Simplicial homotopy's exponential law The following was cited from Simplicial Homotopy Theory by John F. Jardine and Paul Gregory Goerss. We have an adjunction $$\hom_{\mathbf{S}}(X\times Y,Z)\simeq \hom_{\mathbf{S}}(Y,\mathbf{Hom}(X,Z))$$
The part I don't understand is the equivalence of the second and the third squares. How could I use adjunction to transform it?


 A: I'll suppress superscripts and subscripts, write $[-,-]$ for $\mathbf{Hom}(-,-)$, and write $i^*$ as $[i,Y]$, etc.
The first diagram involves the inclusion $r$ of the horn and a map $a:\Lambda\to [L,X]$, maps $b,c:\Delta\to [K,X],[L,Y]$, and $([i,X],[L,p])$. We have $([i,X],[L,p])a=(b,c)r$, so by the universal property of the pullback $$(1)\quad [i,X]a=br$$and $[L,p]a=cr$. Finally since we have maps into a pullback, $[i,Y]c=[K,p]b$ and $[i,Y][L,p]=[K,p][i,X]$.
In the second diagram, we have $c':\Delta\times L\to Y, p:X\to Y,$ maps $r\times L:\Lambda\to\Delta, \Delta\times I:\Delta\times K\to \Delta\times L$ combining to $j$, and maps $a':\Lambda\times L\to X,b':\Delta\times K\to X$ forming the top row, such that $pb'=c'(\Delta \times i),pa'=c'(r\times L),$ and from the pushout property, $(r\times L)(\Lambda \times i )=(\Delta \times i)(r\times K)$ and $$(2)\quad b'(r\times K)=a'(\Lambda\times i)$$
It's clear from the notation how all the maps in the two diagrams induce each other-transposition across the exponential-hom adjunction in the case of $a,b,c,a',b',c'$, or applying the functors $\times -$ or $[-,-]$. So we have only to check that these two sets of equations are equivalent.
Let's compute the transpose of equation (1). $\overline{[i,X]a}$ is going to be $\bar a\circ(\Lambda\times i)$, where $\bar a$ is the transpose of $a$. This may be nonobvious: we have $[\Lambda,[K,X]]\cong [\Lambda\times K,X]$, naturally in $K$ (and $\Lambda$ and $X$), so at our map $i:K\to L$ we should get that the two routes $[\Lambda,[L,X]]\to [\Lambda\times K,X]$ via $[\Lambda\times L,X]$, respectively, $[\Lambda,[K,X]]$ coincide. The first route sends $a\mapsto \bar a\mapsto \bar a\circ \Lambda\times i$, and the second sends $a\mapsto [i,X]\circ a\mapsto \overline{[i,X]\circ a}$, so the claim is shown. On the other hand, $\overline{br}=\bar b\circ (r\times K)$. So equation (1) corresponds to equation (2). Continuing in this way, you'll find that all the equations characterizing the two diagrams pair up as desired.
