Skip to content

Commit

Permalink
Misc edits
Browse files Browse the repository at this point in the history
Signed-off-by: zeramorphic <[email protected]>
  • Loading branch information
zeramorphic committed May 26, 2024
1 parent 98db1c7 commit 0dd5ba3
Show file tree
Hide file tree
Showing 7 changed files with 40 additions and 38 deletions.
4 changes: 2 additions & 2 deletions iii/cat/02_yoneda_lemma.tex
Original file line number Diff line number Diff line change
Expand Up @@ -11,7 +11,7 @@ \subsection{Statement and proof}
We can also define
\[ \mathcal C(-, A) : \mathcal C^\cop \to \mathbf{Set} \]
by
\[ B \mapsto \mathcal C(B, A);\quad (B \xrightarrow f C) \mapsto ((B \xrightarrow g A) \mapsto g f) \]
\[ B \mapsto \mathcal C(B, A);\quad (B \xrightarrow f C) \mapsto ((C \xrightarrow g A) \mapsto g f) \]
\begin{lemma}[Yoneda lemma]
Let \( \mathcal C \) be a locally small category.
Let \( A \in \ob \mathcal C \), and let \( F : \mathcal C \to \mathbf{Set} \) be a functor.
Expand Down Expand Up @@ -127,7 +127,7 @@ \subsection{Statement and proof}
Then we have two functors
\[ \mathcal C \times [\mathcal C, \mathbf{Set}] \to \mathbf{Set} \]
The first is the evaluation functor
\[ (A, F) = FA \]
\[ (A, F) \mapsto FA \]
The second is the composite
\[ \mathcal C \times [\mathcal C, \mathbf{Set}] \xrightarrow{Y \times 1} [\mathcal C, \mathbf{Set}]^\cop \times [\mathcal C, \mathbf{Set}] \xrightarrow{[\mathcal C, \mathbf{Set}](-, -)} \mathbf{Set} \]
The naturality condition is that \( \Phi \) and \( \Psi \) are natural transformations between these two functors, and thus are natural isomorphisms.
Expand Down
3 changes: 2 additions & 1 deletion iii/cat/03_adjunctions.tex
Original file line number Diff line number Diff line change
Expand Up @@ -328,7 +328,8 @@ \subsection{Reflections}
If \( \mathcal D \) is coreflective, there is a best possible way to get \emph{out of} \( \mathcal D \) to some object in \( \mathcal C \).
\begin{example}
\begin{enumerate}
\item \( \mathbf{AbGp} \) is reflective in \( \mathbf{Gp} \); the left adjoint to the inclusion map sends a group \( G \) to its abelianisation \( G^{\mathrm{ab}} = \faktor{G}{H} \), the quotient of \( G \) by its commutator subgroup \( H = \qty{aba^{-1}b^{-1} \mid a, b \in G} \trianglelefteq G \).
\item \( \mathbf{AbGp} \) is reflective in \( \mathbf{Gp} \); the left adjoint to the inclusion map sends a group \( G \) to its abelianisation \( G^{\mathrm{ab}} = \faktor{G}{H} \), the quotient of \( G \) by its commutator subgroup
\[ H = \qty{aba^{-1}b^{-1} \mid a, b \in G} \trianglelefteq G \]
Note that any homomorphism \( G \to A \) where \( A \) is abelian factors uniquely through the quotient map \( G \to G^{\mathrm{ab}} \), giving the adjunction as required.
\item Recall that an abelian group is called \emph{torsion} if all of its elements have finite order, and \emph{torsion-free} if all of its nonzero elements have infinite order.
For an abelian group \( A \), its set of torsion elements forms a subgroup \( A_t \), which is a torsion group.
Expand Down
1 change: 1 addition & 0 deletions iii/cat/04_limits.tex
Original file line number Diff line number Diff line change
Expand Up @@ -313,6 +313,7 @@ \subsection{Preservation and creation}
\end{tikzcd}\]
is a pullback square.
Thus, if \( \mathcal D \) has pullbacks, any monomorphism in \( [\mathcal C, \mathcal D] \) is a pointwise monomorphism, because the pullback in \( [\mathcal C, \mathcal D] \) is constructed pointwise by the previous lemma.
In particular, the monomorphisms and epimorphisms in \( [\mathcal C, \mathbf{Set}] \) are precisely the pointwise monomorphisms and pointwise epimorphisms respectively.
\end{remark}

