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This commit is contained in:
Bokuan Li
2026-03-06 14:06:15 -05:00
parent 173727665b
commit 5034bc4220
109 changed files with 1184 additions and 410 deletions
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1
src/cat/cat/cat-func.tex
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@@ -7,6 +7,7 @@
\[
\mor{A, B} \times \mor{B, C} \to \mor{A, C}
\]
that satisfies the following axioms:
\begin{enumerate}
\item[(CAT1)] For any $A, B, A', B' \in \obj{\catc}$, $\mor{A, B}$ and $\mor{A', B'}$ are disjoint or equal, where $\mor{A, B} = \mor{A', B'}$ if and only if $A = A'$ and $B = B'$.

4
src/cat/cat/index.tex
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@@ -3,5 +3,5 @@
\textit{I do not know much about categories, however some concepts from it are useful in phrasing certain properties.}
\input{./src/cat/cat/cat-func.tex}
\input{./src/cat/cat/universal.tex}
\input{./cat-func.tex}
\input{./universal.tex}

18
src/cat/cat/universal.tex
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@@ -22,7 +22,7 @@
\begin{enumerate}
\item $P \in \obj{\catc}$.
\item For each $i \in I$, $\pi_i \in \mor{P, A_i}$.
\item[\textbf{(U)}] For any pair $(C, \seqi{f})$ satisfying (1) and (2), there exists a unique $f \in \mor{C, P}$ such that the following diagram commutes
\item[(U)] For any pair $(C, \seqi{f})$ satisfying (1) and (2), there exists a unique $f \in \mor{C, P}$ such that the following diagram commutes
\[
\xymatrix{
@@ -31,6 +31,7 @@
}
\]
for all $i \in I$.
\end{enumerate}
\end{definition}
@@ -50,6 +51,7 @@
}
\]
for all $i \in I$.
\end{enumerate}
\end{definition}
@@ -101,6 +103,7 @@
}
\]
\item[(U)] For any pair $(B, \bracsn{g^i_B}_{i \in I})$ satisfying (1) and (2), there exists a unique $g \in \mor{A, B}$ such that the following diagram commutes
\[
@@ -110,6 +113,7 @@
}
\]
for all $i \in I$.
\end{enumerate}
\end{definition}
@@ -128,6 +132,7 @@ Let $\catc$ be a category and $(\seqi{A}, \bracsn{f^i_j| i, j \in I, i \lesssim
}
\]
\item[(U)] For any pair $(B, \bracsn{g^B_i}_{i \in I})$, there exists a unique $g \in \mor{B, A}$ such that the following diagram commutes
\[
@@ -137,6 +142,7 @@ Let $\catc$ be a category and $(\seqi{A}, \bracsn{f^i_j| i, j \in I, i \lesssim
}
\]
for all $i \in I$.
\end{enumerate}
\end{definition}
@@ -155,6 +161,7 @@ Let $\catc$ be a category and $(\seqi{A}, \bracsn{f^i_j| i, j \in I, i \lesssim
}
\]
\item[(U)] For any pair $(B, \bracsn{S^i_B}_{i \in I})$ satisfying (1) and (2), there exists a unique $S \in \hom({A, B})$ such that the following diagram commutes
