Slight adjustment.

src/topology/main/compactify.tex

@@ -27,7 +27,7 @@
 
     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:
+        \item[(U2)] For any compact Hausdorff space $Y$ and $\varphi \in C(X; Y)$, there exists a unique $\beta \varphi \in C(\beta X; Y)$ such that the following diagram commutes:
         \[
         \xymatrix{
         \beta X \ar@{->}[r]^{\beta \varphi} & Y \\
@@ -89,6 +89,7 @@
         X \ar@{->}[u]^{e} \ar@{->}[ru]_{f} & 
         }
         \]
+        \item[(F)] For any discrete space $Y$, $f: X \to Y$, and $\fU \in \beta X$, $\beta f(\fU)$ is the ultrafilter generated by $f(\fU)$. 
     \end{enumerate}
 
     In other words, $(\beta X, e)$ as constructed above is the Stone-Čech compactification of $X$. 
@@ -103,7 +104,7 @@
     \bigcap_{A \in F}U_{A^c} = \bigcap_{A \in F}U_A^c = \braks{\bigcup_{A \in F}U_A}^c \ne \emptyset
     \]
     
-    In particular, $\bigcap_{A \in F}A^c \ne \emptyset$. As such, $\bracsn{A^c|A \in \mathcal{S}}$ generates an ultrafilter $\fU \in \beta X$. Since $\fU \not\in U_A$ for all $A \in \mathcal{S}$, $\mathcal{U}$ does not cover $X$, which is a contradiction. Thus $\beta X$ is compact. 
+    In particular, $\bigcap_{A \in F}A^c \ne \emptyset$. By the \hyperref[ultrafilter lemma]{lemma:ultrafilter}, $\bracsn{A^c|A \in \mathcal{S}}$ generates an ultrafilter $\fU \in \beta X$. Since $\fU \not\in U_A$ for all $A \in \mathcal{S}$, $\mathcal{U}$ does not cover $X$, which is a contradiction. Thus $\beta X$ is compact. 
 
     Finally, since $X$ is discrete, $e$ is automatically continuous. On the other hand, for each $x \in X$, $U_{\bracsn{x}} = e(\bracsn{x})$ is open in $\beta X$. In addition, for any $\emptyset \ne A \subset X$, $e(x) \in U_A$ for all $x \in A$, so $U_A \cap e(X) \ne \emptyset$, and $e(X)$ is dense in $\beta X$. Therefore $e(X)$ is embedded as a dense subspace of $\beta X$. 
   

src/topology/main/stonean.tex

@@ -52,4 +52,29 @@
     Given that $\one_U \le F \le \one_{\ol{U}}$ and $F$ is continuous, $\bracs{F = 1} \supset \ol{U}$, so $F = \one_{\ol{U}}$ is continuous. Hence $\ol{U} = F^{-1}(1)$ must be open. 
 \end{proof}
 
+\begin{theorem}[Gleason, 1958]
+\label{theorem:gleason-extremally-disconnected}
+    Let $X$ be a compact Hausdorff space, then the following are equivalent:
+    \begin{enumerate}
+        \item $X$ is extremally disconnected. 
+        \item[(P)] For any compact Hausdorff spaces $Y, Z$, surjective continuous map $\pi \in C(Y; Z)$, and $f \in C(X; Z)$, there exists a lifting $\tilde f \in C(X; Y)$ such that the following diagram commutes: 
+        \[
+        \xymatrix{
+        & Y \ar@{->}[d]^{\pi} \\
+        X \ar@{->}[ru]^{\tilde f} \ar@{->}[r]^{f} & Z
+        }
+        \]
+        \item[(I)] For every pair of operator systems $S \subset T$ and completely positive map $\phi: S \to C(X; \complex)$, there exists a completely positive map $\Phi: T \to C(X; \complex)$ such that the following diagram commutes: 
+        \[
+        \xymatrix{
+        T \ar@{->}[r]^{\Phi} & C(X; \complex) \\
+        S \ar@{^{(}->}[u] \ar@{->}[ru]_{\phi} & 
+        }
+        \]
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    See \cite{GleasonProjective}. 
+\end{proof}
+