Let $\mathcal{C}^*$ be the category defined as follows:


$2.$ For $A,B\in\mathcal{C}^*$ set $\text{Hom}_{\mathcal{C}^*}(A,B)=\{\varphi:A\to B\;|\; \varphi \text{ is a $*$-homomorphism} \}$

This is a category (it can be easily verified) and we know that $\text{Hom}_{\mathcal{C}^*}(A,B)$ is exactly the set of norm-decreasing homomorphisms (also easily proved). With these homomorphisms, $\mathcal{C^*}$ actually is NOT additive, as we cannot simply add $*$-homomorphisms (the result is not a $*$-homomorphism). However, we do have kernels. My question is: does $\mathcal{C}^*$ have cokernels? I believe so, and here is my work:

Let $A\xrightarrow{\varphi}B$ be a morphism in $\mathcal{C}^*$. Let $I$ denote the closed (double) ideal generated by $\varphi(A)$. Then $B/I$ (with quotient norm) is a $C^*$-algebra and we have the quotient $*$-homomorphism $\pi:B\to B/I$. I claim that $\text{cokernel}(A\xrightarrow{\varphi}B)=B\xrightarrow{\pi}B/I$.

The description of $I$ is the closed linear span of the set $S:=\{\varphi(a),\varphi(a)b, b\varphi(a), b\varphi(a)b': a\in A, b,b'\in B\}$ (maybe this can be simplified if we consider the unitization of $B$, but I'd like to avoid it right now). If $B\xrightarrow{\psi}D$ is a $*$-homomorphism such that $\psi\circ\varphi=0$, then we define $\bar{\psi}:B/I\to D$ by $\bar{\psi}(b+I)=\psi(b)$.

First we need to show that $\bar{\psi}$ is a well defined $*$-homomorphism. The only non-trivial thing is that $\bar{\psi}$ is well defined. But if $b-b'\in I$, we need to show that $\psi(b-b')=0$. But $b-b'=\displaystyle{\lim_{n\to\infty}u_n}$ where $u_n\in\overline{\text{span}(S)}$. By $\psi$'s continuity and preservation of sum and scale, it suffices to show that $\psi\equiv0$ on $S$. but this is true, since $\psi(\varphi(a))=0, \psi(\varphi(a)b)=\psi(\varphi(a))\psi(b)=0,\dots$ and so on.

Obviously $\bar{\psi}\circ\pi=\psi$ and if $\bar{\psi}':B/I\to D$ was another such morphism, then $\bar{\psi}'(b+I)=\bar{\psi}'\circ\pi(b)=\psi(b)=\bar{\psi}(b+I)$, so we have uniqueness.

This is the universal property of cokernels, so we are done. Does my proof contain mistakes? If someone can confirm that there are no mistakes, I suggest editing the title and taking the question mark off.

A comment: Note that $\varphi(A)$ is not equal to the image in the categorical setting: an obvious reason: the image is defined as the kernel of the cokernel. but kernels are always ideals, $\varphi(A)$ is not necessarily an ideal. Also, we know that $\varphi(A)\cong A/\ker(\varphi)$, the co-image of $\varphi$.

  • $\begingroup$ It looks fine for me. $\endgroup$
    – Berci
    Mar 17, 2020 at 21:27
  • $\begingroup$ @Berci Thank you for your reply $\endgroup$ Mar 18, 2020 at 0:45
  • 5
    $\begingroup$ I don't think that $\text{Hom}_{\mathcal{C}^*}(A,B)$ is an abelian group under any operations. Additionally, you can show that the ideal generated by $\varphi(A)$ is the closed span of $\{b\varphi(a)b':a\in A,b\in B\}$ using an approximate identity argument. $\endgroup$
    – Aweygan
    Mar 22, 2020 at 1:51
  • $\begingroup$ @Aweygan I see, I rushed on this and didnt notice that the sum of algebra homomorphisms is not an algebra homomorphism. Thank you for noticing, I will edit my post soon. $\endgroup$ Mar 22, 2020 at 2:02

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