A Moore-Penrose pseudoinverse of a morphism $f: V \rightarrow W$ between Euclidean vector spaces is a map $g: W \rightarrow V$ in the other direction satisfying the identities

$fgf = f$

$gfg = g$

$(fg)^\ast = fg$

$(gf)^\ast = gf$

where $\phi^\ast$ denotes the adjoint of a linear map $\phi$.

Now the first two identities obviously resemble the triangle equalities of an adjunction. My question is: Can one actually understand the Moore-Penrose inverse as an adjoint? One possibility would be to find a "nice" (compatible with composition) partial order on Hom-Sets $\text{Hom}(V,W)$ making the category of Euclidean vector spaces into a 2-category, where the notion of adjunction is defined and the triangle equality in fact would imply $fgf = f$. So a more precise question would be: Does there exist such an order?

symmetryof the definition. The definition of a Moore-Penrose pseudoinverse is completly symmetric for $f$ and $g$. This means that if $f$ had been left (right) adjoint to $g$, then it would have automatically been also right (resp. left) adjoint to $g$. $\endgroup$ – Michal R. Przybylek Jun 5 '12 at 9:41nothingto do with your equations; it turns out that if we restrict our 2-categories to 2-posets, then your equations are satisfied, butnotbecause of the triangle equalities, but because of a mere existence of 2-morphisms $\mathit{id} \to gf$ and $fg \to \mathit{id}$ $\endgroup$ – Michal R. Przybylek Jun 5 '12 at 18:35definition. $\endgroup$ – Michal R. Przybylek Jun 8 '12 at 13:006more comments