Is there any literature on representable presheaves which are fibrations, or categories such that all representable presheaves are fibrations?

A representable presheaf $$\mathcal{C}(-,X):\mathcal{C}^{op}\to{\bf Set}$$ being a fibration would mean that for any function $$f':Y'\to \mathcal{C}(A,X)$$ there exists a Cartesian arrow $$f:A\to Y$$ in $\mathcal{C}$ such that $$\mathcal{C}(-,X)(f)=\circ f=f'$$ so $f'$ is just precomposition with $f$, and in particular $Y'=\mathcal{C}(Y,X)$. This in turn means that for any coinitial arrow $$g:A\to Z$$ (coterminal in $\mathcal{C}^{op}$) such that there exists a function $v':\mathcal{C}(Z,X)\to\mathcal{C}(Y,X)$ satisfying $$\mathcal{C}(-,X)(f)\circ v'=\mathcal{C}(-,X)(g)$$ there exists a unique arrow $$v:Y\to Z$$ such that $v'$ is precomposition with $v$ and $$v\circ f=g.$$

All of this seems to be saying that we have nice representability properties for functions into Hom-sets such that the representable presheaf of the codomain object in the Hom-set is a fibration. This seems like something interesting, but a brief search for literature only turned up the notion of a fibrant object in a model category.

The Yoneda embedding of a category $\mathcal{C}$ being a fibration also yields a nice representability condition for natural transformations $\alpha:F\Rightarrow\mathcal{C}(-,X)$ from arbitrary presheaves into representable ones ($F$ is also strictly representable, and $\alpha$ is just pointwise postcomposition with some arrow in $\mathcal{C}$), but the universal property of Cartesian arrows is immediately satisfied for all arrows in $\mathcal{C}$ under the Yoneda embedding by faithful fullness of Yoneda, so the universal property isn't adding anything new unless I'm mistaken. Any references or insights are appreciated.

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    $\begingroup$ This is somewhat confusing, because presheaves are equivalent to discrete fibrations via the category of elements construction… $\endgroup$
    – Zhen Lin
    Commented Feb 4, 2022 at 7:15
  • $\begingroup$ @ZhenLin I've never really encountered the category of elements construction, and the nlab page on the subject says a bit about how to get at the equivalence you mention, but if you could elaborate it would be appreciated. (if this stuff is too basic for MO I can take the question down and move it to MSE) $\endgroup$
    – Alec Rhea
    Commented Feb 4, 2022 at 7:43
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    $\begingroup$ It's not actually related to the question you ask, except for using the same concepts. I was remarking that seeing a question of the form "when is a presheaf a fibration?" and yet not involving the category of elements is likely to be disorienting. $\endgroup$
    – Zhen Lin
    Commented Feb 4, 2022 at 9:11

1 Answer 1


There are some subtleties involving equality of objects I will gloss over here. It is also inconvenient to have to reverse arrows half the time so I will replace $\mathcal{C}^\textrm{op}$ with $\mathcal{C}$ itself.

Consider a functor $P : \mathcal{C} \to \textbf{Set}$. If we think of it as a forgetful functor of some kind, then the property of being a fibration is the property of being able to contravariantly transport "structure" along (not necessarily bijective!) maps. Once you realise this, some examples immediately suggest themselves. The most prominent example is the forgetful functor $\textbf{Top} \to \textbf{Set}$. It is easy to verify that a continuous map $f : X \to Y$ is cartesian if and only if $X$ has the initial topology with respect to $f$. Furthermore, the forgetful functor $\textbf{Top} \to \textbf{Set}$ is representable. Similarly, the "objects" functor $\textbf{Cat} \to \textbf{Set}$ is a fibration and is representable.

Yet, I struggle to think of any natural example of a category $\mathcal{C}$ with even two non-isomorphic functors $\mathcal{C} \to \textbf{Set}$ that are fibrations and representable, let alone a category where every representable functor is a fibration. The property of $P : \mathcal{C} \to \textbf{Set}$ being a fibration is very strong – for instance, if there is even one object $X$ in $\mathcal{C}$ such that $P X$ is non-empty, then (up to isomorphism) every set and every map is in the image of $P$. If $P$ is representable then there is certainly an $X$ such that $P X$ has an element, but $P$ rarely satisfies this surjectivity condition. For example, the functor represented by an initial object will never be surjective like this. So I think there is basically no hope of finding a natural example of a category where every representable functor is a fibration.


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