Added the algebraic tensor product.

src/cat/cat/index.tex

@@ -5,3 +5,4 @@
 
 \input{./cat-func.tex}
 \input{./universal.tex}
+\input{./tensor.tex}

src/cat/cat/tensor.tex

@@ -0,0 +1,55 @@
+\section{The Tensor Product}
+\label{section:tensor-product}
+
+
+
+\begin{definition}[Tensor Product]
+\label{definition:tensor-product}
+    Let $R$ be a commutative ring and $\seqf{E_j}$ be $R$ modules, then there exists a pair $(\bigotimes_{j = 1}^n E_j, \iota)$ such that:
+    \begin{enumerate}
+        \item $\bigotimes_{j = 1}^n E_j$ is an $R$-module. 
+        \item $\iota: \prod_{j = 1}^n E_j \to \bigotimes_{j = 1}^n E_j$ is a $n$-linear map. 
+        \item[(U)] For any pair $(F, \lambda)$ satisfying (1) and (2), there exists a unique $\Lambda \in \hom(\bigotimes_{j = 1}^n E_j; F)$ such that the following diagram commutes:
+        \[
+        \xymatrix{
+        \prod_{j = 1}^n E_j \ar@{->}[rd]_{\lambda} \ar@{->}[r]^{\iota} & \bigotimes_{j = 1}^n E_j \ar@{->}[d]^{\Lambda} \\
+        & F
+        }
+        \] 
+
+        \item $\bigotimes_{j = 1}^n E_j$ is the linear span of $\iota(\prod_{j = 1}^n E_j)$. 
+    \end{enumerate}
+
+    The module $\bigotimes_{j = 1}^n E_j$ is the \textbf{tensor product} of $\seqf{E_j}$, and $\iota: \prod_{j = 1}^n E_j \to \bigotimes_{j = 1}^n E_j$ is the \textbf{canonical embedding}. For any $(x_1, \cdots, d_n) \in \prod_{j = 1}^n E_j$, the image
+    \[
+    x_1 \otimes \cdots \otimes x_n = \iota(x_1, \cdots, x_n)
+    \]
+
+    is its \textbf{tensor product}.
+     
+\end{definition}
+\begin{proof}
+    Let $M$ be the free module generated by $\prod_{j = 1}^nE_j$, and $N \subset M$ be the submodule generated by elements of the following form:
+    \begin{enumerate}
+        \item For any $1 \le j \le n$, $(x_1, \cdots, x_n) \in \prod_{k = 1}^n E_k$, and $x_j \in E_j$, 
+        \[
+        (x_1, \cdots, x_j + x_j', \cdots, x_n) - (x_1, \cdots, x_j, \cdots, x_n) - (x_1, \cdots, x_j', \cdots, x_n)
+        \]
+
+        \item For any $(x_1, \cdots, x_n) \in \prod_{k = 1}^n E_k$ and $\alpha \in R$, 
+        \[
+        (x_1, \cdots, \alpha x_j, \cdots, x_n) - \alpha(x_1, \cdots, x_n)
+        \]
+    \end{enumerate}
+    
+    (1), (2): Let $\bigotimes_{j = 1}^n E_j = M/N$ and
+    \[
+    \iota: \prod_{j = 1}^n E_j \to \bigotimes_{j = 1}^n E_j \quad (x_1, \cdots, x_n) \mapsto (x_1, \cdots, x_n) + N
+    \]
+
+    then by definition of $N$, $\iota$ is $n$-linear. 
+
+    (U): Let $(F, \lambda)$ be a pair satisfying (1) and (2), then $\lambda$ admits a unique extension to a linear map $\ol \lambda: M \to F$. Since $\lambda$ is $n$-linear, $\ker \ol \lambda \supset N$. By the first isomorphism theorem, there exists a unique $\Lambda \in \hom(\bigotimes_{j = 1}^n E_j; F)$ such that the given diagram commutes. 
+
+    (4): Since $M$ is the free module generated by $\prod_{j = 1}^n E_j$, $M/N$ is generated by $\iota(\prod_{j = 1}^n E_j)$.  
+\end{proof}

src/cat/cat/universal.tex

@@ -209,7 +209,7 @@ Let $\catc$ be a category and $(\seqi{A}, \bracsn{f^i_j| i, j \in I, i \lesssim
 
