Is the category, $\textbf{FHILB}$, of finite dimensional Hilbert spaces and linear maps locally regular, where `locally regular' is defined like this
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Yes. Since it has a terminal object (namely the zero-dimensional space 0=$\{0\}$), the slice category $\mathbf{FHilb} / 0$ is isomorphic to $\mathbf{FHilb}$ itself. Because all slices of a locally regular category are regular, local regularity is equivalent to regularity.
And in fact, $\mathbf{FHilb}$ is regular. It is finitely complete and cocomplete, so definitely has coequalizers of kernel pairs. Regular epimorphisms are surjections, which are stable under pullback, because pullbacks are computed as in $\mathbf{Set}$.
answered Apr 28, 2015 at 9:35
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1$\begingroup$ How close are we to concluding that $\mathbf{FHilb}$ is regular because it's aglebraic? (It isn't quite algebraic, but seems "mostly" so.) $\endgroup$ Commented Apr 28, 2015 at 12:42
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$\begingroup$ Thanks Chris. Don't know why I had it in my head that $\textbf{FHILB}$ wasn't regular. But I am right in thinking that $\textbf{HILB}$ (where you include infinite dimensional spaces) isn't regular, aren't I? If that's the case, then I guess that the same argument shows that $\textbf{HILB}$ isn't locally regular? $\endgroup$ Commented Apr 29, 2015 at 8:08
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1$\begingroup$ Yes, for $\mathbf{Hilb}$ regularity and local regularity are also equivalent. $\mathbf{Hilb}$ is still finitely complete and finitely cocomplete. I'm not sure about pullbacks of regular epis, but if you replace regular epis by so-called zero epis you do get a good factorization system and everything works beautifully, see arxiv.org/abs/0902.2355. $\endgroup$ Commented Apr 29, 2015 at 20:51
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$\begingroup$ Thanks Chris. I'll have a look at that paper. But one more question, I thought Cartesian products didn't always exist in $\textbf{HILB}?$ $\endgroup$ Commented May 1, 2015 at 12:06
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$\begingroup$ That's true, but finite ones do (i.e. products of finitely many objects) $\endgroup$ Commented May 1, 2015 at 16:02