\[
@@ -164,6 +171,7 @@ Let $\catc$ be a category and $(\seqi{A}, \bracsn{f^i_j| i, j \in I, i \lesssim
}
\]
for all $i \in I$.
\end{enumerate}
\end{proposition}
@@ -176,6 +184,7 @@ Let $\catc$ be a category and $(\seqi{A}, \bracsn{f^i_j| i, j \in I, i \lesssim
0 &k \ne i, j
\end{cases}
\]
Let $N \subset M$ be the submodule generated by $\bracs{x_{i, j}|i, j \in I, i \lesssim j, x \in A_i}$, $A = M/N$, and $\pi: M \to M/N$ be the canonical map.
(1): For each $i \in I$, let
@@ -185,6 +194,7 @@ Let $\catc$ be a category and $(\seqi{A}, \bracsn{f^i_j| i, j \in I, i \lesssim
0 &k \ne i
\end{cases}
\]
and $T^i_A = \pi \circ T^i_M$.
(2): Let $i, j \in I$ with $i \lesssim j$, then for any $x \in A_i$, $T^i_Mx - T^j_M T^i_j x \in N$. Hence $T^i_Ax = T^j_A T^i_jx$.
@@ -193,6 +203,7 @@ Let $\catc$ be a category and $(\seqi{A}, \bracsn{f^i_j| i, j \in I, i \lesssim
\[
S_0: M \to B \quad x \mapsto \sum_{i \in I}S^i_B \pi_i x
\]
then $S_0$ is the unique linear map such that $S_0 \circ T^i_M = S^i_B$ for all $i \in I$. For any $i, j \in I$ with $i \lesssim J$, $S^i_B x = S^j_B T^i_j x$, so $\ker S_0 \supset N$. By the first isomorphism theorem, there exists a unique $S \in \hom(A; B)$ such that $S_0 = S \circ \pi$.
\end{proof}
@@ -208,6 +219,7 @@ Let $\catc$ be a category and $(\seqi{A}, \bracsn{f^i_j| i, j \in I, i \lesssim
A \ar@{->}[u]^{T^A_i} \ar@{->}[ru]_{T^A_j} &
}
\]
\item[(U)] For any pair $(B, \bracsn{S^B_i}_{i \in I})$ satisfying (1) and (2), there exists a unique $S \in \hom(B; A)$ such that the following diagram commutes
\[
@@ -217,6 +229,7 @@ Let $\catc$ be a category and $(\seqi{A}, \bracsn{f^i_j| i, j \in I, i \lesssim
}
\]
for all $i \in I$.
\end{enumerate}
\end{proposition}
@@ -225,15 +238,18 @@ Let $\catc$ be a category and $(\seqi{A}, \bracsn{f^i_j| i, j \in I, i \lesssim
\[
A = \bracs{x \in \prod_{i \in I}A_i \bigg | \pi_j(x) = T^i_j\pi_i(x) \forall i, j \in I, i \lesssim j}
\]
For each $i \in I$, let $T^A_i = \pi_i$, then $(A, \bracsn{T^A_i}_{i \in I})$ satisfies (1) and (2) by definition of $A$.
(U): Let $(B, \bracsn{S^B_i}_{i \in I})$ satisfying (1) and (2). Let
\[
S: B \to \prod_{i \in I}A_i \quad \pi_i(Sx) = S^B_i
\]
then for any $x \in B$ and $i, j \in I$ with $i \lesssim j$,
\[
\pi_j (Sx) = S^B_jx = T^i_j S^B_ix = T^i_j \pi_i(S x)
\]
so $S \in \hom(B; A)$, and the diagram commutes. Since any map $f: B \to A$ is uniquely determined by its composition with the projections, $S$ is unique.
\end{proof}

