Passing pullbacks through adjunction I'm having trouble following the proof of Proposition I.5.2 in Goerss-Jardine (Simplicial Homotopy Theory).  After establishing the adjunction $\hat\Delta(X\times K,Y) \simeq \hat\Delta(K,[X,Y])$, they claim that diagrams of the form

correspond via the exponential adjunction to diagrams of the form

where $i\colon K\hookrightarrow L$ is an inclusion, and $p\colon X\to Y$ is a fibration, and the pullback is induced by
I imagine this is at the level of trivial manipulations with the adjunction, but I can't get it to come out right (mainly not sure how to go from the pullback to the pushout).
 A: Halfway through typing my answer I saw that t.b. already gave it.
Since I seem to be unable to comment on anything, I post this comment as
an answer. The original problem is certainly not trivial because there is
more than one pair of adjoint functors: you have $(-)\times K \dashv [K,-]$
and $(-)\times L\dashv [L,-]$.
The missing concept is that of conjugate natural transformations as explained
in Mac Lane's CWM (Section IV-7): suppose
$F,F':\mathcal{X}\rightarrow\mathcal{A}$ and
$G,G':\mathcal{A}\rightarrow\mathcal{X}$ are functors with
$F\dashv G$ and $F'\dashv G'$.
Two natural maps $\alpha: F\rightarrow F'$ and $\beta: G'\rightarrow G$
(note the opposite directions) are said to be conjugate if the diagrams
$$
\matrix{
& \mathcal{A}(F'x,a) & \cong & \mathcal{X}(x,G'a) & \cr
\mathcal{A}(\alpha_x,a) &\Big\downarrow& &\Big\downarrow& \mathcal{X}(x,\beta_a)\cr
& \mathcal{A}(Fx,a) & \cong & \mathcal{X}(x,Ga) & }
$$
commute for all objects $x\in\mathcal{X}$, $a\in\mathcal{A}$, where the
horizontal maps are the isomorphisms given by adjointness. In fact, any given
$\alpha: F\rightarrow F'$ determines a unique conjugate
$\beta=\alpha^*: G'\rightarrow G$ (take $x=G'a$ and start with $id_{G'a}$ in the
upper right corner).
In your situation the inclusion map $i:K\hookrightarrow L$ gives a natural
map $(-)\times i:(-)\times K\rightarrow (-)\times L$ and the corresponding
conjugate is $i^*:[L,-]\rightarrow [K,-]$ given by precomposing with $i$.
This is conveniently already named $i^*$ in your diagram, although with
objects dropped from the notation.
I always wondered why this notion of conjugate transformations is never
made explicit in books on algebraic topology although it is used all the
time.
A: Apart from performing a somewhat lengthy diagram chase, the only thing
to do here is to notice that giving a map
$(a,b) : A \to [K,X] \mathrel{\times_{[K,Y]}} [L,Y]$
amounts to giving
a commutative diagram as on the left by the universal property of the pull-back
$\DeclareMathOperator{\Hom}{Hom}$

while the diagram on the right is equivalent to it by the adjunction
between the cartesian product and the internal $\Hom$ and the
definitions of $p_\ast$ and $i^\ast$.
Let me give names to the maps in the first commutative diagram of the
proof you ask about:

The map $f: \Lambda_{k}^n \to [L,X]$ corresponds to a map
$\tilde{f}: \Lambda_{k}^n \times L \to X$ while giving the bottom map
$(g,h) : \Delta^n \to  [K,X] \mathrel{\times_{[K,Y]}} [L,Y]$ amounts to either of
the two commutative diagrams

Using the universal property of the pull-back defining 
$[K,X] \mathrel{\times_{[K,Y]}} [L,Y]$ 
we see that asserting the commutativity of the square we started with
is equivalent to giving the map $f: \Lambda_{k}^n \to [L,X]$, the
commutative square $(1)$ and requiring the two squares

to be commutative.
Passing to the adjoint side using the map $\tilde{f} : \Lambda_{k}^n \times L \to X$ and the square
$\widetilde{(1)}$, the commutativity of the squares $(2)$ and $(3)$ is equivalent to the two commutative squares

At this point I think I can leave it to you to contemplate the
commutative diagrams $\widetilde{(1)}$, $\widetilde{(2)}$ and
$\widetilde{(3)}$ and the push-out square
defining
$(\Lambda_{k}^n \times L) \mathrel{\cup_{(\Lambda_{k}^n \times K)}}
(\Delta^n \times K)$ and to think about what it means to define a map from $(\Lambda_{k}^n \times L) \mathrel{\cup_{(\Lambda_{k}^n \times K)}}
(\Delta^n \times K)$
in order to see that giving the squares $\widetilde{(1)}$, $\widetilde{(2)}$ and
$\widetilde{(3)}$
and requiring their commutativity is
equivalent to giving the commutative square

as claimed by Goerss and Jardine.
