))$ such that $\ol V$ is a compact subset of $U$, then by \autoref{proposition:existence-curves}, there exists closed rectifiable curves $\seqf{\gamma_j}$ on $V \setminus \sigma_A(x)$ such that 
     \begin{enumerate}[label=(\alph*)]
         \item For all $f \in H(V; \complex)$ and $z_0 \in \sigma_A(x)$, 
         \[
@@ -89,7 +95,6 @@
     \]
 
     (Uniqueness): By (2), the homomorphism extends uniquely to $\complex(z) \cap H(\sigma_A(x); \complex)$. By \hyperref[Runge's Theorem]{corollary:runge-rational-approximation}, it extends uniquely to $H(\sigma_A(x); \complex)$ by continuity. 
-
 \end{proof}
 
 \begin{theorem}[Spectral Mapping Theorem (Holomorphic)]
@@ -103,7 +108,7 @@
 \begin{proof}
     (1): Let $\lambda \in \complex \setminus f(\sigma_A(x))$, then $1/(\lambda - f) \in H(\sigma_A(x); \complex)$. Since the holomorphic functional calculus is a homomorphism, $[1/(\lambda - f)](x) = (\lambda - f(x))^{-1}$. Thus $\sigma_A(f(x)) \subset f(\sigma_A(x))$. 
 
-    On the other hand, let $\lambda \in \sigma_A(x)$, then the function $h(\mu) = [f(\mu) - f(\lambda)]/(\mu - \lambda)$ extends to an element of $H(\sigma_A(x); \complex)$ by \autoref{proposition:zero-finite-multiplicity}. Therefore by the homomorphism property, 
+    On the other hand, let $\lambda \in \sigma_A(x)$, then the function $h(\mu) = [f(\mu) - f(\lambda)]/(\mu - \lambda)$ extends to an element of $H(\sigma_A(x); \complex)$ by \autoref{proposition:zero-finite-multiplicity}. By the homomorphism property, 
     \[
     f(x) - f(\lambda) = (x - \lambda)h(x) = h(x)(x - \lambda)
     \]
@@ -120,7 +125,7 @@
     
     By \hyperref[Runge's Theorem]{corollary:runge-rational-approximation}, there exists $h \in U \cap \complex(z)$. Thus $\norm{g(f(x)) - (g \circ f)(x)}_A < \eps$ as well. As this holds for all $\eps > 0$, $g(f(x)) = (g \circ f)(x)$ for any $f \in H(\sigma_A(x); \complex)$ and $g \in \complex(x) \cap H(f(\sigma_A(x)); \complex)$. 
     
-    Through a second application of \hyperref[Runge's Theorem]{corollary:runge-rational-approximation} again and continuity of the holomorphic functional calculus, the above also holds for all $f \in H(\sigma_A(x); \complex)$ and $g \in H(f(\sigma_A(x)); \complex)$. 
+    Through a second application of \hyperref[Runge's Theorem]{corollary:runge-rational-approximation} and continuity of the holomorphic functional calculus, the above also holds for all $f \in H(\sigma_A(x); \complex)$ and $g \in H(f(\sigma_A(x)); \complex)$. 
 \end{proof}
 
