replace bad hack

2021_Algebra_I.tex

@@ -2474,8 +2474,8 @@ The following is somewhat harder than in the affine case:
 \end{example}
 \subsubsection{Subcategories}
 \begin{definition}[Subcategories]
-    A \vocab{subcategory}  of $\mathcal{A}$ is a category $\cB$ such that $\Ob(\cB) \subseteq \Ob(\mathcal{A})$, such that $\Hom_\cB(X,Y) \subseteq \Hom_\mathcal{A}(X,Y)$ for objects $X$ and $Y$ of $\cB$, such that for every object $X \in \Ob(\cB)$, the identity $\Id_X$ of $X$ is the same in $\cB$ as in $\mathcal{A}$, and such that for composable morphisms in $\cB$, their compositions in $\mathcal{A}$ and $\cB$ coincide.
-    We call $\cB$ a \vocab{full subcategory}  of $\mathcal{A}$ if in addition $\Hom_\cB(X,Y) = \Hom_\mathcal{A}(X,Y)$ for arbitrary $X,Y \in \Ob(\cB)$.
+    A \vocab{subcategory}  of $\mathcal{A}$ is a category $\mathcal{B}$ such that $\Ob(\mathcal{B}) \subseteq \Ob(\mathcal{A})$, such that $\Hom_\mathcal{B}(X,Y) \subseteq \Hom_\mathcal{A}(X,Y)$ for objects $X$ and $Y$ of $\mathcal{B}$, such that for every object $X \in \Ob(\mathcal{B})$, the identity $\Id_X$ of $X$ is the same in $\mathcal{B}$ as in $\mathcal{A}$, and such that for composable morphisms in $\mathcal{B}$, their compositions in $\mathcal{A}$ and $\mathcal{B}$ coincide.
+    We call $\mathcal{B}$ a \vocab{full subcategory}  of $\mathcal{A}$ if in addition $\Hom_\mathcal{B}(X,Y) = \Hom_\mathcal{A}(X,Y)$ for arbitrary $X,Y \in \Ob(\mathcal{B})$.
 \end{definition}
 \begin{example}
     \begin{itemize}
@@ -2489,12 +2489,12 @@ The following is somewhat harder than in the affine case:
 
 \subsubsection{Functors and equivalences of categories}
 \begin{definition}
-    A \vocab[Functor!covariant]{(covariant) functor} (resp.  \vocab[Functor!contravariant]{contravariant functor}) between categories $\mathcal{A} \xrightarrow{F} \cB$ is a map $\Ob(\mathcal{A}) \xrightarrow{F} \Ob(\cB)$ with a family of maps $\Hom_\mathcal{A}(X,Y) \xrightarrow{F} \Hom_\cB(F(X),F(Y))$ (resp. $\Hom_\mathcal{A}(X,Y) \xrightarrow{F} \Hom_\cB(F(Y),F(X))$ in the case of contravariant functors), where $X$ and $Y$ are arbitrary objects of $\mathcal{A}$, such that the following conditions hold:
+    A \vocab[Functor!covariant]{(covariant) functor} (resp.  \vocab[Functor!contravariant]{contravariant functor}) between categories $\mathcal{A} \xrightarrow{F} \mathcal{B}$ is a map $\Ob(\mathcal{A}) \xrightarrow{F} \Ob(\mathcal{B})$ with a family of maps $\Hom_\mathcal{A}(X,Y) \xrightarrow{F} \Hom_\mathcal{B}(F(X),F(Y))$ (resp. $\Hom_\mathcal{A}(X,Y) \xrightarrow{F} \Hom_\mathcal{B}(F(Y),F(X))$ in the case of contravariant functors), where $X$ and $Y$ are arbitrary objects of $\mathcal{A}$, such that the following conditions hold:
      \begin{itemize}
          \item $F(\Id_X) = \Id_{F(X)}$ 
          \item For morphisms $X \xrightarrow{f} Y \xrightarrow{g} Z$ in $\mathcal{A}$, we have $F(gf) = F(g)F(f)$ ( resp. $F(gf) = F(f)F(g)$)
     \end{itemize}
-    A functor is called \vocab[Functor!essentially surjective]{essentially surjective}  if every object of $\cB$ is isomorphic to an element of the image of $\Ob(\mathcal{A}) \xrightarrow{F}  \Ob(\cB)$.
+    A functor is called \vocab[Functor!essentially surjective]{essentially surjective}  if every object of $\mathcal{B}$ is isomorphic to an element of the image of $\Ob(\mathcal{A}) \xrightarrow{F}  \Ob(\mathcal{B})$.
     A functor is called \vocab[Functor!full]{full} (resp. \vocab[Functor!faithful]{faithful}) if it induces surjective (resp. injective) maps between sets of morphisms.
     It is called an \vocab{equivalence of categories} if it is full, faithful and essentially surjective.
 \end{definition}
@@ -2635,10 +2635,10 @@ The following is somewhat harder than in the affine case:
 \subsubsection{Affine open subsets are a topology base}
 
 \begin{definition}
-    A set $\cB$ of open subsets of a topological space $X$ is called a \vocab{topology base} for $X$ if every open subset of $X$ can be written as a (possibly empty) union of elements of $\cB$.
+    A set $\mathcal{B}$ of open subsets of a topological space $X$ is called a \vocab{topology base} for $X$ if every open subset of $X$ can be written as a (possibly empty) union of elements of $\mathcal{B}$.
 \end{definition} 
 \begin{fact}
-If $X$ is a set, then $\cB \subseteq \mathcal{P}(X)$ is a base for some topology on $X$ iff $X = \bigcup_{U \in \cB} U$ and for arbitrary $U, V \in \cB, U \cap V$ is a union of elements of $\cB$.
+If $X$ is a set, then $\mathcal{B} \subseteq \mathcal{P}(X)$ is a base for some topology on $X$ iff $X = \bigcup_{U \in \mathcal{B}} U$ and for arbitrary $U, V \in \mathcal{B}, U \cap V$ is a union of elements of $\mathcal{B}$.
 \end{fact}
 \begin{definition}
     Let $X$ be a variety.