# The category of C* algebras has cokernels

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

$$1.\text{Ob}(\mathcal{C}^*)=\{C^*\text{-algebras}\}$$

$$2.$$ 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 $$\varphi$$.

• It looks fine for me. Mar 17, 2020 at 21:27
• @Berci Thank you for your reply Mar 18, 2020 at 0:45
• 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. Mar 22, 2020 at 1:51
• @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. Mar 22, 2020 at 2:02