 \begin{proposition}

src/op/banach/gelfand.tex

@@ -0,0 +1,60 @@
+\section{The Gelfand Transform}
+\label{section:gelfand-transform}
+
+\begin{definition}[Gelfand Transform]
+\label{definition:gelfand-transform}
+    Let $A$ be a unital Banach algebra, then the \textbf{Gelfand transform} is the homomorphism
+    \[
+    \Gamma = \Gamma_A: A \to C(\Omega(A); \complex) \quad (\Gamma_Ax)(\varphi) = \varphi(x)
+    \]
+\end{definition}
+
+\begin{remark}
+\label{remark:gelfand-transform}
+    The Gelfand transform is limited in studying arbitrary Banach algebras, as they may admit no multiplicative functionals. However, these functionals come in abundance in the commutative case. 
+\end{remark}
+
+
+\begin{proposition}
+\label{proposition:gelfand-transform-gymnastics}
+    Let $A$ be a commutative unital Banach algebra and $x \in A$, then:
+    \begin{enumerate}
+        \item $\Gamma_A$ is a contractive homomorphism. 
+        \item $\Gamma_A(1) = 1$. 
+        \item $x \in G(A)$ if and only if $\Gamma_A x \in G(C(\Omega(A); \complex))$. 
+        \item $(\Gamma_Ax)(\Omega(A)) = \sigma_A(x)$. 
+        \item $\norm{\Gamma_Ax}_u = [x]_{sp}$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Theorem 1.1.13]{FollandHarmonic}}}. ]
+    (2): For each $\phi \in \Omega(A)$, $\phi(1) = 1$, so $\Gamma_A(1) = 1$. 
+
+    (3): Since $A$ is commutative, $x \not\in G(A)$ if and only if the ideal generated by $x$ is proper, if and only if there exists a maximal ideal containing $x$, if and only if there exists $\phi \in \Omega(A)$ with $\phi(x) = 0$. 
+
+    (4): By (1) and (3), 
+    \[
+    (\Gamma_Ax)(\Omega(A)) = \sigma_{C(\Omega(A); \complex)}(\Gamma x) = \sigma_A(x)
+    \]
+\end{proof}
+
+\begin{proposition}
+\label{proposition:gelfand-isometric}
+    Let $A$ be a commutative unital Banach algebra, then the following are equivalent:
+    \begin{enumerate}
+        \item For each $x \in A$, $\normn{x^2}_A = \norm{x}_A^2$. 
+        \item $\Gamma_A$ is an isometry. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1) $\Rightarrow$ (2): For each $x \in A$, by the \hyperref[spectral radius formula]{proposition:spectral-radius-hadamard} and (5) of \autoref{proposition:gelfand-transform-gymnastics},
+    \[
+    \norm{\Gamma_A x}_u = [x]_{sp} = \norm{x}_A
+    \]
+
+    (2) $\Rightarrow$ (1): For each $x \in A$, by (5) of \autoref{proposition:gelfand-transform-gymnastics}, 
+    \[
+    \normn{x^2}_A \ge [x^2]_{sp} = \normn{\Gamma_A x^2}_u = \normn{\Gamma_A x}_u^2 = \normn{x}_A^2
+    \]
+\end{proof}
+
+

src/op/banach/index.tex

@@ -8,3 +8,4 @@
 \input{./spectrum.tex}
 \input{./fc.tex}
 \input{./multiplicative.tex}
+\input{./gelfand.tex}

src/op/banach/spectrum.tex

@@ -77,7 +77,7 @@
     Let $x \in A$. By \autoref{proposition:spectrum-non-empty}, there exists $\lambda \in \sigma_A(x)$. Since $\lambda \cdot 1 - x$ is not invertible and every non-zero element of $A$ is invertible, $x = \lambda \cdot 1$. Therefore the mapping $\complex \to A$ defined by $\lambda \mapsto \lambda \cdot 1$ is an isometric isomorphism. 
 \end{proof}
 
-\begin{proposition}
+\begin{proposition}[Spectral Radius Formula]
 \label{proposition:spectral-radius-hadamard}
     Let $A$ be a unital Banach algebra and $x \in A$, then $[\cdot]_{sp} = \limsup_{n \to \infty}\normn{x^n}_A^{1/n}$. 
 \end{proposition}
@@ -126,4 +126,24 @@
     Let $x \in A$ with $\sigma_A(x) \subset U$, and $\lambda \in U^c$. By \autoref{proposition:banach-algebra-inverse}, for any $y \in A$ with $\norm{y}_A \le \normn{(\lambda - x)^{-1}}_A^{-1}$, $\lambda - x - y \in G(A)$ as well. Since the mapping $\lambda \mapsto \normn{(\lambda - x)^{-1}}_A$ vanishes at infinity and $U^c$ is closed, $\delta = \inf_{\lambda \in U^c}\normn{(\lambda - x)^{-1}}_A^{-1} > 0$. Therefore for every $z \in B_A(x,\delta)$, $\sigma_A(x) \subset U$. 
 \end{proof}
 
