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There is a known construction of pushout of $C^*$-algebras, or rather, the amalgamated free product, which is universal for commutative squares of $*$-homomorphisms. Jensen and Thomsen in their book Elements of KK-theory give, in an appendix, a detailed treatment for the free product (i.e the coproduct), for instance, and Pedersen gives a treatment of pushouts in Pullback and Pushout Constructions in $C^*$-Algebra Theory, J. Func. Analysis 1999, doi:10.1006/jfan.1999.3456, or Core pdf.

However, $*$-homomorphisms are not the only morphisms between $C^*$-algebras that are reasonable to consider. For instance, a non-proper map $X\to Y$ between locally compact Hausdorff spaces induces a $*$-homomorphism $C_0(Y) \to M(C_0(X)) \simeq C_b(X)$, which is continuous for the strict topology on $C_0(Y)$. This motivates somewhat the more general type of morphism, where we can take a map $A\rightsquigarrow B$ between $C^*$-algebras to be a strictly continuous $*$-homomorphism $A\to M(B)$, where $M(B)$ is the multiplier algebra of $B$. Any $*$-homomorphism $A\to B$ gives one of these more general morphisms, namely the composite $A\to B \hookrightarrow M(B)$. It is a fun fact that as a set $M(B)$ is the completion in the strict topology of $B$, though we regard it as being equipped with both the strict topology and its (Banach) $C^*$-topology. These generalised morphisms compose by using the universal property of the strict completion, and so we have a category.

I'm interested in the pushout, in this category, of an arbitrary generalised morphism and a generalised morphism arising from an injective $*$-homomorphism, in particular the inclusion of a full corner.

Conjecture: A full corner inclusion pushes out to also give (the generalised morphism arising from) a full corner inclusion.

Is this true?

Added I thought it worthwhile to spell out the details somewhat. Let $A$ and $B$ be arbitrary $C^*$-algebras, $A_0 = pAp \subseteq A$ be a full corner and $\phi\colon A\to M(B)$ a strictly continuous $*$-homomorphism. Then I am after a $C^*$-algebra $A\sharp_{A_0} B$ and strictly continuous $*$-homomorphisms $i\colon B\to M(A\sharp_{A_0} B)$ and $\psi\colon A\to M(A\sharp_{A_0} B)$ making the diagram $\require{AMScd}$ \begin{CD} A_0 @>>> A \\ @V \phi V V @VV\psi V\\ M(B) @>>I> M(A\sharp_{A_0} B) \end{CD} commute, where $I$ is the unique extension of $i$ to $M(B)$. My conjecture is that $B$ is a full corner of $A\sharp_{A_0} B$, and $i$ factors as the composite $B\hookrightarrow A\sharp_{A_0} B \to M(A\sharp_{A_0} B)$ of the two inclusions. [Added 2 I forgot to explicitly mention that this should be universal: given a $C^*$-algebra $C$, and $B\to M(C)$ and $A\to M(C)$ making the analogous square commute, there should be a unique $A\sharp_{A_0} B \to M(C)$ making the usual triangles commute.]

If this diagram only commutes up to intertwining by an element of $M(A\sharp_{A_0} B)$, then I would be happy with that, too.

Improve this question
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  • $\begingroup$ I guess morphisms need to be non-degenerate in order for the composition to work. $\endgroup$
    – Ruy
    Commented Sep 29, 2020 at 2:17
  • $\begingroup$ Strictly continuous is sufficient, since then the maps extend as needed. We don't need non-degenerate; strictly continuous means the image of the approximate identity of $A$ in $M(B)$ converges to a projection that is not the identity map on $B$, and this is enough. $\endgroup$
    – David Roberts ♦
    Commented Sep 29, 2020 at 3:07
  • $\begingroup$ I think $A \sharp_{A_0} B$ can be defined, but $B$ need not be a corner there. Let's consider the simplest case where $A=M_2(\mathbf{C})$ and $B=\mathbf{C}\oplus\mathbf{C}$ with $A_0=\mathbf{C}e_{1,1}$ identified with $\mathbf{C}\oplus 0\subset B$. They have an obvious representation into $M_2(\mathbf{C})$. Since $B$ is not a corner there, it cannot be a corner of $A\ast_{A_0}B$. $\endgroup$
    – Narutaka OZAWA
    Commented Oct 1, 2020 at 3:00
  • $\begingroup$ @NarutakaOZAWA I just realised I forgot to ask that this square satisfies a universal property in the edit, though it's mentioned in the first part of the question $\endgroup$
    – David Roberts ♦
    Commented Oct 1, 2020 at 4:17
  • $\begingroup$ @David Roberts: $A\ast_{A_0}B$ satisfies the universal property and surjects onto $M_2(\mathbf{C})$ by a *-homomorphism $\pi$. If $B$ is a corner in $A\ast_{A_0}B$, then $\pi(B)$ has to be a corner in $\pi(A\ast_{A_0}B)$. $\endgroup$
    – Narutaka OZAWA
    Commented Oct 1, 2020 at 4:30

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