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Maximilian Keßler 2022-02-16 02:12:13 +01:00
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2021_Algebra_I.tex
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@ -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.