+\begin{proposition}
+\label{proposition:commutative-spectrum-gymnastics}
+    Let $A$ be a unital Banach algebra and $x, y \in A$ with $x = y$, then
+    \begin{enumerate}
+        \item $\sigma_A(x + y) \subset \sigma_A(x) + \sigma_A(y)$.
+        \item $\sigma_A(xy) \subset \sigma_A(x)\sigma_A(y)$.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    For any $z \in A$, denote $R(z) = \bracs{(\lambda - z)^{-1}|\lambda \in \sigma_A(z)}$. Let $B \subset A$ be the closed subalgebra generated by $1$, $x$, $y$, $R(x)$, $R(y)$, $R(xy)$, $R(x + y)$, then $B$ is a commutative algebra with $\sigma_B(x) = \sigma_A(x)$, $\sigma_B(y) = \sigma_A(y)$, $\sigma_B(xy) = \sigma_A(xy)$, and $\sigma_B(x + y) = \sigma_A(x + y)$. 
+
+    By \autoref{proposition:gelfand-transform-gymnastics}, for each $z \in B$, $(\Gamma_Bz)(\Omega(B)) = \sigma_B(z)$. Since for any $u, v \in C(\Omega(B); \complex)$, 
+    \begin{enumerate}
+        \item $\sigma_{C(\Omega(B); \complex)}(u + v) \subset \sigma_{C(\Omega(B); \complex)}(u) + \sigma_{C(\Omega(B); \complex)}(v)$.
+        \item $\sigma_{C(\Omega(B); \complex)}(uv) \subset \sigma_{C(\Omega(B); \complex)}(u)\sigma_{C(\Omega(B); \complex)}(v)$.
+    \end{enumerate}
+    
+    The above holds for $x$ and $y$ with respect to $\sigma_A$. 
+\end{proof}
+
 

src/op/example/bc.tex

@@ -0,0 +1,36 @@
+\section{$BC(X)$}
+\label{section:bounded-continuous-functions-algebra}
+
+\begin{definition}[Algebra of Bounded Continuous Functions]
+\label{definition:bc-algebra}
+    Let $X$ be a topological space, then $BC(X; \complex)$ is the \textbf{space of bounded continuous functions} on $X$. 
+    
+    Equipped with pointwise operations and the uniform norm, $BC(X; \complex)$ forms a commutative Banach algebra. 
+\end{definition}
+
+\begin{theorem}
+\label{theorem:multiplicative-functional-bc}
+    Let $X$ be a completely regular space and $\beta X$ be its \hyperref[Stone-Čech compactification]{definition:stone-cech}. For each $f \in BC(X; \complex)$, let $\beta f \in BC(\beta X; \complex)$ be its unique extension to $\beta X$, then the mapping
+    \[
+    E: \beta X \to \Omega(BC(X; \complex)) \quad E(x)(f) = (\beta f)(x)
+    \]
+
+    is a homeomorphism. Under the identification $\beta X = \Omega(BC(X; \complex))$, $\Gamma_{BC(X; \complex)} = \beta$. 
+\end{theorem}
+\begin{proof}
+    Let $\phi \in BC(X; \complex)^* \setminus \ol{E(X)}$, then there exists $\seqf{f_k} \subset BC(X; \complex)$ and $\eps > 0$ such that for every $x \in X$, 
+    \[
+    f(x) = \sum_{k = 1}^n |f_k(x) - \dpn{f_k, \phi}{BC(X; \complex)}|^2 \ge \eps^2
+    \]
+
+    In which case, if $\phi \in \Omega(BC(X; \complex))$, then 
+    \[
+        \dpn{f, \phi}{BC(X; \complex)} = \sum_{k = 1}^n \phi(f_k - \phi(f_k))\phi(\ol{f_k - \phi(f_k)}) = 0
+    \]
+
+    Since $f(x) > 0$ for all $x \in X$, $f \in G(BC(X; \complex))$. As $\phi \in \Omega(BC(X; \complex))$, the above contradicts (3) of \autoref{proposition:multiplicative-unit}. Thus $\Omega(BC(X; \complex)) \setminus \ol{E(X)}$ is empty, and $E(X)$ is dense in $\Omega(BC(X; \complex))$. 
+
+    By (U) of the \hyperref[Stone-Čech compactification]{definition:stone-cech}, $E|_X$ extends uniquely to a surjective continuous map. Since $BC(\beta X; \complex)$ and $BC(X; \complex)$ are in bijection via the restriction map, $E$ is injective. Therefore $E$ is bijective, and $E$ is a homeomorphism by \autoref{proposition:compact-hausdorff-homeomorphism}. 
+\end{proof}
+
+

