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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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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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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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