# Moore-Penrose Inverse as an adjoint

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?

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Chris, it will make things even worse --- adjoint morphisms are unique up to 2-isomorphisms, and only 2-isomorphisms in a 2-poset are identities. The real problem is in the symmetry of 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 $g$. – Michal R. Przybylek Jun 5 '12 at 9:41
Now let us assume that we are in a 2-poset and $f$ is both left and right adjoint to $g$. From the first adjunction we have inequalities $\mathit{id} \le gf$, $fg \le \mathit{id}$, and from the second $\mathit{id} \le fg$, $gf \le \mathit{id}$. Thus $fg = \mathit{id}$ and $gf = \mathit{id}$, so the concept of adjoint sequence $f \dashv g \dashv f$ collapses to the concept of inverse. – Michal R. Przybylek Jun 5 '12 at 9:41
Nicolas, I am really skeptical about any such construction. Let me summarize two reasons: a) the triangle equalities say that the composition of the obvious morphisms $f \to fgf \to f$ and $g \to gfg \to g$ are identities; this has about nothing to do with your equations; it turns out that if we restrict our 2-categories to 2-posets, then your equations are satisfied, but not because of the triangle equalities, but because of a mere existence of 2-morphisms $\mathit{id} \to gf$ and $fg \to \mathit{id}$ – Michal R. Przybylek Jun 5 '12 at 18:35
Nicolas, I do not buy your explanation. I do not know much about "indempotent adjunction", but have found the following page: ncatlab.org/nlab/show/idempotent+adjunction. It seems that the idea behind the indempotent adjunction is to impose morphisms $f \rightarrow fgf$ and $g \rightarrow gfg$ to be isomorphisms. If you assume furthermore that these isomorphisms are identities, then you get your equations --- not "for free", but directly from your definition. – Michal R. Przybylek Jun 8 '12 at 13:00
Here is the final answer --- in any 2-category adjunctions composes, i.e. if $f \dashv g$ and $h \dashv k$ for compatible $f$ and $h$ then $f h \dashv k g$, whreas Moore-Penrose pseudoinverses generally do not. – Michal R. Przybylek Jun 9 '12 at 11:20

1. Equations $g = gfg$ and $fgf$ do not reaseble triangle equalities. Let me focus on the first equation. The "corresponding" triangle equality says that the composition $g \to^{\eta_g} gfg \to^{g\epsilon}g$ is the identity on $g$. Obviously $\eta_g$ is an inclusion, but generally there are no reasons for $\eta_g$ to be an isomorphism. However, in the world of 2-posets (i.e. categories enriched over posets) there is a reason. If we have a pair of 2-morphisms $\mathit{id} \le gf$ and $fg \le \mathit{id}$, then we may compose the first one on the right and the second on the left with $g$, obtaining $g \le gfg$ and $gfg \le g$, therefore $g = gfg$. Notice that I have not used any triangle equality here (just the existence of an appropriate pair of 2-morphisms).
2. The concept of an adjunction is inherently asymmetric --- a left adjoint is the best approximation of the identity from the left, and the right adjoint is the best approximation of the identity from the right; whilst the concept of Moore-Penrose pseudoinverse is perfectly symmetric. This means that if we had a 2-category, buit upon the category of vector spaces, where every pseudoinverse was a part of an adjunction (satisfying, perhaps, some other conditions), then every morphism would have both left and right adjoint, furthermore these adjoint functors would be isomorphic to the pseudoinverse (so isomorphic to each other). For a morphism $f$ in a 2-poset being both a left and right adjoint to $g$ is just being the inverse of $g$ --- the unit and counit from one adjunction give inequalities $\mathit{id} \le gf$ and $fg \le \mathit{id}$, whereas the unit and counit of the other adjunction give inequalities $\mathit{id} \le fg$ and $gf \le \mathit{id}$, hence $\mathit{id} = gf$ and $\mathit{id} = fg$. This fully answers your precise question, since in the category of vector spaces not every pesudoinverse is an inverse.
3. In any 2-category adjunctions compose --- if $f \colon A \to B \dashv f^+ \colon B \to A$ and $g \colon B \to C \dashv g^+ \colon C \to B$ then $g f \colon A \to C \dashv f^+ g^+ \colon C \to A$, but Moore-Penrose pseudoinverses --- generally --- do not. This almost answers your main question, because if $f \colon A \to B$, $g \colon B \to C$ are any maps between vector spaces then from the above property: $(gf)^+ \approx f^+ g^+$, so from this point of view pseudoinverses are not stable under isomorphisms, thus are not categorical.
I'm not sure what you mean when you say that pseudoinverses aren't categorical. The definition can be phrased in a purely categorical manner in any dagger category so ought to be invariant under unitary isomorphisms (isomorphisms $\phi$ such that $\phi^{\dagger} = \phi^{-1}$). – Qiaochu Yuan Jun 9 '12 at 18:51