Suppose that $F: A \rightarrow B$ is a functor, and $- \circ F: \widehat{B} \rightarrow \widehat{A}$ is the functor on the presheaf categories induced by precomposition. If $- \circ F$ is full and faithful, is $F$ full? Is it faithful? What if $- \circ F$ is an equivalence? What if $F$ is already essentially surjective?

Help would be much appreciated! Many thanks.

  • $\begingroup$ I'm a little confused about the last part of the question; I'm not sure exactly how it's supposed to be added to the other conditions. $\endgroup$ – Qiaochu Yuan Apr 11 '14 at 4:19
  • $\begingroup$ It was kind of an after thought. What I meant was: if $F$ is essentially surjective, do we more information from the functor between presheaf categories? $\endgroup$ – David Myers Apr 11 '14 at 4:46
  1. If $F : \mathcal{C} \to \mathcal{D}$ is essentially surjective on objects, then the induced functor $F^* : [\mathcal{D}^\mathrm{op}, \mathbf{Set}] \to [\mathcal{C}^\mathrm{op}, \mathbf{Set}]$ is conservative, but not necessarily fully faithful.
  2. $F^* : [\mathcal{D}^\mathrm{op}, \mathbf{Set}] \to [\mathcal{C}^\mathrm{op}, \mathbf{Set}]$ is an equivalence of categories if and only if $F : \mathcal{C} \to \mathcal{D}$ becomes an equivalence of categories after Cauchy-completion; in particular, if $\mathcal{C}$ and $\mathcal{D}$ are Cauchy-complete (= have splittings for all idempotents), then $F^*$ is an equivalence if and only if $F$ is. This yields counterexamples to the converse of (1).
  3. $F^* : [\mathcal{D}^\mathrm{op}, \mathbf{Set}] \to [\mathcal{C}^\mathrm{op}, \mathbf{Set}]$ can be fully faithful when $F : \mathcal{C} \to \mathcal{D}$ is neither full nor faithful: for example, if $\mathcal{C} = \mathbf{sSet}$, $\mathcal{D} = \operatorname{Ho} \mathbf{sSet}$ (in the sense of Quillen), and $F : \mathcal{C} \to \mathcal{D}$ is the localisation functor, then $F^* : [\mathcal{D}^\mathrm{op}, \mathbf{Set}] \to [\mathcal{C}^\mathrm{op}, \mathbf{Set}]$ is fully faithful (by the universal property of localisation) but $F : \mathcal{C} \to \mathcal{D}$ is neither full nor faithful. (It is, however, a bijection on objects.)

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