Let $\mathcal C \to \mathcal D \to \mathcal E$ be a pair of functors. Assume that $\mathcal C$ is dense in $\mathcal E$. Is then $\mathcal D$ also dense in $\mathcal E$? One definition of denseness is that the nerve, that is the restricted Yoneda embedding $\mathcal E \to \mathrm{PSh}(\mathcal C)$ is fully faithful. We want that the same holds with $\mathcal D$ instead of $\mathcal C$. Everything fits into a commutative diagram

$$\require{AMScd} \begin{CD} \mathcal E @>>> \mathrm{PSh}(\mathcal D)\\ @. {_{}\style{display: inline-block; transform: rotate(30deg)}{{\xrightarrow[\rule{4em}{0em}]{}}}} @VVV\\ @. \mathrm{PSh}(\mathcal C), \end{CD}$$

where the vertical functor is the restriction. Obviously the diagonal functor is faithful because the top is faithful but from there I would need something like the restriction being faithful for example, which is not always the case. So is there a counterexample to the question or some (reference to a) proof?

  • $\begingroup$ To be clear, you take the composite $C\to E$ to be a functor realising $C$ as dense in $E$, and want to know if this implies $D\to E$ realises $D$ as dense in $E$? $\endgroup$
    – FShrike
    Jun 28, 2023 at 1:36
  • $\begingroup$ @FShrike Correct. If you mean that the functors are inclusions, then this would be sufficient for me. Or you can just take the definition of a dense functor which doesn‘t necessarily require subcategories as in ncatlab.org/nlab/show/dense+functor. $\endgroup$
    – HDB
    Jun 28, 2023 at 9:06

1 Answer 1


A useful reference for cancellability of dense functors is §5.2 of Kelly's Basic Concepts of Enriched Category Theory. For instance, in your example, if $\mathcal C \to \mathcal D$ is essentially surjective, or if $\mathcal D \to \mathcal E$ is fully faithful, then density of $\mathcal C \to \mathcal D \to \mathcal E$ implies density of $\mathcal D \to \mathcal E$ (Proposition 5.11 and Theorem 5.13 ibid.).


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