Let $\mathcal{C}$ be a small category and $\mathrm{Cat}$ be the 2-category of small categories.

Let $F,G : \mathcal{C} \to \mathrm{Cat}$ be two functors and $\theta : F \to G$ be a natural transformation such that for any object $X$ of $\mathcal{C}$, the functor $\theta_X : F(X) \to G(X)$ is an equivalence of categories (in other words, there exists a functor $\eta_X : G(X) \to F(X)$ and isomorphisms of functors $\eta_X \theta_X \cong 1_{F(X)}$ and $\theta_X \eta_X \cong 1_{G(X)}$).

Then, is $\theta$ an equivalence of natural transformations? In other words, is there a natural transformation $\eta : G \to F$ and isomorphisms of natural transformations $\eta \theta \cong 1_F$ and $\theta \eta \cong 1_G$?

  • 2
    $\begingroup$ In general it will only be a pseudonatural transformation. $\endgroup$
    – Zhen Lin
    Nov 13, 2015 at 12:24

2 Answers 2


A high-tech version of the answer: for fixed $C$, the functors $C\to Cat$ are the algebras for a 2-monad $T$ on $\mathrm{Cat}^{\mathrm{ob}(C)}$, the natural transformations are the $T$-morphisms, and the pseudonatural transformations are the pseudo $T$-morphisms. For any 2-monad, if a $T$-morphism is an equivalence in the underlying 2-category, then it is an equivalence in the 2-category of $T$-algebras and pseudo $T$-morphisms. The proof is just a more abstract version of the argument given by Mattia in this special case: follow your nose.


You can do this by hand, by keeping track of isomorphisms and verifying all compatibilities. It's highly possible there's a high-tech explanation that I don't know about.

So presumably you want to allow $\theta$ to be a pseudo-natural transformation (as Zhen Lin points out), i.e.

  • for every $X$ you have a functor $\theta_X\colon F(X)\to G(X)$, and
  • for every arrow $f\colon X\to Y$ in $\mathcal{C}$ you have a natural isomorphism of functors $\alpha(f)\colon G(f)\circ \theta_X\to \theta_Y\circ F(f)$ (draw the obvious square diagram)

and these data satisfy some compatibility property on compositions of maps $X\to Y$ and $Y\to Z$ in $\mathcal{C}$.

Using these data, you can promote your family of quasi-inverses $\eta_X\colon G(X)\to F(X)$ to a psuedo-natural transformation.

Specifically, for every arrow $f\colon X\to Y$ you want a natural isomorphism of functors $\beta(f)\colon F(f)\circ \eta_X\to \eta_Y\circ G(f)$, and you want these to satisfy the same compatibilities that the $\alpha$'s have to satisfy (and that I haven't written down!).

You get $\beta(f)$ like this: pick $\xi\in G(X)$, and define an arrow $$F(f)(\eta_X(\xi))\to \eta_Y(G(f)(\xi))$$ in the category $F(Y)$ as follows: choose an object $\zeta$ in $F(X)$ such that $\theta_X(\zeta)\cong \xi$, and plug in $\theta_X(\zeta)$ in the above (using the chosen isomorphism). Now you need to produce an arrow $$F(f)(\eta_X(\theta_X(\zeta))))\to \eta_Y(G(f)(\theta_X(\zeta)))$$ but now the second object is isomorphic (via $\alpha(f)$) to $\eta_Y(\theta_Y(F(f)(\zeta)))$, and by using the two isomorphisms $\eta_X\circ \theta_X\cong id$ and $\theta_X\circ \eta_X\cong id$, you are reduced to producing an arrow $F(f)(\zeta)\to F(f)(\zeta)$, and now you have a very canonical one, namely the identity.

I think you can check (it might not be pleasant) that this will give you a well defined pseudo-natural transformation $\eta$.

I hope I'm not overlooking some subtlety here, if I'm wrong I'm sure someone will correct me.

  • $\begingroup$ I suspect that we have to turn $(\theta_X,\eta_X)$ into an adjoint equivalence in order to make this work. $\endgroup$
    – HeinrichD
    Sep 12, 2016 at 15:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.