FactorizationSystem 2008-10-15 14:21:15 2008-10-15 17:12:48 Definition \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}} Recall that any function $f:A\to B$ can be factored as $h\circ g$ where $g:A\to f(A)$ is a surjection and $h:f(A)\to B$ is an injection. This phenomenon is true in many mathematical systems: homomorphisms between groups, rings, lattices, continuous maps between topological spaces, etc... However, in the setting of category theory, while it is still true that a morphism can be factored into the composition of two morphisms (one of them, say, being the identity morphism), the fact that one factor is an epimorphism and the other a monomorphism no longer holds in general. Categories where such kinds of factorizations exist is of great interest. Factorization of morphisms in a category can be formalized as follows: \textbf{Definition}. Let $\mathcal{C}$ be a category. An ordered pair $(\mathcal{E},\mathcal{M})$ of classes of morphisms in $\mathcal{C}$ is called a \emph{factorization system} if \begin{itemize} \item every morphism $f$ in $\mathcal{C}$ can be ``factored'' as $f=m\circ e$ where $m\in \mathcal{M}$ and $e\in \mathcal{E}$, \item $\mathcal{E}$ is orthogonal to $\mathcal{M}$: $\mathcal{E} \perp \mathcal{M}$, \item every isomorphism is in both $\mathcal{E}$ and $\mathcal{M}$, and \item both $\mathcal{E}$ and $\mathcal{M}$ are closed under composition; in other words, if $x,y$ are both in one class and $x\circ y$ is defined, then $x\circ y$ is in that class too. \end{itemize} When there is a factorization system $(\mathcal{E},\mathcal{M})$ on a category $\mathcal{C}$, we say that $\mathcal{C}$ \emph{has $(\mathcal{E},\mathcal{M})$-factorization}, and an \emph{$(\mathcal{E},\mathcal{M})$-factorization} of a morphism $f$ is a factorization of $f$: $f=m\circ e$, such that $m\in \mathcal{M}$ and $e\in \mathcal{E}$. One of the first properties of having a factorization system $(\mathcal{E},\mathcal{M})$ is that the $(\mathcal{E},\mathcal{M})$-factorization of a morphism $f$ is unique up to isomorphism: \begin{prop} If we have a commutative diagram $$\xymatrix@+=1.5cm{A \ar[dr]^f \ar[r]^s \ar[d]_t & C \ar[d]^u \\ D \ar[r]_v & B}$$ where $s,t\in \mathcal{E}$ and $u,v\in \mathcal{M}$, then $C\cong D$. \end{prop} \begin{proof} Because $\mathcal{E}\perp \mathcal{M}$, there are unique morphisms $g:C\to D$ and $h:D\to C$ such that the diagram $$\xymatrix@+=1.5cm{A \ar[r]^s \ar[d]_t & C \ar[d]^u \ar@<-0.5ex>[dl]_g \\ D \ar[r]_v \ar@<-0.5ex>[ur]_h & B}$$ is commutative. Then $h\circ g\circ s = h \circ t = s$ and $u \circ h\circ g = v \circ g = u$, which means we have a commutative diagram $$\xymatrix@+=1.5cm{A \ar[r]^s \ar[d]_s & C \ar[d]^u \ar[dl]|{h\circ g} \\ C \ar[r]_u & B}$$ But $s\perp u$, so the morphism $h\circ g: C\to C$ making the above diagram commute is uniquely determined. Since $1_C:C \to C$ is another morphism making the diagram commute, we must have $h\circ g=1_C$. Similarly, one sees that $g\circ h=1_D$. This implies that $C\cong D$. \end{proof} More to come... \begin{thebibliography}{9} \bibitem{fb} F. Borceux \emph{Basic Category Theory, Handbook of Categorical Algebra I}, Cambridge University Press, Cambridge (1994) \end{thebibliography}