 \begin{proposition}
 \label{proposition:module-inverse-limit}
-    Let $R$ be a ring and $(\seqi{A}, \bracs{T^i_j|i, j \in I, i \lesssim j)}$ be a downward-directed system of $R$-modules, then there exists $(A, \bracsn{T^A_i}_{i \in I})$ such that:
+    Let $R$ be a ring and $(\seqi{A}, \bracs{T^i_j|i, j \in I, i \lesssim j})$ be a downward-directed system of $R$-modules, then there exists $(A, \bracsn{T^A_i}_{i \in I})$ such that:
     \begin{enumerate}
         \item For each $i \in I$, $T^A_i \in \hom(A; A_i)$.
         \item For any $i, j \in I$ with $i \lesssim j$, the following diagram commutes:
@@ -253,3 +253,6 @@ Let $\catc$ be a category and $(\seqi{A}, \bracsn{f^i_j| i, j \in I, i \lesssim
 
     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}
+
+
+

src/measure/bochner-integral/bochner.tex

@@ -44,6 +44,8 @@
     
 \end{proof}
 
+
+
 \begin{theorem}[Dominated Convergence Theorem]
 \label{theorem:dct-bochner}
     Let $(X, \cm, \mu)$ be a measure spacs, $E$ be a Banach space over $K \in \RC$, $\seq{f_n} \subset L^1(X; E)$, and $f \in L^1(X; E)$. If
@@ -58,3 +60,48 @@
     By the classical \hyperref[Dominated Convergence Theorem]{proposition:dct-lp}, $f_n \to f$ in $L^1(X; E)$. Since $h \mapsto \int h d\mu$ is a bounded linear operator, $\int f d\mu = \limv{n}\int f_n d\mu$. 
 \end{proof}
 
+\subsection{Vector Measure Version}
+\label{subsection:bochner-vector}
+
+
+
+\begin{definition}[Bochner Integral]
+\label{definition:bochner-integral-vector}
+    Let $(X, \cm)$ be a measurable space, $E, F$ be normed spaces over $K \in \RC$, $G$ be a Banach space over $K$,
+    \[
+    \lambda: E \times F \to G \quad (x, y) \mapsto xy
+    \]
+
+    be a bounded bilinear map, and $\mu: \cm \to F$ be a vector measure, then there exists a unique $I_\lambda \in L(L^1(X, |\mu|; E); G)$ such that:
+    \begin{enumerate}
+        \item For any $x \in E$ and $A \in \cm$, $I_\lambda(x \cdot \one_A) = x \mu(A)$.
+        \item For any $f \in L^1(X, |\mu|; E)$, $\normn{I_\lambda f}_{G} \le \norm{\lambda}_{L^2(E, F; G)} \cdot \norm{f}_{L^1(X, |\mu|; E)}$.
+    \end{enumerate}
+    
+    For any $f \in L^1(X; E)$, $I_\lambda f = \int f d\lambda\mu$ is the \textbf{Bochner integral} of $f$ with respect to $\mu$ and $\lambda$. 
+\end{definition}
+\begin{proof}
+    Same as \autoref{definition:bochner-integral}.
+\end{proof}
+
+
+\begin{theorem}[Dominated Convergence Theorem]
+\label{theorem:dct-bochner-vector}
+    Let $(X, \cm, \mu)$ be a measure space, $E, F$ be normed spaces over $K \in \RC$, $G$ be a Banach space over $K$, 
+    \[
+    \lambda: E \times F \to G \quad (x, y) \mapsto xy
+    \]
+
+    be a bounded bilinear map, $\mu: \cm \to F$ be a vector measure, $\seq{f_n} \subset L^1(X, |\mu|; E)$, and $f \in L^1(X, |\mu|; E)$. If
+    \begin{enumerate}
+        \item[(a)] $f_n \to f$ strongly pointwise.
+        \item[(b)] There exists $g \in L^1(X) \cap L^+(X)$ such that $\norm{f_n}_E \le g$ for all $n \in \natp$. 
+    \end{enumerate}
+    
+    then $\int f d\lambda\mu = \limv{n}\int f_n d\lambda\mu$. 
+\end{theorem}
+\begin{proof}
+    By the classical \hyperref[Dominated Convergence Theorem]{proposition:dct-lp}, $f_n \to f$ in $L^1(X, |\mu|; E)$. Since $h \mapsto \int h d\lambda\mu$ is a bounded linear operator, $\int f d\lambda\mu = \limv{n}\int f_n d\lambda\mu$. 
+\end{proof}
+
+