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.

I've been trying to figure out proposition 2.8 of Gambino and Kock 1.

The precise statemnet is (probably) not imporant but the context a locally cartesian closed category $\mathbf C$ where they are constructing an $A$-morphism $w : B \to B'$ in a slice $\mathbf C / A$. The statement confusing me is the following

... Since w must be an A-map, we can construct it fiberwise, so
we need for each a in A a map B'_a -> B_a. This allows
reduction to the case A = 1, ...

I'm unfamiliar with the notion of constructing such morphisms (in LCCCs other than Set) fibrewise. To me it seems they suggest that in order to specify a morphisms in $\mathbf C / A$ it is enough to specify all its images through the pullback functors along global elements (i.e. morphisms $1 \to A$), though I can't quite understand how I could have missed this.

Though my guess is that I am either being stupid and missing something utterly obvious or that I am to ignorant about reasoning using internal languages, I'm wondering if such morphisms can be defined by specifying their fibres (i.e. images of pullbacks along global elements).

Eagerly awaiting facepalm :)

share|improve this question

1 Answer 1

up vote 1 down vote accepted

It is not literally true that we can reduce to the case $A = 1$. For example, let $\mathcal{C}$ be the topos of sheaves on a non-trivial topological space, and let $A$ be any non-empty open set that isn't the whole space – then there is no morphism $1 \to A$ at all! Rather, what is true is that any slice of a locally cartesian closed category is again a locally cartesian closed category, and perhaps that is what they mean by reduction to the case $A = 1$. Or maybe they mean an argument using generalised elements – in which case the ordinary category-theoretic statement follows by taking the generalised element $a = \textrm{id}_A : A \to A$.

However, sometimes something more sophisticated can be said. Suppose $\mathcal{C}$ is a locally cartesian closed category with finite disjoint coproducts and coequalisers (e.g. any elementary topos), then the codomain fibration $\textrm{codom} : [\mathbb{2}, \mathcal{C}] \to \mathcal{C}$ is a stack for the extensive coverage on $\mathcal{C}$. This means it is literally true a morphism in the slice category $(\mathcal{C} \downarrow A)$ can be specified by a compatible family of morphisms in $(\mathcal{C} \downarrow U_1) \times \cdots \times (\mathcal{C} \downarrow U_n)$, where $U_1, \ldots, U_n$ constitute a covering family for $A$.

Let me explain what this means in the case of a regular epimorphism $\alpha : U \twoheadrightarrow A$. Form the kernel pair of $\alpha$, so that we have a coequaliser diagram: $$U \times_A U \rightrightarrows U \rightarrow A$$ Now, consider the triple adjunction induced by $\alpha : U \to A$ on slice categories: $$\Sigma_\alpha \dashv \alpha^* \dashv \Pi_\alpha : (\mathcal{C} \downarrow U) \to (\mathcal{C} \downarrow A)$$ This exists because $\mathcal{C}$ is locally cartesian closed. As usual, we get an induced monad $\mathbb{T} = (\alpha^* \Sigma_\alpha, \eta, \mu)$. Let $\epsilon : \Sigma_\alpha \alpha^* \Rightarrow \textrm{id}$ be the counit of the adjunction $\Sigma_\alpha \dashv \alpha^*$, and consider the diagram $$\Sigma_\alpha \alpha^* \Sigma_\alpha \alpha^* X \underset{\epsilon_{\Sigma_\alpha \alpha^* X}}{\overset{\Sigma_\alpha \alpha^* \epsilon_X}{\rightrightarrows}} \Sigma_\alpha \alpha^* X \overset{\epsilon_X}{\rightarrow} X$$ for each object $p : X \to A$ in $(\mathcal{C} \downarrow A)$. The pullback pasting lemma implies that this diagram is just $$U \times_A U \times_A X \underset{\pi_{2,3}}{\overset{\pi_{1,3}}{\rightrightarrows}} U \times_A X \overset{\pi_2}{\rightarrow} X$$ which is a coequaliser because $p^* : (\mathcal{C} \downarrow A) \to (\mathcal{C} \downarrow X)$ preserves all colimits. Thus $\alpha^*$ is a right adjoint of descent type. In particular, to specify a morphism $f : X \to Y$ in $(\mathcal{C} \downarrow A)$, it is enough to specify a morphism $\tilde{f} : U \times_A X \to U \times_A Y$ in $(\mathcal{C} \downarrow U)$ making the diagram $$\begin{array}{ccc} U \times_A U \times_A X & \overset{\pi_{1,3}}{\longrightarrow} & U \times_A X \\ {\scriptsize U \times_A \tilde{f}} {\Large \downarrow} & & {\Large \downarrow} {\scriptsize \tilde{f}} \\ U \times_A U \times_A Y & \underset{\pi_{1,3}}{\longrightarrow} & U \times_A Y \end{array}$$ commute. (Strictly speaking, this is monadic descent, but this is equivalent to descent for fibred categories in the case of the codomain fibration by e.g. the Bénabou–Roubaud theorem.)

share|improve this answer
    
Very nice writeup, thanks! I figured the statement couldn't be true (in general). Now I just need to figure out what they mean by fibrewise :) –  Tilo Wiklund Aug 22 '12 at 11:21

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.