quasi-inverse of a function

Let $f:X\to Y$ be a function from sets $X$ to $Y$. A quasi-inverse $g$ of $f$ is a function $g$ such that

1. $g:Z\to X$ where $\mathrm{ran}(f)\subseteq Z\subseteq Y$, and

2. $f\circ g\circ f=f$, where $\circ$ denotes functional composition operation.

Note that $\mathrm{ran}(f)$ is the range of $f$.

Examples.

1. If $f$ is a real function given by $f(x)=x^2$. Then $g(x)=\sqrt{x}$ defined on $[0,\mathrm{\infty })$ and $h(x)=-\sqrt{x}$ also defined on $[0,\mathrm{\infty })$ are both quasi-inverses of $f$.

2. If $f(x)=1$ defined on $[0,1)$. Then $g(x)=\frac{1}{2}$ defined on $ℝ$ 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 $f$.

3. 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 $f$.

Remarks.

• 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 every subset $Z$ of $Y$ such that $\mathrm{ran}(f)\subseteq Z$, there is a quasi-inverse $g$ of $f$ whose domain is $Z$.

• 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:

  If $\mathrm{ran}(f)\ne Y$, then there are at least two quasi-inverses, one with domain $\mathrm{ran}(f)$ 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 $x_1=x_2$.

• 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.

• 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 $hfh(0)=hf(0)=h(1)=1\ne h(0)$.

• Let $g$ be a quasi-inverse of $f$, then the restriction of $g$ to $\mathrm{ran}(f)$ is one-to-one. If $g$ and $f$ are quasi-inverses of one another, and $\mathrm{g}$ strictly includes $\mathrm{ran}(f)$, then $g$ is not one-to-one.

• 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.