MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

A very simple question, I just totally forgot how it was called, and google is not helping.

There's a pair of functions $f:X\to Y$, $g:Y\to X$.

$fgf = f$, $gfg = g$, but $fg$ and $gf$ don't need to be identities (and usually are not in interesting cases).

A simple example would be $f(a,b,c)=(a,b)$, $g(a,b)=(a,b,0)$

What were $f$ and $g$ called?

share|cite|improve this question
The notion of a von Neumann inverse or of a von Neumann regular element has some resemblance to what you're looking for. – Matthias Künzer Apr 26 '11 at 8:56
up vote 10 down vote accepted

It is called "generalized inverse". In that case $fg$ and $gf$ are idempotents. In particular, if you have a semigroup of maps $X\to X$ (i.e. a set of maps closed under composition) such that every $f$ has a generalized inverse, the semigroup is called regular. If the generalized inverse is unique, the semigroup is called inverse. See Clifford and Preston "Algebraic theory of semigroups".

share|cite|improve this answer
Yes, $fg$ and $gf$ would be idempotents. I'm sure the name was something else than "generalized inverse" back when I learned it. It was in context of computer science and program analysis, $f$ would normally project to a simpler domain, $g$ would expand back. If $fgf=f$ and $gfg=g$, this allows every function of tuples of $X$ to be extended to tuples of $Y$. Or something like that. Does it help? – user14613 Apr 23 '11 at 1:25
I do not know the application you have in mind, unfortunately. In the case of matrices (i.e. linear maps from ${\mathbb R}^n$ to ${\mathbb R}^m$), it is sometimes called pseudoinverse (see, say, – Mark Sapir Apr 23 '11 at 2:02
In your case, by the way, you have two linear transformations $(a,b,c)\to (a,b)$ and $(a,b)\to (a,b,0)$ which are Moore-Penrose pseudoinverses of each other. – Mark Sapir Apr 23 '11 at 3:41
@t-a-w: In domain theory you'd have seen Galois connections - is that it? – Colin McQuillan Apr 23 '11 at 9:19
Thanks, "galois connections" is just the name I've been looking for. Too bad I cannot accept a comment as answer, so I'll accept this one and upvote the comment. – user14613 Apr 23 '11 at 13:30

Linear case

In the linear case, these identities are part of the definition of the Moore-Penrose pseudo-inverse, which exists and is unique. Given $A\in M_{p\times q}(\mathbb C)$, its MPpi is the matrix $A^\dagger\in M_{q\times p}(\mathbb C)$ that satisfies $$AA^\dagger A=A,\qquad A^\dagger AA^\dagger=A^\dagger,\qquad(AA^\dagger)^H=AA^\dagger,\qquad(A^\dagger A)^H=A^\dagger A,$$ where the superscript $H$ stands for the Hermitian adjoint.

If $A\in GL_n(\mathbb C)$, then $A^\dagger=A^{-1}$. But otherwise, $AA^\dagger$ and $A^\dagger A$ are only unitary projections.

Nonlinear case

The situation where $f=g$ is amazing: one looks at functions $h$ such that $h\circ h\neq {\rm id}$, whereas $h\circ h\circ h=h$. Then we have $h^{(2k)}=h^2$ and $h^{(2k-1)}=h$ for every $k\ge1$.

Such an $h$ can be obtained by the following construction, when we are given $f,g$ such that $fgf=f$, $gfg=g$ and at least one of $fg$ or $gf$ is not the identity. Just define $h(x,y)=(f(x),g(y))$ on the cartesian product.

Application: take for $f$ the backward shift on $\ell^p({\mathbb N})$ and for $g$ the forward shift.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.