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Consider the category whose objects are $C^\ast$-algebras and for which the morphisms between $A$ and $B$ are the $\ast$-homomorphisms $\phi$ from $A$ into the multiplier algebra $M(B)$ such that $\overline{\operatorname{span}\phi(A)B}$ is dense in $B$. Note that these morphisms can be extended in a unique way to strictly continuous $\ast$-homomorphisms from $M(A)$ to $M(B)$ so that composition makes sense.

If one restricts to commutative $C^\ast$-algebras, this category is dual to the category of locally compact Hausdorff spaces with all (not only proper) continuous maps as morphisms, which makes it natural from the viewpoint of "noncommutative topology". For example, this notion of morphisms is used in the definition of locally compact quantum groups.

Question: Does this category have finite coproducts?

It does not seem like the usual free product $A\ast B$ does the job since this comes with $\ast$-homomorphisms $A,B\to A\ast B$ (not into the multiplier algebra), and I see no reason to assume this would be true for the coproduct in the category described above.

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Given two locally compact spaces $X$ and $Y$ then the product $X \times Y$ is an open subset inside $\overline{X} \times \overline{Y}$, where $\overline{X}$ and $\overline{X}$ are the one point compactification of $X$ and $Y$ so you can build this product of locally compact spaces as an open inside a product of compact spaces. I believe this can adapted to the non-commutative case to build the coproduct you are looking for.

Disclaimer: I hadn't thought about it before and I haven't checked all the details, so take this with a grain of salt - some checking might be required. But this seem to be working:

Let $A^+$ and $B^+$ the unitarization of $A$ and $B$. Take the free product $A^+ * B^+$ and the two sided ideal inside of it generated by the $a * b$ for $a \in A$ and $b\in B$. Call this $A \star B \subset A^+ * B^+$.

The two sided ideal inclusion $A \star B $ into $A^+ * B^+$ gives you a map $A^+ * B^+ \to M(A \star B)$ which in turn give you maps $A \to M(A \star B)$ and $B \to M(A \star B)$ that are in your category.

Now, any map $A \to M(C)$ in your category extend to a map unital map $A^+ \to M(C)$.

So given two $A \to M(C)$ and $B \to M(C)$, you get maps $A^+ \to M(C)$ and $B^+ \to M(C)$ which induces $A^+ * B^+ \to M(C)$ and restrict to $A \star' B \to M(C)$, which also belongs to your category.

Uniqueness seems clear.

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    $\begingroup$ Thank you! I've been wondering about something like this for a long time, though what I really want is a certain class of pushouts. I think I'll ask another question about that one. $\endgroup$
    – David Roberts
    Commented 8 hours ago
  • $\begingroup$ Thank you for the answer. It seems to me like the details work out. This construction matches nicely with the commutative picture in that one has products with a factor from each algebra in order to have decay to zero at infinity in all variables. $\endgroup$
    – MaoWao
    Commented 3 hours ago
  • $\begingroup$ One question: Did I miss something or can one work with the multiplier algebra instead of the unitization throughout? $\endgroup$
    – MaoWao
    Commented 3 hours ago
  • $\begingroup$ I suspect you can, but I felt like the unitarization gave you a more concrete answer I was more confortable with. $\endgroup$ Commented 2 hours ago

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