Consider a situation where there is a pseudofunctor from some category $C$. Due to it's geometrical properties, it might induce some sort of an internal hom on the original category (making it a closed category) and then this functor becomes $C$-enriched.

Now, to describe the construction of the internal hom there is no need for any property of the pseudofunctor.

Therefore we can understand exactly what geometrical properties of the pseudofunctor determine this internal hom structure.

This is the construction:

Let $C$ be a category equipped with a pseudofunctor $P:C^{op}\to Cat$

Then it induces another pseudofunctor: $$P^*:[C^{op}\times C, C] \to [C^{op} \times C, Cat]$$ There is also the pseudofunctor $$[P-, P-] : C^{op}\times C \to Cat$$ As an object in the 2-category $[C^{op}\times C, Cat]$ it can be uniquely identified as a 2-functor from the initial 2-category: $$i_{[P-, P-]}:I\to [C^{op}\times C, Cat]$$

The unique object in the image of the right 2-Kan lift of $i_{[P-, P-]}$ through $P^*$ is the internal hom of $C$ through $P$


  1. If $C=Poset$ and $P$ is the interpretation of an ordered set as a category then we get the usual hom of order preserving functions.

  2. If $C=Group$ and $P=$B then I have no idea how to describe this Kan extension - if we manage to have a closed category in this case, then the forgetful functor to $Set$ would give all the members of the codomain that induce a conjugacy relation between the image subgroups.

  3. If $C=Top$ with $P=\pi_0$ I think this reconstructs the usual internal hom in $Htpy$ but I haven't looked at this one deeply yet.


  1. Have you seen this construction elsewhere?

  2. What is this structure in example 2 (does it even exist)?

  3. What are the properties of $P$ needed to make this an internal hom in a closed category?

  4. Can this be easily generalized to $\infty$-category theory (the last example shows it might be interesting)?

  • $\begingroup$ Kan extension? Isn't it a Kan lift? It is very different because there is no theorem that ensures the existence of a Kan lift. $\endgroup$ – Ivan Di Liberti Aug 18 '18 at 10:21
  • $\begingroup$ @Ivan You're right, it is now edited. Note I never claimed the lift always exists - a big part of the question is to determine when it exists. $\endgroup$ – Omer Rosler Aug 18 '18 at 11:16
  • $\begingroup$ I have the feeling that this is but a convolute way to say that $P^*$ has a right adjoint, and in this case this right adjoint $P_*$ determines an enrichment of $C$ over itself. $\endgroup$ – Fosco Loregian Aug 18 '18 at 14:03
  • $\begingroup$ @Fosco Following your comment: On the face of it, a right adjoint for $P^*$ is a stronger requirement then this Kan lift for $[P-, P-]$. But, if these are as strong as you suspect - then this Kan lift would determine the adjoint completely. From this point comes a speculation: maybe the category $[-, K]$ where $K: I \to [C^{op}\times C, C]$ is the Kan lift can be given an extra structure to make it a right adjoint to $P^*$. $\endgroup$ – Omer Rosler Aug 18 '18 at 15:13

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