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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?

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    $\begingroup$ The notion of a von Neumann inverse or of a von Neumann regular element has some resemblance to what you're looking for. $\endgroup$ Apr 26, 2011 at 8:56

3 Answers 3

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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".

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  • $\begingroup$ 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? $\endgroup$
    – user14613
    Apr 23, 2011 at 1:25
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    $\begingroup$ 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, en.wikipedia.org/wiki/Moore%E2%80%93Penrose_pseudoinverse). $\endgroup$
    – user6976
    Apr 23, 2011 at 2:02
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    $\begingroup$ @t-a-w: In domain theory you'd have seen Galois connections - is that it? $\endgroup$ Apr 23, 2011 at 9:19
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    $\begingroup$ 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. $\endgroup$
    – user14613
    Apr 23, 2011 at 13:30
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    $\begingroup$ @t-a-w: Perhaps Colin can upgrade his comment to an answer and then you will accept this answer. I think it is a common practice here. $\endgroup$
    – user6976
    Apr 23, 2011 at 13:42
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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.

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  • $\begingroup$ One way to construct such $h$ is to take idempotents. This seems to be the case for the examples you construct (I think you meant $h(x,y) = (f(y),g(x))$). $\endgroup$ May 1, 2020 at 15:24
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This is also naturally related to adjointness:

if you consider both $X$ and $Y$ to be poset categories, and add the conditions

  • both $f$ and $g$ order-preserving
  • $x \leq_X g(f(x))$ and $f(g(y)\leq_Y y$ for all $(x,y)\in X\times Y$

to your conditions (this is admittedly a little less general, but satisfied in many natural situations), then:

the conditions are met

$\Longleftrightarrow$

$f:X\leftrightarrow Y:g$

is an adjunction.

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