src/op/example/convolution.tex

@@ -0,0 +1,87 @@
+\section{$\ell^1(\integer)$}
+\label{section:convolution-algebra-integer}
+
+\begin{definition}[$\ell^1(\integer)$]
+\label{definition:convolution-algebra-integer}
+    Let $\ell^1(\integer)$ be the $\ell^1$ sequence space on $\integer$. For each $f, g \in \ell^1(\integer)$, let
+    \[
+    (f * g)(n) = \sum_{k \in \integer}f(k)g(n - k)
+    \]
+
+    then:
+    \begin{enumerate}
+        \item $\ell^1(\integer)$ is a commutative Banach algebra.
+        \item The multiplicative unit of $\ell^1(\integer)$ is $\delta_0 = \one_{\bracs{n = 0}}$. 
+    \end{enumerate}
+
+    The space $\ell^1(\integer)$ is the \textbf{convolution algebra} on $\integer$. 
+\end{definition}
+\begin{proof}
+    For each $f, g \in \ell^1(\integer)$, by \hyperref[Fubini's Theorem]{theorem:fubini-tonelli},
+    \begin{align*}
+        \normn{f * g}_{\ell^1(\integer)} &= \sum_{n \in \integer} \abs{\sum_{k \in \integer}f(k)g(n - k)} \\
+        &\le \sum_{n, k \in \integer}|f(k)| \cdot |g(n-k)| \le \sum_{k \in \integer}|f(k)| \cdot \sum_{n \in \integer}|g(n - k)| \\
+        &= \norm{f}_{\ell^1(\integer)} \cdot \norm{g}_{\ell^1(\integer)}
+    \end{align*}
+\end{proof}
+
+
+\begin{proposition}
+\label{proposition:convolution-integer-gelfand}
+    The Gelfand transform of $\ell^1(\integer)$ is not isometric. 
+\end{proposition}
+\begin{proof}
+    Let $f = \one_{\bracs{n = 1}} - \one_{\bracs{2 \le n \le 3}}$, then
+    \[
+    f^2(n) = \begin{cases}
+        -1 &n \in \bracs{1, 5} \\
+        -2 &n = 2 \\
+        -1 &n = 3 \\
+        2 &n = 4 \\
+        0 &n \not\in [1, 5] 
+    \end{cases}
+    \]
+
+    so $\normn{f^2}_{\ell^1(\integer)} = 7 < \normn{f}_{\ell^1(\integer)}^2$. By \autoref{proposition:gelfand-isometric}, the Gelfand transform is not isometric. 
+\end{proof}
+
+\begin{theorem}
+\label{theorem:convolution-integer-gelfand}
+    Let $\ell^1(\integer)$ be the convolution algebra on $\integer$, then the mapping
+    \[
+    F: \mathbf{S}^1 = \partial B_\complex(0, 1) \to \Omega(\ell^1(\integer)) \quad
+    (F(z))(f) = \sum_{n \in \integer}f(n)z^n
+    \]
+
+    is a homeomorphism. Under the identification $\partial B_\complex(0, 1) = \Omega(\ell^1(\integer))$, the Gelfand transform has the explicit form
+    \[
+    \Gamma(f)(z) = \sum_{n \in \integer} f(n)z^n
+    \]
+\end{theorem}
+\begin{proof}{Proof, {{\cite[Theorem 6.3]{Zhu}}}. }
+    Let $z \in \mathbf{S}^1$ and $f, g \in \ell^1(\integer)$, then by \hyperref[Fubini's Theorem]{theorem:fubini-tonelli},
+    \begin{align*}
+        F(z)(f * g) &= \sum_{n \in \integer}z^n \sum_{k \in \integer}f(k)g(n-k) \\
+        &= \sum_{k \in \natp}f(k)z^k \sum_{n \in \integer}z^{n-k}g(n-k) \\
+        &= F(z)(f) \cdot F(z)(g)
+    \end{align*}
+
+    so $F(z)$ is a multiplicative functional for all $z \in \mathbf{S}^1$. 
+
+    For each $n \in \integer$, let $\delta_n = \one_{\bracs{k = n}}$, then for any $m, n \in \integer$, $\delta_m * \delta_n = \delta_{m + n}$. Let $\phi \in \Omega(\ell^1(\integer))$ and $z = \phi(\delta_1)$, then $\phi(\delta_n) = z^n$ for all $n \in \integer$. Since the span of $\bracsn{\delta_n|n \in \integer}$ is dense in $\ell^1(\integer)$, $\phi = F(z)$ by the \hyperref[Dominated Convergence Theorem]{theorem:dct}.
+
+    Finally, by the \hyperref[Dominated Convergence Theorem]{theorem:dct}, $F: \textbf{S}^1 \to \Omega(\ell^1(\integer))$ is continuous. Since $\textbf{S}^1$ is compact, $\Omega(\ell^1(\integer))$ is Hausdorff, and $F$ is a bijection, $F$ is a homeomorphism by \autoref{proposition:compact-hausdorff-homeomorphism}. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:convolution-integer-spectrum}
+    Let $\ell^1(\integer)$ be the convolution algebra on $\integer$ and $f \in \ell^1(\integer)$, then
+    \[
+    \sigma(f) = \bracs{\sum_{n \in \integer}f(n)z^n \bigg | z \in \partial B_\complex(0, 1)}
+    \]
+\end{proposition}
+\begin{proof}
+    By \autoref{theorem:convolution-integer-gelfand} and (4) of \autoref{proposition:gelfand-transform-gymnastics}.
+\end{proof}
+
+

