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In Jacob Lurie's book Higher Algebra, for an object $M$ of a monoidal $\infty$-category $\mathcal{C}$, he constructs a category $\mathcal{C}[M]$ which can be thought of as "maps in $\mathcal{C}$ of the form $A\otimes M\to M$ with associated coherence data". $\mathcal{C}[M]$ is shown to be monoidal, and algebra objects of this category are shown to be precisely the algebras which coherently act on $M$ (though one algebra that acts on $M$ in two different ways will be thought of as two different objects of $\mathcal{C}[M]$). Moreover, he shows that if $\mathcal{C}[M]$ has a final object, which we'll denote $End(M)$, this object is an algebra object of $\mathcal{C}$ and $M$ is an object of $LMod_{End(M)}$. Thus any algebra that acts on $M$ admits a morphism to $End(M)$.

I'm in particular thinking about the quasicategory of $\infty$-groupoids, $\mathcal{C}=Top$, so $M$ and $End(M)$ and everything else will just be spaces for the rest of this. We have another more down-to-earth notion of "endomorphisms of $M$." That is, the space of morphisms $Map_{Top}(M,M)$. This is a monoid object of $Top$ whose monoid structure is given by composition. Moreover there is an evaluation map $ev:Map_{Top}(M,M)\times M\to M$ so there is a corresponding object of $Top[M]$. Is this object clearly final in $Top[M]$?

In other words, if $M$ is an object of the quasicategory $LMod_A$ for $A$ some loop space in $Top$, is there induced a map $A\to Map_{Top}(M,M)$? Certainly given a map of topological spaces $A\times M\to M$ one can produce, by adjunction, a map $A\to Map_{Top}(M,M)$, but it's not clear to me that this is necessarily a map of algebras or that it carries all the necessary structure (though I suspect it does). In other words this would require a more complex kind of adjunction than just the one for tensor/hom.

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  • $\begingroup$ In other examples I've thought about, to produce the map $A \to Map_{Top}(M,M)$ qua algebras requires more than just the action $A \times M \to M$, but also the associativity data of that action. Does that work here? $\endgroup$ – Theo Johnson-Freyd Jun 21 '15 at 23:55
  • $\begingroup$ @TheoJohnson-Freyd yes you definitely also need that data. I'm not really mentioning it. I mean, quasicategorically you need a LOT of data (all the higher associativity morphisms), so it'd be nice to get that somehow without specifying an infinite list of $n$-morphisms $\endgroup$ – Jonathan Beardsley Jun 21 '15 at 23:59
  • $\begingroup$ It may also be fruitful to think about the map $A\times M \to M$ as being part of a simplicial object that Lurie refers to as a "left action object." I'm trying to fiddle with that right now. $\endgroup$ – Jonathan Beardsley Jun 22 '15 at 0:00
  • $\begingroup$ In fact, since an $A$-module is given by a map $N(\Delta)^{op}\times \Delta^1\to Top$, I suspect that the relevant map of algebras is precisely the adjoint $N(\Delta)^{op}\to Map(\Delta^1,Top)$ which picks out the map from $A$ to $End(M)$ as algebras. $\endgroup$ – Jonathan Beardsley Jun 22 '15 at 0:09
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Suppose $\mathcal{C}$ is any closed monoidal $\infty$-category and $X$ is an object of $\mathcal{C}$. Then $\mathcal{C}[X]$ can be described as the $\infty$-category of pairs $(C, C \otimes X \to X)$, so it shouldn't be hard to see that the universal property of a final object in $\mathcal{C}[X]$ is precisely that of the internal Hom, say $X^{X}$, from $X$ to $X$. So what Lurie proves (4.7.2.40) is that there is a canonical associative algebra structure on $X^{X}$ - intuitively this is given by composing morphisms.

Now I think what you're asking about is the case where you've somehow constructed another algebra structure on $X^{X}$ (I suppose you could for instance do this if the $\infty$-category $\mathcal{C}$ came from a nice closed monoidal model category). Then by the uniqueness of this algebra, "all" you have to do to show it's the same as Lurie's endomorphism object is to promote it to an algebra object of $\mathcal{C}[X]$. That sounds hard, but it's actually not so bad since Theorem 4.7.2.34 tells you that giving an algebra in $\mathcal{C}[X]$ is the same as giving an algebra $A$ in $\mathcal{C}$ together with an $A$-module structure on the fixed object $X$. So all you need to do is check that your algebra structure on $X^{X}$ gives you a map $X^{X} \otimes X \to X$ compatible with composition, which you obviously ought to have if your algebra structure really was given by "composing maps"

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  • $\begingroup$ I think maybe I don't really understand the nature of $C[X]$. I guess I was thinking the objects of $C[X]$ were the data $(C, C\otimes X\to X)$ AND a bunch of higher coherence data. But I guess that that higher coherence data is only put in when you consider algebras of this category. $\endgroup$ – Jonathan Beardsley Jun 22 '15 at 18:49
  • $\begingroup$ In other words, an arbitrary map $M\to M$ is not necessarily an algebra action of $1_C$ on $M$ (obviously). So, ok. I guess that settles it. $\endgroup$ – Jonathan Beardsley Jun 22 '15 at 18:50
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    $\begingroup$ But yeah, I'm not being complicated at all with the algebra structure on the internal hom object, I'm truly just working through your first paragraph. I guess the point is that I should do it all with universal properties, in this case the universal property of $X^X$. $\endgroup$ – Jonathan Beardsley Jun 22 '15 at 18:56

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