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

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