\subsection{Interaction with adjunctions}
Expand Down
64 changes: 32 additions & 32 deletions iii/cat/05_monads.tex
Original file line number Diff line number Diff line change
Expand Up @@ -150,41 +150,41 @@ \subsection{Kleisli categories}
\arrow["Tg", from=1-2, to=1-3]
\arrow["{\mu_C}", from=1-3, to=1-4]
\end{tikzcd}\]
These satisfy the unit and associativity laws.
% https://q.uiver.app/#q=WzAsNCxbMCwwLCJBIl0sWzEsMCwiVEIiXSxbMiwwLCJUVEIiXSxbMiwxLCJUQiJdLFswLDEsImYiXSxbMSwyLCJUXFxldGFfQiJdLFsyLDMsIlxcbXVfQiJdLFsxLDMsIjFfe1RCfSJdXQ==
\[\begin{tikzcd}
A & TB & T^2B \\
&& TB
\arrow["f", from=1-1, to=1-2]
\arrow["{T\eta_B}", from=1-2, to=1-3]
\arrow["{\mu_B}", from=1-3, to=2-3]
\arrow["{1_{TB}}"', from=1-2, to=2-3]
\end{definition}
These satisfy the unit and associativity laws.
% https://q.uiver.app/#q=WzAsNCxbMCwwLCJBIl0sWzEsMCwiVEIiXSxbMiwwLCJUVEIiXSxbMiwxLCJUQiJdLFswLDEsImYiXSxbMSwyLCJUXFxldGFfQiJdLFsyLDMsIlxcbXVfQiJdLFsxLDMsIjFfe1RCfSJdXQ==
\[\begin{tikzcd}
A & TB & T^2B \\
&& TB
\arrow["f", from=1-1, to=1-2]
\arrow["{T\eta_B}", from=1-2, to=1-3]
\arrow["{\mu_B}", from=1-3, to=2-3]
\arrow["{1_{TB}}"', from=1-2, to=2-3]
\end{tikzcd}\quad\quad\begin{tikzcd}
A & TA \\
TB & T^2B \\
& TB
\arrow["{\eta_A}", from=1-1, to=1-2]
\arrow["Tf", from=1-2, to=2-2]
\arrow["{\mu_B}", from=2-2, to=3-2]
\arrow["f"', from=1-1, to=2-1]
\arrow["{\eta_{TB}}", from=2-1, to=2-2]
\arrow["{1_{TB}}"', from=2-1, to=3-2]
\end{tikzcd}\]
\[\begin{tikzcd}
A & TB & {T^2C} & {T^3D} & {T^2D} \\
&& TC & {T^2D} & TD
\arrow["f", from=1-1, to=1-2]
\arrow["Tg", from=1-2, to=1-3]
\arrow["{T^2h}", from=1-3, to=1-4]
\arrow["{T\mu_D}", from=1-4, to=1-5]
\arrow["{\mu_D}", from=1-5, to=2-5]
\arrow["{\mu_{TD}}", from=1-4, to=2-4]
\arrow["{\mu_D}"', from=2-4, to=2-5]
\arrow["{\mu_C}", from=1-3, to=2-3]
\arrow["Th"', from=2-3, to=2-4]
A & TA \\
TB & T^2B \\
& TB
\arrow["{\eta_A}", from=1-1, to=1-2]
\arrow["Tf", from=1-2, to=2-2]
\arrow["{\mu_B}", from=2-2, to=3-2]
\arrow["f"', from=1-1, to=2-1]
\arrow["{\eta_{TB}}", from=2-1, to=2-2]
\arrow["{1_{TB}}"', from=2-1, to=3-2]
\end{tikzcd}\]
\[\begin{tikzcd}
A & TB & {T^2C} & {T^3D} & {T^2D} \\
&& TC & {T^2D} & TD
\arrow["f", from=1-1, to=1-2]
\arrow["Tg", from=1-2, to=1-3]
\arrow["{T^2h}", from=1-3, to=1-4]
\arrow["{T\mu_D}", from=1-4, to=1-5]
\arrow["{\mu_D}", from=1-5, to=2-5]
\arrow["{\mu_{TD}}", from=1-4, to=2-4]
\arrow["{\mu_D}"', from=2-4, to=2-5]
\arrow["{\mu_C}", from=1-3, to=2-3]
\arrow["Th"', from=2-3, to=2-4]
\end{tikzcd}\]
