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).

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    $\begingroup$ The first construction goes by the name "Dorroh extension by $\mathbb{Z}$." [In ring theory this is actually a very nice, natural, and interesting construction. So in the future, you might be careful about using judgemental words.] The second construction [which is also nice, natural, and interesting] does not embed $A$ in a unital ring in general. Consider the case where $A=\{0,x\}$ has trivial multiplication and $x+x=0$. $\endgroup$ – Pace Nielsen Mar 8 '17 at 19:05
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    $\begingroup$ The same problem Pace points out already comes up for embedding semigroups in monoids (so the one-object case). If the semigroup does not act faithfully on the right of itself, then taking the endomorphisms of the right regular representation do not give an embedding. $\endgroup$ – Benjamin Steinberg Mar 8 '17 at 19:20
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    $\begingroup$ @PaceNielsen I agree the one I find more interesting is not always an embedding, although there is always a map. I don't apologize for this choice of valuative word --- of course it is intended to (slightly) provoke, but mostly I think that by clearly broadcasting my own biases, I can get answers that are more useful for my application. In my case, I have some objects which have identities and some which (as far as I can tell) do not, and it is important that I unitalize without destroying the identities I already have. $\endgroup$ – Theo Johnson-Freyd Mar 8 '17 at 20:13
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    $\begingroup$ It's not exactly what you asked for, but maybe you will find arxiv.org/abs/1210.0212 relevant. $\endgroup$ – Yonatan Harpaz Mar 9 '17 at 21:33
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    $\begingroup$ Concerning the pre-question, I suggest the names "unitalization" and "multiplier algebra". $\endgroup$ – HeinrichD Mar 10 '17 at 11:35

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