equalizer Equalizer 2004-10-22 17:50:23 2008-10-01 17:00:52 Definition coequalizer parallel morphisms have equalizers % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb,amscd} \usepackage{amsmath} \usepackage{amsfonts} % 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} % there are many more packages, add them here as you need them % define commands here Let $\mathcal{C}$ be a category. A family of morphisms in $\mathcal{C}$ is said be \PMlinkescapetext{\emph{parallel}} if they belong to $\operatorname{Hom}(A,B)$, for some objects $A,B$ in $\mathcal{C}$. Let $f,g$ be a pair of parallel morphisms in $\operatorname{Hom}(A,B)$. A morphism $d\colon X\to A$ is said to equalize $f$ and $g$ if $fd=gd$. In other words, the following diagrams are equal: $$\xymatrix@1{X\ar[r]^d&A\ar[r]^f&B}=\xymatrix@1{X\ar[r]^d&A\ar[r]^g&B}$$ \par An \emph{equalizer} of $f$ and $g$ is a morphism $d$ from an object $X \in \mathcal{C}$ to $A$, such that \begin{enumerate} \item $d$ \emph{equalizes} $f$ and $g$ \item $d$ is universal among all morphisms that equalize $f$ and $g$. Specifically, if $e$ is a morphism from an object $Y\in\mathcal{C}$ to $A$ such that $e$ equalizes $f$ and $g$, then there exists a unique morphism $h:Y\to X$ and a commutative diagram: $$\xymatrix@1{Y \ar[d]_h \ar[dr]^e \\ X \ar[r]_d & A}$$ \end{enumerate} Reversing all the arrows in the previous paragraphs, we have the dual notion of an equalizer: that of a coequalizer. To make this statement explicitly, let there be given two morphisms $f,g\in\operatorname{Hom}(A,B)$, a \emph{coequalizer} is a morphism $c$ from $B$ to an object $Z\in\mathcal{C}$ such that \begin{enumerate} \item $\xymatrix@1{A\ar[r]^f&B\ar[r]^c&Z}=\xymatrix@1{A\ar[r]^g&B\ar[r]^c&Z}$. Such a morphism is said to \emph{coequalize} $f$ and $g$. \item $c$ is universal among all morphisms that coequalizes $f$ and $g$. This means that given a morphism $r$ from $B$ to an object $Y\in\mathcal{C}$, there exists a unique morphism $r\in\operatorname{Hom}(Z,Y)$ so the following diagram commutes: $$\xymatrix@1{B \ar[dr]_e \ar[r]^c & Z \ar[d]^r \\ & Y}$$ \end{enumerate} \par \textbf{Remarks} \begin{itemize} \item An equalizer is a monomorphism (but not the other way around, a monomorphism that is also an equalizer is called a \emph{regular monomorphism}). A coequalizer is an epimorphism (and conversely, an epimorphism that is also a coequalizer is called a \emph{regular epimorphism}). This follows directly from the above definitions and definitions of monomorphisms and epimorphisms. \item If $X\to A$ is an equalizer of $f,g\colon A\to B$, then $[X\to A]$ is a subobject of $A$. Furthermore, by the universality of the equalizer, it is the ``largest'' such subobject. Similarly, If $B\to Z$ is a coequalizer of $f,g$, then $[B\to Z]$ is the ``largest" quotient object of $B$. \item From the above discussion, we can safely say \emph{the} equalizer of $f$ and $g$ and \emph{the} coequalizer of $f$ and $g$. \item The equalizer of a morphism $f:A\to B$ and itself is the identity morphism $1_A$ on $A$. \item A category is said to \emph{have equalizers} if every pair of parallel morphisms has an equalizer. \end{itemize} One can also define an equalizer of an arbitrary set of morphisms with a common domain and a common codomain: if $\lbrace f_i:A\to B\mid i\in I\rbrace$ is a set of morphisms from $A$ to $B$, indexed by a set $I$, then an equalizer of the $f_i$'s is a morphism $d$ from an object $X$ to $A$ such that $d$ equalizes every pair of morphisms $f_i$ and $f_j$ and that $d$ is universal among all morphisms with such a property. \textbf{Remark}. An equalizer (coequalizer) is also known as a \emph{difference kernel} (\emph{difference cokernel}). This name is justifiably given as we recognize that a kernel of a morphism $f$ is, in a way, the ``difference" between $f$ and $o$, the zero morphism.