where in the last diagram, the upper composite is \( (hg)f \) and the lower composite is \( h(gf) \) in \( \mathcal C_{\mathbb T} \).
\end{definition}
\begin{proposition}
There is an adjunction \( F_{\mathbb T} \dashv G_{\mathbb T} \) where \( F_{\mathbb T} : \mathcal C \to \mathcal C_{\mathbb T} \) and \( G_{\mathbb T} : \mathcal C_{\mathbb T} \to \mathcal C \) that induces the monad \( \mathbb T \).
\end{proposition}
Expand Down
2 changes: 1 addition & 1 deletion iii/commalg/01_chain_conditions.tex
Original file line number Diff line number Diff line change
Expand Up @@ -86,7 +86,7 @@ \subsection{Noetherian and Artinian modules}
Note that every Noetherian module is finitely generated.
Let \( R = \mathbb Z[T_1, T_2, \dots] \), and let \( M = R \) as an \( R \)-module.
\( M \) is generated by \( 1_R \), so in particular it is finitely generated.
But it has a submodule \( \langle T_1, T_2, \dots \rangle \) that is not finitely generated.
But it has a \( \mathbb Z \)-submodule \( \langle T_1, T_2, \dots \rangle \) that is not finitely generated.
So in the above lemma we indeed must check every submodule.
\begin{definition}
A ring \( R \) is Noetherian (respectively Artinian) if \( R \) is Noetherian (resp.\ Artinian) as an \( R \)-module.
Expand Down
2 changes: 1 addition & 1 deletion iii/commalg/02_tensor_products.tex
Original file line number Diff line number Diff line change
Expand Up @@ -18,7 +18,7 @@ \subsection{Introduction}
\end{example}
\begin{example}
Now consider \( {\mathbb R}^n \otimes_{\mathbb R} {\mathbb R}^\ell \).
We will show later that this is isomorphic to \( \mathbb R^{n+\ell} \).
We will show later that this is isomorphic to \( \mathbb R^{n\ell} \).
\end{example}

\subsection{Definition and universal property}
Expand Down
Original file line number Diff line number Diff line change
Expand Up @@ -243,7 +243,7 @@ \subsection{Integral closure}
\begin{enumerate}
\item \( \mathbb Z\qty[\sqrt{5}] \) is not integrally closed, because \( \alpha = \frac{1 + \sqrt{5}}{2} \in FF\qty(\mathbb Z\qty[\sqrt{5}]) = \mathbb Q\qty[\sqrt{5}] \), and \( \alpha^2 - \alpha - 1 = 0 \) so it is \( \mathbb Z\qty[\sqrt{5}] \)-integral.
\item \( \mathbb Z \) is integrally closed.
\item If \( k \) is a field, \( k[T_1, \dots, T_n] \) are integrally closed.
\item If \( k \) is a field, \( k[T_1, \dots, T_n] \) is integrally closed.
\end{enumerate}
\end{example}
Examples (ii) and (iii) are special cases of the following result.
Expand Down

0 comments on commit 0dd5ba3

Please sign in to comment.