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\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}