Does the unitization of a Banach algebra satisfy some universal property?

Question

Let $A$ be a Banach algebra, and $A^e = A \times \Bbb C$ be its unitization, with multiplication defined as $$(a,p)\cdot(b,q) = (ab + pb + qa, pq)$$ $A^e$ is also a Banach algebra with $(0,1)$ as its unit. Does the unitization of a Banach algebra satisfy some universal property? I am hoping to define $A^e$ in a "categorical way", whatever that means. Thanks for any insights!

Answer 1

Say $B$ is a complex unital Banach algebra with an injective linear homomorphism $\varphi:A\to B$. Claim: there is a unique morphism of unital Banach algebras $\psi:A^e\to B$ with $\psi(a,0)=\varphi$ for all $a\in A$ (i.e. $\psi$ uniquely takes the inclusion $A\hookrightarrow A^e$ into $\varphi$).

Well that's not too hard to show. Because $\psi(0,1)=\mathbf{1}_B$ the unit is demanded as well as linearity, so that $\psi(a,\lambda)=\psi(a,0)+\psi(0,\lambda)=\varphi(a)+\lambda\mathbf{1}_B$ is forced. So there is one and only one way to define $\psi$. Probably this is compatible with continuity considerations too. We could actually drop quite a lot here, e.g. the injectivity and the "homomorphism" aspect are irrelevant.

Idea: adjoining a unit $e$ demands also adjoining at least $\{a+\operatorname{span}(e):a\in A\}\sim A\times\Bbb C$. The most efficient way to do so is to just take $A\times\Bbb C$ itself.

Disclaimer: I kinda made this up. I don't know if this is the canonical universal property associated to $A^e$.