Given an associative nonunital algebra $A$, there are (at least) two standard ways to produce a unital algebra $A'$ together with a map $A \to '$. The boring way is to simply add to $A$ a new element which you declare to be a unit. The more interesting way is to notice that $A$ acts on itself (from the right, say) and to consider the algebra of all $A$-linear endomorphisms of $A$. In C*-land, nonunital algebras are thought of noncompact spaces, the boring way corresponds to the one-point compactification, and the interesting way to the Stone-Cech compactification. In addition to being more interesting, another reason to prefer the latter method is that if $A$ happened to already have a unit, then the latter unitalization doesn't change $A$, whereas the former does.

Pre-question: Do these operations have standard names? Not knowing any, I will call them the "boring" and "interesting" unitalizations.

There is a similar construction in categories. Suppose $C$ is a non-unital category (meaning it has an associative composition, but not necessarily identities morphisms). The boring unitalization of $C$ is produced by adding to $C$ a new morphism for each object in $C$, declaring that morphism to be the identity on that object. The interesting unitalization of $C$ is given by studying natural transformation as between the representable functors $\hom_C(-,c)$ for $c$ ranging over the objectin $C$s, and to declare that $C'$ is the category with objects $\mathrm{ob}(C)$ and morphisms $\hom_{C'}(c,c') = \text{natural transformations}(\hom_{C}(-,c), \hom_C(-,c'))$.

Main question: Have these unitalizations, and in particular the more interesting one, been studied for $\infty$-categories?

For example, I find myself in the following situation. I have a semisimplicial (no degeneracies) space satisfying the Segal condition. I think of it as a "nonunital $(\infty,1)$-category". The boring unitalization is just the output of freely creating degeneracy maps, thereby producing a complete Segal space. For my application, I want the interesting unitalization (as Segal space, complete or not I can handle).

Given an associative nonunital algebra $A$, there are (at least) two standard ways to produce a unital algebra $A'$ together with a map $A \to A'$. Following the discussion in the comments below, these are:

• The (free) unitalization of $A$ is the unital algebra achieved by simply adding to $A$ a new element which you declare to be the unit.

• The multiplier algebra of $A$ is the algebra of all $A$-linear endomorphism of $A$-as-a-right-$A$-module.

In C*-land, nonunital algebras are thought of noncompact spaces, free unitalization corresponds to the one-point compactification, and the multiplier algebra to the Stone-Cech compactification. I remark that if $A$ already has a unit, then passing to the multiplier algebra doesn't change $A$, whereas free unitalization does.

There is a similar construction in categories. Suppose $C$ is a non-unital category (meaning it has an associative composition, but not necessarily identities morphisms). The free unitalization of $C$ is produced by adding to $C$ a new morphism for each object in $C$, declaring that morphism to be the identity on that object. The multiplier category of $C$ is given by studying natural transformation as between the representable functors $\hom_C(-,c)$ for $c$ ranging over the objectin $C$s, and to declare that $C'$ is the category with objects $\mathrm{ob}(C)$ and morphisms $\hom_{C'}(c,c') = \text{natural transformations}(\hom_{C}(-,c), \hom_C(-,c'))$.

Main question: Have these unitalizations, and in particular the multiplier category, been studied for $\infty$-categories?

For example, I find myself in the following situation. I have a semisimplicial (no degeneracies) space satisfying the Segal condition. I think of it as a "nonunital $(\infty,1)$-category". The free unitalization is just the output of freely creating degeneracy maps, thereby producing a complete Segal space. For my application, I want the multiplier category (as Segal space, complete or not I can handle).