src/op/example/disk.tex

@@ -13,10 +13,17 @@
 
 \begin{proposition}
 \label{proposition:disk-algebra-multiplicative-functional}
-    Let $\phi \in A(D)^*$ be a multiplicative linear functional, then there exists $z_0 \in \ol D$ such that for every $f \in A(D)$, $\phi(f) = f(z_0)$. 
+    Let $D = B_\complex(0, 1)$ and $A(D)$ be the disk algebra, then the mapping
+    \[
+    E: \ol D \to \Omega(D) \quad E(z_0)(f) = f(z_0)
+    \]
+
+    is a homeomorphism. 
 \end{proposition}
 \begin{proof}
-    By \autoref{proposition:multiplicative-unit}, $\norm{\phi}_{A(D)^*} = 1$. Let $p$ be the identity polynomial, and $z_0 = \phi(p)$, then for every $q \in \complex[x]$, $\phi(q) = q(x_0)$. By density of polynomials in $A(D)$, $\phi(f) = f(z_0)$ for all $f \in A(D)$. 
+    Let $\phi \in \Omega(D)$. By \autoref{proposition:multiplicative-unit}, $\norm{\phi}_{A(D)^*} = 1$. Let $p$ be the identity polynomial, and $z_0 = \phi(p)$, then for every $q \in \complex[x]$, $\phi(q) = q(x_0)$. By density of polynomials in $A(D)$, $\phi(f) = f(z_0)$ for all $f \in A(D)$. Therefore $\ol D$ is in bijection with $\Omega(D)$. 
+
+    Since $A(D) \subset C(\ol D)$, $E$ is continuous. As $\ol D$ is compact and $\Omega(D)$ is Hausdorff, $E$ is a homeomorphism by \autoref{proposition:compact-hausdorff-homeomorphism}. 
 \end{proof}
 
 

src/op/example/hardy.tex

@@ -3,7 +3,7 @@
 
 \begin{definition}[Hardy Space]
 \label{definition:hardy-space}
-    Let $D = B_\complex(0, 1)$, then the \textbf{Hardy space} $H^\infty(D)$ is the space of all bounded holomorphic functions on $D$, equipped with the uniform norm. 
+    Let $D = B_\complex(0, 1)$, then the \textbf{Hardy space} $H^\infty(D)$ is the algebra of all bounded holomorphic functions on $D$, equipped with the uniform norm. 
 \end{definition}
 
 \begin{proposition}
@@ -23,12 +23,11 @@
         \item If $\fU \to x_0 \in \ol{D}$, then $f \in A(D)$, $\phi_{\fU}(f) = f(z_0)$. 
         \item $\phi_{\fU}$ is a multiplicative linear functional on $H^\infty(D)$. 
     \end{enumerate}
-    
 \end{proposition}
 \begin{proof}
     Let $f \in H^\infty(D)$, then by \autoref{proposition:imagefilterbase}, $f(\fU)$ is an ultrafilter base. Since $f$ is bounded, $f(\fU)$ converges to exactly one element of $\complex$. Hence the limit is well-defined. 
 
