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This question is motivated by Yemon Choi's answer here: Epimorphisms have dense range in TopHausGrp?

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).

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  • $\begingroup$ In the nonunital case, we have to restrict continuous maps between locally compact Hausdorff spaces to only the proper ones. To get a categorical duality, I think we also have to restrict -homomorphisms between (nonunital) C-algebras to nondegenerate ones, i.e. those $f \colon A \to B$ for which $\{f(a)b \mid a \in A, b \in B\}$ is dense in $B$. So we really want to know what epimorphisms in this category are, which should hopefully make the question easier. $\endgroup$
    – Chris Heunen
    Commented Mar 16, 2011 at 20:58
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    $\begingroup$ Chris, I may be missing something, but isn't the whole point of Matthew's question that he doesn't want to restrict only to proper continuous maps between LCH spaces, but instead consider all continuous maps between such? $\endgroup$
    – Yemon Choi
    Commented Mar 17, 2011 at 1:09
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    $\begingroup$ Oh, I see. So the question is equivalent to what the monomorphisms in the category of locally compact Hausdorff spaces with continuous maps are? $\endgroup$
    – Chris Heunen
    Commented Mar 17, 2011 at 6:32
  • $\begingroup$ Yes, exactly-- if you restrict to commutative C*-algebras, then you can equivalently ask: "What are monomorphisms and epimorphisms in the category of locally compact Hausdorff spaces with continuous maps as morphisms?" $\endgroup$
    – Matthew Daws
    Commented Mar 17, 2011 at 10:18

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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. (EuDML)

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  • $\begingroup$ I wasn't aware of the paper; it looks interesting. (It seems not to have been noticed or cited by the authors of the paper I mentioned in the answer Matthew links to -- a case of reinventing the wheel?) $\endgroup$
    – Yemon Choi
    Commented Mar 14, 2011 at 19:41
  • $\begingroup$ Yes, so this gives a nice answer to the question with *-homomorphisms-- as Yemon says, this paper also pre-dates the reference Yemon found). But, I'll leave this open, as I think this doesn't answer my actual question. $\endgroup$
    – Matthew Daws
    Commented Mar 14, 2011 at 22:08
  • $\begingroup$ Btw, I regard $*$-homomorphisms as "too restrictive" by comparison with the commutative case-- $*$-homomorphisms $C_0(X)$ to $C_0(Y)$ only correspond to "proper" (inverse image of compact is compact) continuous maps $X\rightarrow Y$. If you want all continuous maps, then you need this more general notion of morphism. $\endgroup$
    – Matthew Daws
    Commented Mar 15, 2011 at 8:15
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    $\begingroup$ In fact, the algebra homomorphisms seem to be too restrictive when one deals with tensor products (for instance Haagerup tensor product). The point is that sometimes seems more natural to consider B as an A-bimodule, and then to obtain the morphism of category of Hilbert C*-modules over A into the one over B. But then is seems more natural to consider the maps A$\to$End<sub>B</sub>B = M(B). This picture has very nice analogies in the operator algebra theory. $\endgroup$
    – Kolya Ivankov
    Commented Mar 17, 2011 at 16:32

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