Unitization

Remarks

One writes $(\lambda, a)$ as $\lambda 1 + a$. Every algebra homomorphism $\varphi$ from $A$ to a unital algebra $B$ extends uniquely to a unital homomorphism $A^+ \to B$, by $\lambda 1 + a \mapsto \lambda 1_B + \varphi(a)$. If $A$ is a normed algebra, then $\norm{(\lambda, a)} = \abs{\lambda} + \norm{a}$ makes $A^+$ a normed algebra, complete when $A$ is complete. Indeed $\abs{\lambda \mu} + \norm{\lambda b + \mu a + a b} \le (\abs{\lambda} + \norm{a})(\abs{\mu} + \norm{b})$. For a C*-algebra $A$ one uses another norm on $A^+$, which makes it a C*-algebra. The construction adds a new unit even when $A$ already has one.

Example: for $A = C_0(\R)$, the complex continuous functions that tend to $0$ at infinity, $A^+$ is isomorphic to the functions $f + \lambda$ with $f \in A$ and $\lambda \in \C$. These are the continuous functions on $\R$ with one common limit at $+\infty$ and $-\infty$. They are the continuous functions on the one-point compactification of $\R$.

Non-example: the componentwise product $(\lambda, a)(\mu, b) = (\lambda \mu, a b)$ on $\mathbb{K} \times A$ gives no unitization when $A \neq 0$. For it, $(1, 0)(\mu, b) = (\mu, 0)$, so $(1, 0)$ is not a unit.