PointedSet 2006-05-29 20:38:58 2007-03-18 12:20:10 Definition basepoint pointed subset \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} % define commands here \subsection{Definition} A \emph{pointed set} is an ordered pair $(A,a)$ such that $A$ is a set and $a\in A$. The element $a$ is called the \emph{basepoint} of $(A,a)$. At first glance, it seems appropriate enough to call any non-empty set a pointed set. However, the basepoint plays an important role in that if we select a different element $a^{\prime}\in A$, the ordered pair $(A,a^{\prime})$ forms a different pointed set from $(A,a)$. In fact, given any non-empty set $A$ with $n$ elements, $n$ pointed sets can be formed from $A$. A function $f$ between two pointed sets $(A,a)$ and $(B,b)$ is just a function from $A$ to $B$ such that $f(a)=b$. Whereas there are $|B|^{\mid A\mid}$ functions from $A$ to $B$, only $|B|^{\mid A\mid-1}$ of them are from $(A,a)$ to $(B,b)$. Pointed sets are mainly used as illustrative examples in the study of universal algebra as algebras with a single constant operator. This operator takes every element in the algebra to a unique constant, which is clearly the basepoint in our definition above. Any \PMlinkname{homomorphism}{HomomorphismBetweenAlgebraicSystems} between two algebras preserves basepoints (taking the basepoint of the domain algebra to the basepoint of the codomain algebra). From the above discussion, we see that a pointed set can alternatively described as any constant function $p$ where the its domain is the underlying set, and its range consists of a single element $p_0\in \operatorname{dom}(p)$. A function $f$ from one pointed set $p$ to another pointed set $q$ can be seen as a function from the domain of $p$ to the domain of $q$ such that the following diagram commutes: $$\xymatrix{ {\operatorname{dom}(p)}\ar[r]^{f}\ar[d]_{p}&{\operatorname{dom}(q)}\ar[d]^{q}\\ {\lbrace p_0\rbrace}\ar[r]_{c}&{\lbrace q_0\rbrace}} $$ \subsection{Creation of Pointed Sets from Existing Ones} \textbf{Pointed Subsets}. Given a pointed set $(A,a)$, a pointed subset of $(A,a)$ is an ordered pair $(A^{\prime},a)$, where $A^{\prime}$ is a subset of $A$. A pointed subset is clearly a pointed set. \textbf{Products of Pointed Sets}. Given two pointed sets $(A,a)$ and $(B,b)$, their product is defined to be the ordered pair $(A\times B,(a,b))$. More generally, given a family of pointed sets $(A_i,a_i)$ indexed by $I$, we can form their Cartesian product to be the ordered pair $(\prod A_i, (a_i))$. Both the finite and the arbitrary cases produce pointed sets. \textbf{Quotients}. Given a pointed set $(A,a)$ and an equivalence relation $R$ defined on $A$. For each $x\in A$, define $\overline{x}:=\lbrace y\in A \mid y R x\rbrace$. Then $A/R:=\lbrace \overline{x}\mid x\in A\rbrace$ is a subset of the power set $2^A$ of $A$, called the quotient of $A$ by $R$. Then $(A/R,\overline{a})$ is a pointed set.