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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

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correspond via the exponential adjunction to diagrams of the form

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where $i\colon K\hookrightarrow L$ is an inclusion, and $p\colon X\to Y$ is a fibration, and the pullback is induced byenter image description here

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).

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A pullback is a limit; right adjoints preserve limits (the right adjoint of a limit is the limit of the right adjoints). If your functor is an appropriate contravariant adjoint, then it will transform limits into colimits. –  Arturo Magidin Dec 18 '11 at 5:02
@ArturoMagidin: But the pushout [K,X] -> [K,Y] <- [L,Y] is not the result of an application of [-,Y] or [K,-] to some diagram of simplicial sets... –  Yuri Sulyma Dec 18 '11 at 5:47
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2 Answers

up vote 4 down vote accepted

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 $\DeclareMathOperator{\Hom}{Hom}$

pull-back and its adjoint

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:

First commutative square of Goerss-Jardine

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

Squares corresponding to maps from Delta to the pull-back

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

The two squares corresponding to commutativity of the first square of GJ

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

The two squares from before, but now on the adjoint side

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

Second commutative square of Goerss-Jardine

as claimed by Goerss and Jardine.

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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.

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Yes, I should probaly have mentioned that. However, I fail to see how this gives the desired correspondence between the diagrams without doing the verifications I outlined. –  t.b. Dec 23 '11 at 10:32
I've seen your comment, thank you for your response to my query. I post it here since it seems to fit a comment better instead of an edit to the post): "note: @t.b: there is nothing missing from your answer. My above remark was meant as a comment to the original question, because the OP wondered if he was missing some basic concept." –  t.b. Dec 23 '11 at 20:43
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