diff --git a/01-intro.tex b/01-intro.tex
index 942f85e..49cfdc0 100644
--- a/01-intro.tex
+++ b/01-intro.tex
@@ -8,16 +8,45 @@
 \section{The $\cC$-Albanese morphism in the presence of rational curves}
 \subversionInfo
 
-\begin{thm}\label{thm:1}
+\begin{thm}\label{thm:1}%
   Let $X$ be a projective manifold and let $x \in X$ be any point. If $C \subset
-  X$ is a rational curve, then the Albanese morphism of the $\cC$-pair $(X,0)$,
+  X$ is a rational curve, then the Albanese morphism\watchOut{Stefan 03Jun24:
+  Need to make assumptions to ensure that the Albanese exists.} of the
+  $\cC$-pair $(X,0)$,
   \[
     \alb_x(X,0) : X \to \Alb_x(X,0),
   \]
   maps the curve $C$ to a point.
 \end{thm}
 \begin{proof}
-  \todo{PENDING}
+  The normalization of $C$ yields a diagram 
+  \[
+    \begin{tikzcd}[column sep=2cm]
+      \bP¹ \ar[r, "n\text{, normalization}"'] \ar[rr, bend left=15, "\eta"] & C  \ar[r, "\text{inclusion}"'] & X.
+    \end{tikzcd}
+  \]
+  Consider the point $y := n(0_{\bP¹}) \in X$.  It follows from the universal
+  property of the Albanese that the Albanese varieties $\Alb_x(X,0)$ and
+  $\Alb_y(X,0)$ are isomorphic. To be more precise, there exists a unique Lie
+  group isomorphism $t$ that makes the following diagram commute,
+  \[
+    \begin{tikzcd}[column sep=2cm]
+      X \ar[r, "\alb_x(X{,}0)"] \ar[d, equal] & \Alb_x(X,0) \ar[d, two heads, hook, "t"] \\
+      X \ar[r, "\alb_y(X{,}0)"'] & \Alb_y(X,0).
+    \end{tikzcd}
+  \]
+  We may therefore assume without loss of generality that $x = y = n(0_{\bP¹})$.
+
+  Given that $X$ is smooth, recall from \cite{orbiAlb1} that $\eta$ is a
+  $\cC$-morphism, between $\cC$-pairs $(\bP¹, 0)$ and $(X, 0)$. Invoking the
+  universal property of the Albanese once more, we find an analogous diagram
+  \[
+    \begin{tikzcd}[column sep=2cm]
+      \bP¹ \ar[r, "\alb_0(\bP¹{,}0)"] \ar[d, equal] & \Alb_0(\bP¹,0) \ar[d] \\
+      X \ar[r, "\alb_x(X{,}0)"'] & \Alb_x(X,0).
+    \end{tikzcd}
+  \]
+  The claim follows immediately once we observe that $\Alb_0(\bP¹,0)$ is a point.
 \end{proof}
 
 \begin{itemize}