PartialOrderingsOfSubobjectsOfAnObject 2008-09-02 14:21:10 2008-09-03 01:02:07 Definition subobject union intersection union of subobjects intersection of subobjects \usepackage{amssymb,amscd} \usepackage{amsmath} \usepackage{amsfonts} \usepackage{mathrsfs} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions \usepackage{amsthm} % making logically defined graphics \usepackage{xypic} \usepackage{pst-plot} % define commands here \newcommand*{\abs}[1]{\left\lvert #1\right\rvert} \newtheorem{prop}{Proposition} \newtheorem{thm}{Theorem} \newtheorem{ex}{Example} \newcommand{\real}{\mathbb{R}} \newcommand{\pdiff}[2]{\frac{\partial #1}{\partial #2}} \newcommand{\mpdiff}[3]{\frac{\partial^#1 #2}{\partial #3^#1}} \newcommand{\Sub}{{\mathrm{Sub}}} Let $\mathcal{C}$ be a category and $A$ an object $\mathcal{C}$. Recall that a subobject of $A$ is just an equivalent class of equivalent monomorphisms into $A$. A subobject is denoted by $[f:B\to A]$, where $f$ is monomorphic, or simply $[B]$, whenever there is no confusion. Furthermore, we write $B\subseteq A$ to mean that $B$ is a subobject of $A$, and write $B\cong C$ whenever $[B]=[C]$. The class of all subobjects of $A$ is denoted by $\Sub(A)$. We wish to put a partial order on $\Sub(A)$. Given any two monomorphisms $f:B \to A$ and $g:C\to A$, define $g\le f$ iff there is a morphism $h:C\to B$ such that $$ \xymatrix{ C\ar[dr]^g\ar[dd]_h\\ &A\\ B\ar[ur]_f } $$ Now, if $f':B'\to A$ is equivalent to $f$ and $g':C\to A$ is equivalent to $g$, then we have morphisms $x:C\to C'$ and $x':C'\to C$ and $y:B\to B'$ and $y':B'\to B$ such that the diagram below consisting of only the solid lines is commutative: $$ \xymatrix{ C\ar[dr]_g\ar[dd]_h\ar[rr]_x & & C'\ar[dl]^{g'}\ar[ll]_{x'} \ar@{.>}[dd]^{h'} \\ &A &\\ B\ar[ur]^f\ar[rr]_y & & B'\ar[ul]_{f'}\ar[ll]_{y'} } $$ We define $h'=y\circ h\circ x'$ (the dotted line above). Then the diagram above including the dotted line is commutative: as $f'\circ h'=f'\circ (y\circ h\circ x')= f\circ (h\circ x')=g \circ x' = g'$. Hence $g'\le f'$. As a result, $\le$ induces a binary relation on $\Sub(A)$: $$C\le B$$ iff there are representing monomorphisms $f:B\to A$ and $g:C\to A$ such that $g\le f$. Let us use the same notation $\le$ for the induced relation. Then $\le$ on $\Sub(A)$ is a partial ordering on $\Sub(A)$: \begin{description} \item[reflexivity:] $B\le B$, because the composition of the identity morphism $1_B$ with any representing monomorphism $f:B\to A$ is $f$ itself. \item[anti-symmetry:] if $B\le C$ and $C\le B$, then there are representing monomorphisms $f:B\to A$ and $g:C\to A$ such that $f\le g$ and $g\le f$. So there are morphisms $h:B\to C$ and $h':C\to B$ such that $$ \xymatrix{ C\ar@<0.5ex>[dr]^g \ar@<0.5ex>[dd]^{h'}\\ &A\\ B\ar@<-0.5ex>[ur]_f \ar@<0.5ex>[uu]^h } $$ is commutative. But this just means that $h$ and $h'$ are inverses of one other, or $B\cong C$, so that they are the same subobject of $A$. \item[transitivity:] if $D\le C$ and $C\le B$, then there are representing monomorphisms $f:B\to A$, $g:C\to A$ and $h:D\to A$ such that $h\le g$ and $g\le f$. This means there are morphisms $r:D\to C$ and $s:C\to B$ such that $g\circ r = h$ and $f\circ s=g$. Since $f\circ (s\circ r)=g\circ r=h$, $h\le f$, and as a result, $D\le B$. \end{description} Thus, $\le$ turns $\Sub(A)$ into a partially ordered class, with top element $A$. With this, we may form the notions of unions and intersections of subobjects. Formally, let $\lbrace B_i \mid i\in I\rbrace$ be a collection of subobjects of $A$, indexed by a set $I$. \begin{itemize} \item The \emph{union} $C$ of $B_i$ is the supremum of the $B_i$'s with respect to $\le$, provided that it exists. In notation, we write $$C=\bigvee_{i\in I} B_i.$$ \item The \emph{intersection} $D$ of $B_i$ is the infimum of the $B_i$'s with respect to $\le$, provided that it exists. In notation, we write $$D=\bigwedge_{i\in I} B_i.$$ \end{itemize} For example, let $f:B \to A$ and $g:C\to A$ be subobjects of $A$. Then the pullback of $f$ and $g$, if it exists, is the intersection of $B$ and $C$. \textbf{Remark}. It can be shown that in any Abelian category, $\Sub(A)$ under $\le$ is a lattice for any object $A$.