I am learning category theory using Basic Category Theory by Tom Leinster as my main source. In the chapter on natural transformations he says that isomorphism of categories is unreasonably strict for the notion of the sameness of two categories. Isomorphism would require functors, $$ F:A\rightarrow B,G:B\rightarrow A $$ such that $$ G\circ F=1_A, F\circ G=1_B $$

Instead he says that for equivalence we loosen the requirement on these functors to be isomorphic, $$ G\circ F\cong 1_A,F\circ G\cong 1_B $$ Then this is better. This section threw me for a loop. I don't understand the difference between the equivalence and the isomorphism statements. Any help clarifying what is trying to be said here is greatly appreciated.

  • 1
    $\begingroup$ Natural isomorphisms provide a bit more wiggle room. Your best bet is to pore over examples, e.g. the natural isomorphism between the identity and double-dual functors on the category of finite-dimensional vector spaces over a fixed field. $\endgroup$
    – Hayden
    Commented May 28, 2019 at 20:17
  • 2
    $\begingroup$ Well, do you understand the difference between $=$ and $\cong$? $\endgroup$ Commented May 29, 2019 at 1:27

2 Answers 2


Well, the actual difference between the two statements is that for an equivalence of categories, we only require that that the composites $F \circ G$ and $G \circ F$ are naturally isomorphic to the identity functors rather than exactly equal. That is, there's a collection of isomorphisms $\eta_x :GF(x) \rightarrow x$ for each object of $A$ such that whenever $f: x \rightarrow y$ is a morphism in $A$, $\eta_y GF(f) = f \eta_x$, and a similar natural isomorphism for $F \circ G$.

As for why we do this... Imagine we're both doing group theory, so we both get ourselves a category of groups and start doing group theory in that category. But then we compare our categories and they're not the same: your category has one object for each isomorphism class of groups, while the objects of my category are given by a set $X$ along with a multiplication $\otimes: X \times X \rightarrow X$ which makes it a group.

Our categories aren't isomorphic, not by a long shot: for every object in your category there's a large class of objects in mine. So if we could only use isomorphisms of categories it would look like we're working on entirely different things.

Fortunately, our two categories are equivalent: using one functor which sends a set and a multiplication to its isomorphism class, and the other functor which takes each isomorphism class and picks a realisation of that group. Therefore, we're justified in calling both categories 'The Category of Groups' and any result you get in your category will also work in mine.


Two functors are said to be isomorphic if there is a natural isomorphism between them, i.e. a natural transformation $\eta: G \circ F \rightarrow 1_A$ such that the components $\eta_X: (G \circ F) (X) \rightarrow 1_A(X)$ are isomorphisms $\forall X \in ob(A)$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .