Let $(\mathcal A,||\cdot||)$ be a normed algebra (with or without a unit). The **unitization** of $\mathcal A$ is the space $\mathcal A\_+:=\mathcal A\oplus \Bbb C$ where the multiplication operation $\cdot$ and norm $|||\cdot |||$ are defined by
$$\begin{align}
(a,\lambda)\cdot(b,\mu) &:=(ab+\mu a+\lambda b,\lambda\mu) \\
|||(a,\lambda)||| &:= ||a||+|\lambda| 
\end{align}$$ 
for any $a,b\in\mathcal A$ and $\mu,\lambda\in \Bbb C$. It can be easily verified that $(0,1)$ is a unit in $\mathcal A\_+$ with norm $1$, and that $\mathcal A$ isometrically embeds into $\mathcal A\_+$. This construction seems to be a very fundamental tool used to augment a unit into a space without one
, it came up in my first class of the chapter on Banach Algebra.


Our professor tried to convince us that unitization is done in the same spirit as completion of metric spaces. However, this construction troubles me for 2 main reasons. 


Firstly, nothing stop me from unitizing a unital normed algebra. This gives me an essentially **different** space than the one I started with. This is not the case for the completion $\hat X$ of a Banach space $X$, in which we have $\hat X \cong X$. To add salt to the wound, the original unit in $\mathcal A$ is no longer a unit in $\mathcal A\_+$.


Secondly, unlike the process of completion, the adjoined unit seems quite artificial. In metric space completion we only fill in "holes" but it seems like in unitization we artificially add genuine new direction into our space. 



> 
> I know that my professor's analogy shouldn't be taken literally but in what sense does unitization resemble metric completion? Conceptually, what is unitization? 
> 
> 
> 


It also occurred to be that I'll learn to appreciate unitization as I encounter more and more results in this field. Nevertheless, I would like to have an intuitive understanding of unitization. 


For example, take $(\mathcal A,+,\cdot,||\cdot||)$ to be $(L^1(\Bbb R),+,\*,||\cdot||\_{1})$, where $\*$ is the convolution. This is a non-unital algebra since the Dirac distribution does not belong to $L^1$. In this case what does $\mathcal A\_+$ look like?



Conceptually, what does unitization do?
Unitization and metric completion are both [left adjoint](https://en.wikipedia.org/wiki/Adjoint_functors) functors, as are may other "-tion" operations in mathematics, such as [localization](https://en.wikipedia.org/wiki/Localization_of_a_ring) or [abelianization](https://en.wikipedia.org/wiki/Commutator_subgroup#Abelianization). Specifically, there is a forgetful functor from unital algebras to nonunital algebras (including norms is not particularly important here), and unitization is its left adjoint. Conceptually this means that it is in some sense the "freest" or "laziest" way of adding a unit to a nonunital algebra. (It's important to keep in mind that "nonunital" means "not necessarily having a unit" here.) 


An important difference between unitization and metric completion is that the inclusion of complete metric spaces into all metric spaces is fully faithful, but the inclusion of unital algebras into nonunital algebras is not. This is responsible for your observation that when you complete a complete thing nothing happens, but when you unitize a unital thing you get something different. One of the relevant keywords here is [idempotent monad](https://ncatlab.org/nlab/show/idempotent+monad). 


One way of thinking geometrically about what unitization accomplishes is to apply the [commutative Gelfand-Naimark theorem](https://en.wikipedia.org/wiki/Gelfand_representation#Statement_of_the_commutative_Gelfand-Naimark_theorem) to nonunital C\*-algebras. This says that every nonunital C\*-algebra is the algebra of functions vanishing at infinity on some locally compact Hausdorff space $X$. Taking the unitization then gives the algebra of functions on the [one-point compactification](https://en.wikipedia.org/wiki/Alexandroff_extension) $\hat{X}$ of $X$. Taking the unitization again gives the algebra of functions on the one-point compactification of $\hat{X}$, which is $\hat{X}$ with a disjoint point added. 


