QuasiInverseOfAFunction 2006-11-03 10:57:37 2009-04-06 17:44:47 Definition function quasi-inverse function \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} \usepackage{pst-plot} \usepackage{psfrag} % define commands here Let $f:X\to Y$ be a function from sets $X$ to $Y$. A \emph{quasi-inverse} $g$ of $f$ is a function $g$ such that \begin{enumerate} \item $g:Z\to X$ where $\operatorname{ran}(f)\subseteq Z\subseteq Y$, and \item $f\circ g\circ f=f$, where $\circ$ denotes functional composition operation. \end{enumerate} Note that $\operatorname{ran}(f)$ is the range of $f$. \textbf{Examples.} \begin{enumerate} \item If $f$ is a real function given by $f(x)=x^2$. Then $g(x)=\sqrt{x}$ defined on $[0,\infty)$ and $h(x)=-\sqrt{x}$ also defined on $[0,\infty)$ are both quasi-inverses of $f$. \item If $f(x)=1$ defined on $[0,1)$. Then $g(x)=\frac{1}{2}$ defined on $\mathbb{R}$ is a quasi-inverse of $f$. In fact, any $g(x)=a$ where $a\in [0,1)$ will do. Also, note that $h(x)=x$ on $[0,1)$ is also a quasi-inverse of $f$. \item If $f(x)=[x]$, the step function on the reals. Then by the previous example, $g(x)=[x]+a$, any $a\in[0,1)$, is a quasi-inverse of $f$. \end{enumerate} \textbf{Remarks.} \begin{itemize} \item Every function has a quasi-inverse. This is just another form of the Axiom of Choice. In fact, if $f:X\to Y$, then for \emph{every} subset $Z$ of $Y$ such that $\operatorname{ran}(f)\subseteq Z$, there is a quasi-inverse $g$ of $f$ whose domain is $Z$. \item However, a quasi-inverse of a function is in general not unique, as illustrated by the above examples. When it is unique, the function must be a bijection: \begin{quote} If $\operatorname{ran}(f)\ne Y$, then there are at least two quasi-inverses, one with domain $\operatorname{ran}(f)$ and one with domain $Y$. So $f$ is onto. To see that $f$ is one-to-one, let $g$ be the quasi-inverse of $f$. Now suppose $f(x_1)=f(x_2)=z$. Let $g(z)=x_3$ and assume $x_3\ne x_1$. Define $h:Y\to X$ by $h(y)=g(y)$ if $y\ne z$, and $h(z)=x_1$. Then $h$ is easily verified as a quasi-inverse of $f$ that is different from $g$. This is a contradition. So $x_3=x_1$. Similarly, $x_3=x_2$ and therefore $x_1=x_2$. \end{quote} \item Conversely, if $f$ is a bijection, then the inverse of $f$ is a quasi-inverse of $f$. In fact, $f$ has only one quasi-inverse. \item The relation of being quasi-inverse is not symmetric. In other words, if $g$ is a quasi-inverse of $f$, $f$ need not be a quasi-inverse of $g$. In the second example above, $h$ is a quasi-inverse of $f$, but not vice versa: $h(0)=0$, but $hfh(0)=hf(0)=h(1)=1\ne h(0)$. \item Let $g$ be a quasi-inverse of $f$, then the restriction of $g$ to $\operatorname{ran}(f)$ is one-to-one. If $g$ and $f$ are quasi-inverses of one another, and $\operatorname{g}$ strictly includes $\operatorname{ran}(f)$, then $g$ is \emph{not} one-to-one. \item The set of real functions, with addition defined element-wise and multiplication defined as functional composition, is a ring. By remark 2, it is in fact a Von Neumann regular ring, as any quasi-inverse of a real function is also its pseudo-inverse as an element of the ring. Any space whose ring of continuous functions is Von Neumann regular is a P-space. \end{itemize} \begin{thebibliography}{9} \bibitem{ss} B. Schweizer, A. Sklar, \emph{Probabilistic Metric Spaces}, Elsevier Science Publishing Company, (1983). \end{thebibliography}