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

*

*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}


*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.
 A: 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)

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 :

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 5.2.2.12 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.
