Suppose we have a split fibration $p : \mathbb{E}\to\mathbb{B}$ with (split) simple products. To fix notation, this means that for every projection $\pi_{I,J} : I\times J \to I$ in the base category, the reindexing functor $\pi_{I,J}^* : \mathbb{E}_{I} \to \mathbb{E}_{I\times J}$ has a right adjoint $\forall_{I,J}$ and these functors satisfy the Beck-Chevalley condition, meaning that the canonical natural transformation $$u^*\circ\forall_{K,J} \Rightarrow \forall_{I,J}\circ (u\times \text{id})^*$$ is identity for every $u : I\to K$ in $\mathbb{B}$. This canonical morphism is the transpose of $(u\times \text{id})^*(\varepsilon_X)$.

My question is, is this enough for the pair $u^*$ and $(u\times \text{id})^*$ to be a map of adjunctions. Actually, it would be enough to show that reindexing functors preserve units of the adjunctions, meaning $$u^*(\eta_X) = \eta'_{u^*(X)}$$ for each $X$ and where $\eta$ is the unit of $\pi_{K,J}^* \dashv \forall_{K,J}$ and $\eta'$ of $\pi_{I,J}^* \dashv \forall_{I,J}$.

It seems to me that this really ought to hold and the proof should use the fact that the "canonical" morphism is the identity, but I couldn't produce one.


The answer is yes, and the Beck–Chevalley condition you cited is exactly what is needed to prove it.

We are assuming $$\forall_{I,J} (u \times \textrm{id})^* \epsilon^{K,J} \bullet \eta^{I,J} u^* \forall_{K,J} = \textrm{id}$$ so: \begin{align} u^* \eta^{K,J} & = \left( \forall_{I,J} (u \times \textrm{id})^* \epsilon^{K,J} \bullet \eta^{I,J} u^* \forall_{K,J} \right) \pi_{K,J}^* \bullet u^* \eta^{K,J} \\ & = \forall_{I,J} (u \times \textrm{id})^* \epsilon^{K,J} \pi_{K,J}^* \bullet \eta^{I,J} u^* \forall_{K,J} \pi_{K,J}^* \bullet u^* \eta^{K,J} \\ & = \forall_{I,J} (u \times \textrm{id})^* \epsilon^{K,J} \pi_{K,J}^* \bullet \forall_{I,J} \pi_{I,J}^* u^* \eta^{K,J} \bullet \eta^{I,J} u^* \\ & = \forall_{I,J} (u \times \textrm{id})^* \epsilon^{K,J} \pi_{K,J}^* \bullet \forall_{I,J} (u \times \textrm{id})^* \pi_{K,J}^* \eta^{K,J} \bullet \eta^{I,J} u^* \\ & = \forall_{I,J} (u \times \textrm{id})^* \left( \epsilon^{K,J} \pi_{K,J}^* \bullet \pi_{K,J}^* \eta^{K,J} \right) \bullet \eta^{I,J} u^* \\ & = \eta^{I,J} u^* \end{align} (The last step uses the triangle identity for adjunctions.)

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  • $\begingroup$ Thank you. It really is a simple, once you find it, application of all the basic assumptions. $\endgroup$ – Aleš Bizjak Aug 29 '12 at 10:32

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