diff --git a/sections/introduction.tex b/sections/introduction.tex
@@ -554,37 +554,37 @@ containing \(\mathfrak{g}\) as a Lie subalgebra. In practice this means\dots
\begin{proof}
Let \(f : \mathfrak{g} \to A\) be a homomorphism of Lie algebras. By the
- universal property of free algebras, there is a homomorphism of algebras \(g
- : T \mathfrak{g} \to A\) such that
+ universal property of free algebras, there is a homomorphism of algebras
+ \(\tilde f : T \mathfrak{g} \to A\) such that
\begin{center}
\begin{tikzcd}
- T \mathfrak{g} \arrow[dotted]{dr}{g} & \\
+ T \mathfrak{g} \arrow[dotted]{dr}{\tilde f} & \\
\mathfrak{g} \uar \rar[swap]{f} & A
\end{tikzcd}
\end{center}
Since \(f\) is a homomorphism of Lie algebras,
\[
- g([X, Y])
+ \tilde f([X, Y])
= f([X, Y])
= [f(X), f(Y)]
- = [g(X), g(Y)]
- = g(X \otimes Y - Y \otimes X)
+ = [\tilde f(X), \tilde f(Y)]
+ = \tilde f(X \otimes Y - Y \otimes X)
\]
for all \(X, Y \in \mathfrak{g}\). Hence \(I = ([X, Y] - (X \otimes Y - Y
- \otimes X) : X, Y \in \mathfrak{g}) \subset \ker g\) and therefore \(g\)
- factors through the quotient \(\mathcal{U}(\mathfrak{g}) = \mfrac{T
- \mathfrak{g}}{I}\).
+ \otimes X) : X, Y \in \mathfrak{g}) \subset \ker \tilde f\) and therefore
+ \(\tilde f\) factors through the quotient \(\mathcal{U}(\mathfrak{g}) =
+ \mfrac{T \mathfrak{g}}{I}\).
\begin{center}
\begin{tikzcd}
- T \mathfrak{g} \rar{g} \dar & A \\
- \mathcal{U}(\mathfrak{g}) \arrow[swap, dotted]{ur}{\bar{g}} &
+ T \mathfrak{g} \rar{\tilde f} \dar & A \\
+ \mathcal{U}(\mathfrak{g}) \arrow[swap, dotted]{ur}{\bar{\tilde f}} &
\end{tikzcd}
\end{center}
- Combining the two previous diagrams, we can see that \(\bar{g}\) is indeed an
- extension of \(f\). The uniqueness of the extension then follows from the
- uniqueness of \(g\) and \(\bar{g}\).
+ Combining the two previous diagrams, we can see that \(\bar{\tilde f}\) is
+ indeed an extension of \(f\). The uniqueness of the extension then follows
+ from the uniqueness of \(\tilde f\) and \(\bar{\tilde f}\).
\end{proof}
We should point out this construction is functorial. Indeed, if
@@ -603,8 +603,8 @@ algebras \(\mathcal{U}(f) : \mathcal{U}(\mathfrak{g}) \to
\end{center}
It is important to note, however, that \(\mathcal{U} : K\text{-}\mathbf{LieAlg}
-\to K\text{-}\mathbf{Alg}\) is not the ``inverse'' of \(\operatorname{Lie} :
-K\text{-}\mathbf{Alg} \to K\text{-}\mathbf{LieAlg}\). For instance, if
+\to K\text{-}\mathbf{Alg}\) is not the ``inverse'' of our functor
+\(K\text{-}\mathbf{Alg} \to K\text{-}\mathbf{LieAlg}\). For instance, if
\(\mathfrak{g} = K\) is the \(1\)-dimensional Abelian Lie algebra then
\(\mathcal{U}(\mathfrak{g}) \cong K[x]\), which is infinite-dimensional.
Nevertheless, proposition~\ref{thm:universal-env-uni-prop} may be restated