# Reference request: The unit of an adjunction of $\infty$-categories in the sense of Riehl-Verity is a unit in the sense of Lurie

I'm looking for a reference (or proof) for the statement given in the title: that when we have an adjunction between quasicategories in the sense of Riehl and Verity (defined e.g. in Section 4 of their paper "The 2-category theory of quasi-categories" or Definition 2.1.1 of their ongoing book project at http://www.math.jhu.edu/~eriehl/elements.pdf), i.e. an adjunction in the homotopy 2-category of quasicategories, that then its unit is a unit transformation in the sense of Lurie, defined in Definition 5.2.2.7 of "Higher Topos Theory".

It is shown in Appendix F of the above mentioned book that their definitions of adjunction agree, but, as far as I could tell, no comparison between the two notions of unit. I also couldn't easily piece it together from other statements. There is Corollary 4.1.3 that goes in that direction, but that still needs an identification of the notions of mapping spaces and maps induced on them, and is less explicit than I would like.

Let $$\mathcal{C},\mathcal{D}$$ two $$\infty$$-categories and $$f:\mathcal{C}\to\mathcal{D}$$ and $$g:\mathcal{D}\to \mathcal{C}$$ two functors.

Recall (HTT.5.2.2.7) that a natural transformation $$u:1_{\mathcal{C}}\to gf$$ is a unit transformation if the natural transformation $$(\ast)\qquad\mathrm{Map}_{\mathcal{D}}(f-,-)\xrightarrow{g} \mathrm{Map}_{\mathcal{C}}(gf-,g-)\xrightarrow{u^*}\mathrm{Map}_{\mathcal{C}}(-,g-)$$ is an equivalence.

Theorem A natural transformation $$u:1_{\mathcal{C}}\to gf$$ is a unit transformation if and only if there is a natural transformation $$e:fg\to 1_{\mathcal{D}}$$ satisfying the triangular identities.

Proof:

Let us first show that if such an $$e$$ exists, then the natural transformation $$(\ast)$$ is an equivalence. Indeed we will show that the natural transformation $$(\ast\ast)\qquad\mathrm{Map}_{\mathcal{C}}(-,g-)\xrightarrow{f} \mathrm{Map}_{\mathcal{D}}(f-,fg-)\xrightarrow{e_*}\mathrm{Map}_{\mathcal{D}}(f-,-)$$ is an inverse. Indeed, if $$l:fc\to d$$ is an arrow in $$\mathrm{D}$$, the transformation $$(\ast)$$ sends it to $$gl\circ u_{c}$$. Then, $$(\ast\ast)$$ sends it to $$e_{d}\circ f(gl\circ u_{c})\cong e_{d}\circ fgl \circ fu_{c}\cong l\circ e_{fc}\circ fu_c \cong l$$ where the second equivalence is the naturality of $$e$$ and the third is one of the triangular identities. So $$(\ast\ast)$$ is a left inverse of $$(\ast)$$. The dual proof (using the other triangular identity) implies that it is a right inverse too, so that $$(\ast)$$ is an equivalence.

Let us now prove the other direction. if $$(\ast)$$ is an equivalence, in particular, the map $$\mathrm{Map}_{\mathcal{D}}(fg-,-)\cong\mathrm{Map}_{\mathcal{C}}(g-,g-)$$ is a natural equivalence. By taking ends, there is an equivalence of the space of natural transformations $$\mathrm{Nat}(fg,1_{\mathcal{D}})\cong \mathrm{Nat}(g,g)\,.$$ Let $$e:fg\to 1_{\mathcal{D}}$$ be the natural transformation corresponding to $$1_g:g\to g$$ under this equivalence. I claim that this satisfies the triangular identities.

In fact, for every $$d\in\mathcal{D}$$, the arrow $$e_d$$ has been chosen so that there is an equivalence $$(ge_d)\circ u_{gd}\cong 1_{gd}$$ naturally in $$d$$. So one triangular identity is satisfied. On the other hand, arguing as in the previous direction, this implies that $$(\ast\ast)$$ is a right inverse of the map $$(\ast)$$. Since that was an equivalence, it must be a left inverse too, Choosing $$l=1_{fc}$$, this shows $$e$$ satisfies the second triangular identity. $$\square$$

If I understood the question in the comments, your doubt can be boiled down to the following construction: if $$\mathcal{C},\mathcal{D}$$ are two $$\infty$$-categories, $$F,G:\mathcal{C}\to\mathcal{D}$$ two functors and $$e:F\to G$$ a natural transformation, there exists a natural commutative diagram of the form $$\require{AMScd} \begin{CD} \mathrm{Map}_{\mathcal{C}}(-,-) @>{F}>> \mathrm{Map}_{\mathcal{D}}(F-,F-)\\ @V{G}VV @V{e_*}VV \\ \mathrm{Map}_{\mathcal{D}}(G-,G-) @>{e^*}>> \mathrm{Map}_{\mathcal{D}}(F-,G-) \end{CD}\,.$$

In order to construct this we notice that the pullback of the right bottom corner of the diagram is just the functor $$\mathrm{Map}_{\mathcal{D}^{\Delta^1}}(e-,e-)$$ where we interpret the natural transformation $$e$$ as a functor $$\mathcal{C}\to \mathcal{D}^{\Delta^1}$$, sending $$c$$ to $$Fc\to Gc$$ (this is just the usual formula for mapping spaces in functor categories using ends, in this case applied to the functor category $$\mathcal{D}^{\Delta^1}$$). But then giving the required square is just equivalent to giving a natural transformation $$\mathrm{Map}_{\mathcal{C}}(-,-)\to \mathrm{Map}_{\mathcal{D}^{\Delta^1}}(e-,e-)\,.$$ And we can simply choose the standard map induced by the functoriality of $$e$$.

• Ok, thanks a lot. Do you know of some place where at least the easy direction is written down, so that I do not have to do it myself? – Robin Stoll Jul 11 '19 at 15:48
• @RobinStoll Sorry, I don't. But if you want I can add it to the answer. Is the level of detail in the "hard direction" ok? – Denis Nardin Jul 11 '19 at 15:51
• Thanks for adding it. It was roughly clear to me how it goes, but I am somewhat uncomfortable with just checking it "on elements", since it needs to be continuous in $l$. It is not hard to formulate it in an element free way, but then one needs that $(e_d \circ) \circ fg \simeq (\circ e_{fc})$ as maps $\operatorname{Map}_{\mathcal D}(fc, d) \to \operatorname{Map}_{\mathcal D}(fgfc, d)$, which seems obvious, but it is unclear to me why it is true formally. (As you can tell, I am not very experienced with these things, so I am sorry if these are stupid questions.) – Robin Stoll Jul 11 '19 at 16:33
• @RobinStoll This starts to depend on exactly what definition you take for precomposition. I don't have time right now, but I'll try to add a short explanation today or tomorrow. Long story short: this is the map induced on fibers of the twisted arrow category fibrations, so it commutes with pushforwards and pullbacks – Denis Nardin Jul 11 '19 at 16:40
• Ok, thanks a lot! – Robin Stoll Jul 11 '19 at 16:52