I'm having troubles proving these two related statements, which are immediate for 1-categories and should of course be true for $\infty$-categories:

  1. Given a natural transformation $\alpha: f \Rightarrow g: \mathcal{C} \to \mathcal{D}$, for every two objects $X,Y \in \mathcal{C}$ there is a homotopy commutative square

$\require{AMScd}$ \begin{CD} \mathcal{C}(X,Y) @>f>> \mathcal{D}(fX,fY)\\ @VgVV & @VV \alpha_{Y*} V\\ \mathcal{D}(gX,gY) @>> \alpha_X^* > \mathcal{D}(fX,gY) \end{CD}

  1. Given two functors $L: \mathcal{C} \leftrightarrows \mathcal{D}: R$ and two natural transformation $\eta: 1_{\mathcal{C}} \Rightarrow RL$ and $\varepsilon: LR \Rightarrow 1_{\mathcal{D}}$, these data constitute an adjunction if and only if the homotopy triangular identities hold, that is, $(R \ast \varepsilon) \circ (\eta \ast R) \simeq 1_R$ and $(\varepsilon \ast L) \circ (L \ast \eta) \simeq 1_L$.

Any insight would be welcome.


1- Your natural transformation can be seen as a functor $C \to D^{\Delta^1}$, which therefore induces a commutative square of $\infty$-categories

$\require{AMScd} \begin{CD} C_{x/} @>>> D^{\Delta^1}_{\alpha_x/} \\ @VVV @VVV \\ C @>>> D^{\Delta^1}\end{CD}$

and thus a morphism of fibers over $y\in C$. The fiber of the leftmost vertical map is $map_C(x,y)$ , and the fiber of the rightmost vertical map over the image of $y$, i.e. $\alpha_y$ sits in a (cartesian) square

$\require{AMScd} \begin{CD} map(\alpha_x,\alpha_y)@>>> map(fx,fy) \\ @VVV @VVV \\ map(gx,gy) @>>> map(gx,fy) \end{CD}$

For your purposes, you don't even need to know the square is cartesian, and then it just comes from the fact that $\Delta^1\times\Delta^1$ is a commutative square.

Here's maybe more detail : consider an arrow $g:x_1\to y_1$ in an $\infty$-category $E$, then $Fun(\Delta^2,E)\times_{Fun(\Delta^1,E)} \{g\}\simeq E_{/g}$, where we look at evaluation at $\Delta^{\{1,2\}}$.

Also recall that the forgetful map $E_{/g}\to E_{/x_1}$ is an equivalence (because the space of fillings for a given inner horn is contractible).

Now consider the following sequence of inclusions $\Delta^1\to \Delta^2 \to\Delta^1\times \Delta^1$ : the first map is inclusion at $\Delta^{\{0,1\}}$,and the second one is the diagonal arrow.

This provides you with functors $Fun(\Delta^1\times\Delta^1,E)\to Fun(\Delta^2,E)\to Fun(\Delta^1,E)$, where the second one induces an equivalence upon taking the fiber over $g$ and $x_1$ respectively. So you have functors $Fun(\Delta^1\times\Delta^1,E)\times_{Fun(\Delta^1, E)} \{g\}\to Fun(\Delta^2,E)\times_{Fun(\Delta^1,E)}\{g\}\to Fun(\Delta^1,E)\times_E \{x_1\}$, where the second one is an equivalence.

Now do the same for the other nondegenerate $2$-simplex in $\Delta^1\times\Delta^1$, where this time you'll look at another arrow $f: x_0\to y_0$ in $E$. This gives you a big diagram as follows,witnessing the fact that $\Delta^1\times\Delta^1$ is a commutative square:

(because AMScd does not support diagonal arrows, this is hard to draw on MO, so I just drew it and took a picture)

enter image description here

It all commutes, and the "wrong way" morphisms become invertible upon taking the appropriate fibers. On fibers (I'll let you figure out which fibers I mean), this then gives you the following, still commutative diagram :

enter image description here

which induces the desired commutative diagram, relating $map(f,g), map(x_0,x_1), map(y_0,y_1)$ and $map(x_0,y_1)$.

For your question 2-, the answer is not as easy because it depends on what your original definition of adjunction is. If you're just following Higher topos theory, then your statement is essentially in that book, and the proof requires a number of preliminaries. If that's not what your definition is, you're going to need to be more precise about what you mean.

  • $\begingroup$ For point 1, I guess the issue then becomes how to show that there is a pullback square like the one you drew, which is again trivial in 1-categories and expected in $\infty$-categories, but not immediate. For point 2, gives me one direction, but I believe I have the other, given point 1. $\endgroup$ Apr 21 at 13:33
  • $\begingroup$ As I said you don't need to know it's a pullback square for the argument, you only need to know it is commutative, and this follows from the fact that $Fun(\Delta^1, D)_{\alpha_x/}$ can be seen as a subcategory of $Fun(\Delta^1, Fun(\Delta^1,D)) \simeq Fun(\Delta^1\times \Delta^1, D)$ $\endgroup$ Apr 21 at 13:49
  • $\begingroup$ I'm sorry, but I still fail to see how the required commutativity is a direct consequence of that $\endgroup$ Apr 22 at 11:44
  • $\begingroup$ Sorry I only now saw this. I've added some details. Because of how AMScd works I decided to draw diagrams myself instead, I don't know how the result looks. There's some implicit things left to you, but hopefully they are clear from the diagrams $\endgroup$ Apr 22 at 19:28
  • $\begingroup$ I had almost drawn the very same diagram myself, and then got stuck at the choice of what fibers to take, but now I made it work. Thanks, this was helpful $\endgroup$ Apr 23 at 8:54

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