-    (2): By \autoref{proposition:operator-space-completeness}. 
+    (2): By \autoref{definition:multiplicative-linear-functional-space}. 
 \end{proof}
 
 

src/op/example/index.tex

@@ -5,4 +5,5 @@
 \input{./bounded.tex}
 \input{./hardy.tex}
 \input{./disk.tex}
-
+\input{./convolution.tex}
+\input{./bc.tex}

src/op/notation.tex

@@ -13,8 +13,12 @@
     $[x]_{sp}$ & The spectral radius of $x$. & \autoref{definition:spectral-radius} \\
     $\Omega(A)$ & Space of multiplicative functionals on $A$. & \autoref{definition:multiplicative-functional} \\
     $\cm(A)$ & Maximal ideal space of $A$. & \autoref{definition:maximal-ideal} \\
+    $\Gamma = \Gamma_A$ & The Gelfand transform on $A$. & \autoref{definition:gelfand-transform} \\ 
+
     $M_n(\complex)$ & Algebra of $n \times n$ matrices over $\complex$.  & \autoref{definition:matrix-algebra} \\
     $B(H)$ & Algebra of bounded operators on a Hilbert space. & \autoref{definition:hilbert-endomorphism} \\
     $A(D)$ & The disk algebra. & \autoref{definition:disk-algebra} \\
     $H^\infty(D)$ & The Hardy space.  & \autoref{definition:hardy-space} \\
+    $\ell^1(\integer)$ & Convolution algebra on $\integer$. &\autoref{definition:convolution-algebra-integer} \\ 
+    $\delta_0$ & Multiplicative unit of $\ell^1(\integer)$. & \autoref{definition:convolution-algebra-integer} \\ 
 \end{tabular}

src/topology/main/compact.tex

@@ -70,6 +70,15 @@
     Let $x \in E^c$. For each $y \in E$, there exists $U_y \in \cn(y)$ and $V_y \in \cn(x)$ such that $U_y \cap V_y = \emptyset$. By compactness, there exists $E_0 \subset E$ finite such that $\bigcup_{y \in E_0}U_y \supset E$. In which case, $E \cap \bigcap_{y \in E_0}V_y = \emptyset$ and $\bigcap_{y \in E_0}V_y \in \cn(x)$. Thus $E^c$ is open.
 \end{proof}
 
+\begin{proposition}
+\label{proposition:compact-hausdorff-homeomorphism}
+    Let $X$ be a compact topological space, $Y$ be a Hausdorff space, and $f \in C(X; Y)$ be a bijection, then $f$ is a homeomorphism. 
+\end{proposition}
+\begin{proof}
+    For each $K \subset X$ closed, $K$ and $f(K)$ are compact by \autoref{proposition:compact-extensions}. By \autoref{proposition:compact-closed}, $f(K)$ is closed. Thus $f^{-1}$ maps closed sets to closed sets, and hence open sets to open sets.
+\end{proof}
+
+
 \begin{proposition}
 \label{proposition:compact-hausdorff-normal}
     Let $X$ be a Hausdorff space, $A, B \subset X$ be compact with $A \cap B = \emptyset$, then there exists $U \in \cn(A)$ and $V \in \cn(B)$ such that $U \cap V = \emptyset$.

src/topology/main/compactify.tex

@@ -25,7 +25,7 @@
         \]
     \end{enumerate}
 
-    Moreover, if $(\beta X, e)$ is \textit{any} pair that satisfies (1) and (U1), then
+    Moreover, if $(\beta X, e)$ is \textit{any} pair that satisfies (U1), then
     \begin{enumerate}
         \item[(U2)] For any pair $(Y, \varphi)$ satisfying (1), there exists a unique $\beta \varphi \in C(\beta X; Y)$ such that the following diagram commutes:
         \[