Mono- and epi-morphisms for C*-algebras

2011-03-14T12:02:14Z

This question is motivated by Yemon Choi's answer here: http://mathoverflow.net/questions/56453/epimorphisms-have-dense-range-in-tophausgrp/56459#56459

It's well-known that the category of unital commutative C*-algebras and $*$ -homomorphisms is dual to the category of compact Hausdorff spaces and continuous maps. One finds that for all C*-algebras (with $*$ -homomorphisms as morphisms) that monomorphisms are just injective maps, and epimorphisms are surjections (the latter point is non-trivial-- see the final paper which Yemon suggests in the link above).

Suppose instead we look at locally compact Hausdorff spaces, with continuous maps as morphisms. Then dually, we get all commutative C*-algebras, but now the notion of a $*$ -homomorphism is too restrictive (it corresponds to proper continuous maps). Instead we say that a morphism between C*-algebras $A$ and $B$ is a non-degenerate $*$ -homomorphism $A\rightarrow M(B)$ form $A$ to the multiplier algebra of $B$. Such a map extends uniquely to a strictly continuous $*$ -homomoprhisms $M(A)\rightarrow M(B)$, and so we can compose such maps. Hence we get a category. A little checking shows that the full subcategory of commutative C*-algebras, with these morphisms, is now dual to the category of locally compact Hausdorff spaces with continuous maps. (I think Woronowicz was the first person to articulate this view).

For C*-algebras, with morphisms as arrows, what are epimorphisms and monomorphisms?

Restricting to the commutative case, we can instead look at locally compact Hausdorff spaces, and reverse the arrows. So working through, a monomorphism remains just an injective map; but I see no simple description of epimorphisms (at the level of algebras-- for spaces, it's just injective continuous maps).

Edit: Maybe this notion of "non-degenerate" is confusing. If $f:X\rightarrow Y$ is a continuous map between locally compact Hausdorff spaces, then we define $f_*:C_0(Y)\rightarrow C^b(X); a \mapsto a\circ f$. Notice that we really do need the codomain to be all bounded continuous functions-- but that's okay, as $C^b(X)$ is just the multiplier algebra of $C_0(X)$, and $f_*$ turns out to be non-degenerate. Conversely, every non-degenerate $*$ -homomorphism $C_0(Y)\rightarrow C^b(X)$ arises in this way (but a general $*$ -homomorphism $C_0(Y)\rightarrow C^b(X)$ can be much more complicated).

Answer by Chris Heunen for Mono- and epi-morphisms for C*-algebras

2011-03-14T16:49:15Z

I'm not sure why *-homomorphisms would be too restrictive a choice, but taking those as morphisms between C*-algebras as objects, the epimorphisms are precisely the surjective *-homomorphisms. This is proposition 2 in G. A. Reid's "Epimorphisms and surjectivity", Inventiones Mathematicae 9:295-307, 1970.

Mono- and epi-morphisms for C*-algebras - MathOverflow

Mono- and epi-morphisms for C*-algebras

Answer by Chris Heunen for Mono- and epi-morphisms for C*-algebras