Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

1 Answer

up vote 2 down vote accepted

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.)

share|improve this answer
    
Thank you. It really is a simple, once you find it, application of all the basic assumptions. –  Aleš Bizjak Aug 29 '12 at 10:32
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.