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Consider a functor $p \colon E \to B$ in $\mathbf{Cat}$. Then there is an induced functor $p^* \colon \mathbf{Cat}/B \to \mathbf{Cat}/E$. It is on objects given by

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But now I wanna ask its possible to investigate using the domain functor $d^*\colon \mathbf{Cat}/E \to \mathbf{Cat}$ that to say that $p^*$ has a right adjoint is equivalent to saying that the functor $- \times_B E \colon \mathbf{Cat}/B \to \mathbf{Cat}$ has a right adjoint. Is this true? Is this just using the composition of adjunctions?

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    $\begingroup$ Just for a reference: dependent product is the right adjoint to pullback. The 'domain functor' d* looks to me to be Cat/E -> Cat/1 = Cat, (composition with E->1) which is the left adjoint to (-)xE: Cat -> Cat/E, hence d* always has a right adjoint. So this is asking whether p* having a right adjoint is equivalent to d^* p^* : Cat/B -> Cat/E -> Cat has a right adjoint, where we know d^* has one. This is a good general question to ponder even before talking about slices of Cat. $\endgroup$
    – David Roberts ♦
    Commented Sep 11 at 12:49

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First, observe that, for any category $\mathscr C$ and object $C \in \mathscr C$ for which $\mathscr C$ admits binary products with $C$, the slice category $\mathscr C/C$ is the category of coalgebras for the comonad given by $({-}) \times C$ (this is sometimes called the coreader comonad). Consequently, for each category $\mathscr E$, the forgetful functor $\mathbf{Cat}/\mathscr E \to \mathscr E$ is comonadic.

Therefore, by the (dual of the) adjoint triangle theorem, since $\mathbf{Cat}/\mathscr E \to \mathscr E$ is comonadic, any functor $\mathscr X \to \mathbf{Cat}/\mathscr E$ from a category $\mathscr X$ with coreflexive equalisers has a right adjoint if and only if the composite $\mathscr X \to \mathbf{Cat}/\mathscr E \to \mathscr E$ does. In particular, since $\mathbf{Cat}/\mathscr B$ is complete (since $\mathbf{Cat}$ is), any functor $\mathbf{Cat}/\mathscr B \to \mathbf{Cat}/\mathscr E$ has a right adjoint if and only if the composite $\mathbf{Cat}/\mathscr B \to \mathbf{Cat}/\mathscr E \to \mathscr E$ does.

(This argument appears (more concisely) in Street's 2001 note Powerful functors, which is presumably what motivates the question.)

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  • $\begingroup$ An absolutely minor comment: the use of curly letters for the objects of $\mathbf{Cat}$, but a curly $\mathscr{C}$ for the ambient category in the first sentence and plain $C$ for an object—and then taking $\mathscr{C}=\mathbf{Cat}$ and $C=\mathscr{E}$—is a little bit triggering for me :-) $\endgroup$
    – David Roberts ♦
    Commented Sep 12 at 0:50

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