1
src/cat/gluing/index.tex
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@@ -37,5 +37,6 @@
\[
T(\lambda x + y) = T_V(\lambda x + y) = \lambda T_Vx + T_Vy = \lambda Tx + Ty
\]
and $T \in \hom(E; F)$.
\end{proof}

6
src/cat/index.tex
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@@ -1,6 +1,6 @@
\part{General Tools}
\label{part:part-categories}
\input{./src/cat/cat/index.tex}
\input{./src/cat/gluing/index.tex}
\input{./src/cat/tricks/index.tex}
\input{./cat/index.tex}
\input{./gluing/index.tex}
\input{./tricks/index.tex}

8
src/cat/tricks/dyadic.tex
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@@ -14,6 +14,7 @@
\[
\text{rk}(q) = \min\bracs{n \in \natp|x \in \mathbb{D}_n}
\]
is the \textbf{dyadic rank} of $q$.
\end{definition}
@@ -32,7 +33,7 @@
\begin{proof}
First suppose that $\text{rk}(x) = 1$. In which case, $x \in \bracs{0, 1/2}$, and either $M(x) = \emptyset$ or $M(y) = \bracs{1}$.
Assume inductively that the proposition holds for all dyadic rationals of rank $n$. Let $x \in \mathbb{D}_{n+1} \setminus \mathbb{D}_n$. By \ref{lemma:dyadic-decompose}, there exists a unique $y \in \mathbb{D}_n$ such that $x = y + 2^{-n-1}$. Since $x > 0$, $x \ge 1/2^{-n-1}$, so $y \in [0, 1)$. By the inductive assumption, there exists a unique $M(y) \subset \natp \cap [1, n]$ such that $y = \sum_{k \in M(y)}2^{-k}$. In which case, $M(x) = M(y) \cup \bracs{n}$ is the desired set.
Assume inductively that the proposition holds for all dyadic rationals of rank $n$. Let $x \in \mathbb{D}_{n+1} \setminus \mathbb{D}_n$. By \autoref{lemma:dyadic-decompose}, there exists a unique $y \in \mathbb{D}_n$ such that $x = y + 2^{-n-1}$. Since $x > 0$, $x \ge 1/2^{-n-1}$, so $y \in [0, 1)$. By the inductive assumption, there exists a unique $M(y) \subset \natp \cap [1, n]$ such that $y = \sum_{k \in M(y)}2^{-k}$. In which case, $M(x) = M(y) \cup \bracs{n}$ is the desired set.
\end{proof}
\begin{proposition}
@@ -48,15 +49,18 @@
\[
\phi(x) + \phi(y) = g_2 + g_2 \le g_1 = \phi(x + y)
\]
In the second, $\phi(x) + \phi(y) = \phi(x + y)$.
Now assume inductively that the proposition holds for all $x, y \in \mathbb{D} \cap (0, 1)$ with $\text{rk}(x) = n$ and $\text{rk}(y) \le n$. Let $x \in \mathbb{D}_{n+1} \setminus \mathbb{D}_n$ and $y \in \mathbb{D}_{n+1}$. By \ref{lemma:dyadic-decompose}, there exists $x_0 \in \mathbb{D}_n$ such that $x = x_0 + 2^{-n-1}$. If $y \in \mathbb{D}_n$, then
Now assume inductively that the proposition holds for all $x, y \in \mathbb{D} \cap (0, 1)$ with $\text{rk}(x) = n$ and $\text{rk}(y) \le n$. Let $x \in \mathbb{D}_{n+1} \setminus \mathbb{D}_n$ and $y \in \mathbb{D}_{n+1}$. By \autoref{lemma:dyadic-decompose}, there exists $x_0 \in \mathbb{D}_n$ such that $x = x_0 + 2^{-n-1}$. If $y \in \mathbb{D}_n$, then
\[
\phi(x) + \phi(y) = \phi(x_0) + \phi(y) + g_{n+1} \le \phi(x_0 + y) + g_{n+1} = \phi(x + y)
\]
by the inductive assumption. Otherwise, there exists $y_0 \in \mathbb{D}_n$ such that $y = y_0 + 2^{-n-1}$, so
\[
\phi(x) + \phi(y) \le \phi(x_0) + \phi(y_0) + g_n = \phi(x_0) + \phi(y_0) + \phi(2^{-n}) \le \phi(x + y)
\]
by the inductive assumption.
\end{proof}

4
src/cat/tricks/index.tex
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@@ -1,5 +1,5 @@
\chapter{Inequalities and Computations}
\label{chap:tricks}
\input{./src/cat/tricks/dyadic.tex}
\input{./src/cat/tricks/product.tex}
\input{./dyadic.tex}
\input{./product.tex}

3
src/cat/tricks/product.tex
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@@ -7,10 +7,12 @@
\[
ab \le \frac{a^p}{p} + \frac{b^q}{q}
\]
and for any $\eps > 0$,
\[
ab \le \eps a^p + \frac{1}{q}(\eps q)^{-q/p}b^q
\]
\end{lemma}
\begin{proof}
Since $x \mapsto \exp(x)$ is convex,
@@ -22,4 +24,5 @@
\[
ab = (\eps p)^{1/p}a \cdot \frac{b}{(\eps p)^{1/p}} \le \eps a^p + \frac{b^q}{q}(\eps p)^{-(1/p)q} = \eps a^p + \frac{1}{q}(\eps q)^{-q/p}b^q
\]
\end{proof}