Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This time I have a conceptual question. Reading Categories for the Working Mathematician, I noticed the author gives two different notions between isomorphisms:

1) An isomorphism $T:B\rightarrow C$ of categories is a functor $T$ from $B$ to $C$, which is bijection both on objects and arrows.

2) If we take two bifunctors, we sometimes can find a bijective natural transformation between them, for example, $\text{Cat}(A\times B,C)\cong \text{Cat}(A,C^N)$ is natural on $A,B$ and $C$.

Does the second notion implies that $\text{Cat}(A\times B,C)$ and $ \text{Cat}(A,C^B)$ are isomorphic regarded as categories? What is the meaning behind the naturality of the bijection, the extra information?

As always, thanks in advance for your answers.

share|cite|improve this question
This is not a good definition. The correct definition is that an isomorphism is a functor with an inverse. (But isomorphism is itself the wrong notion for categories; usually the interesting notion is equivalence.) – Qiaochu Yuan Jul 11 '13 at 5:15
That is the exact definition given on the book. The author says it is equivalent to the existence of a two-sided inverse functor. What is your definition of equivalence? – Felipe Pérez Jul 11 '13 at 5:25
It's a fine definition, but usually too rigid for applications. A more flexible definition is equivalence of categories, which is a pair of adjoint functors with units and counits that are isomorphisms. An equivalence of categories preserves a great deal of structures you typically care about (limits, colimits, etc.) but is much less rigid. We usually don't really care whether or not $FGX$ is exactly $X$. If $FGX$ is isomorphic to $X$, we're usually just as happy. This other notion of of categories being "the same" is also in MacLane's book as "equivalence of categories". – Zach L. Jul 11 '13 at 5:30
@Felipe: it points you in the wrong direction. If you tell students that an isomorphism is a homomorphism that is a bijection and then ask them to prove that a map is an isomorphism, they will do it by proving that it is a homomorphism, an injection, and a surjection. This is sometimes the only strategy, but it is the last strategy you should try, not the first. The first strategy you should try is to find the inverse, since it's usually the easiest, cleanest, and conceptually most satisfying thing to do. So it might as well be the definition. – Qiaochu Yuan Jul 11 '13 at 5:34
@Felipe: for example, I graded for a group theory class in which students had to prove that various maps were isomorphisms, such as the inner automorphisms $x \mapsto gxg^{-1}$ on a group. The students did this by proving that the maps were homomorphisms, that they were injective, and that they were surjective. What they should've done was prove that they were homomorphisms and that $x \mapsto g^{-1} x g$ provides an inverse. – Qiaochu Yuan Jul 11 '13 at 5:35
up vote 4 down vote accepted

$Cat(A,B)$ is the set of functors from the category $A$ to the category $B$. In general, $\mathcal{C}(A,B)$ denotes the hom-set between objects $A,B$ in the category $\mathcal{C}$. Since there is a "category of categories", something like $Cat(A \times B, C) \cong Cat(A, C^B)$ is saying that there is a bijection between functors, sending $F:A\times B \rightarrow C$ to the functor $G: A \times C^B$. (This is exactly like saying $$Maps(A \times B, C) \cong Maps(A, C^B)$$ in $Sets$, except the functions are now functors!) What is $G$? Well, $C^B$ is a functor category. The objects are functors, and so given an object $a$ of $A$, $G(a)$ had better be a functor from $B$ to $C$. What functor do we take? The only option is $F(a,-)$. But $G$ also needs to do something to morphisms. So a morphism $k:a \rightarrow a'$ needs to go to a natural transformation $F(a,-) \rightarrow F(a',-)$. But this comes built in to the definition of $F$, since it's natural in both slots. We can also go back. Given such a $G$, we can build an $F$, and do it in a way that the two constructions are mutually inverse.

Now, you can then view $Cat(A,B)$ as a category (a functor category), with objects the morphisms and morphisms as natural transformations between them. The above construction doesn't "come with" a way to send morphisms (natural transformations) to morphisms. But given a natural transformation between bifunctors $S \Rightarrow T:A \times B \rightarrow C$, I would believe that one could construct a natural transformation between the adjoints, which I'll denote $\hat{S}$ and $\hat{T}$. If $\eta$ is the natural transformation from $S$ to $T$, then define $\hat{\eta}$ at $a$ to be... what? Well, it needs to go from $\hat{S}(a)$ to $\hat{T}(a)$. These are objects in the functor category $Cat(B,C)$, so $\hat{S}(a)$ is a functor from $B$ to $C$, as is $\hat{T}(a)$. So $\eta_a$ should be a natural transformation! What should the natural transformation be? Well, I would use $\eta(a,-)$.

I don't have any reason to think that this $\hat{\cdot}$ operation won't induce a bijection on natural transformations, and so those two functor categories are indeed isomorphic, which you could write as $$C^{A\times B} = (C^B)^A.$$ But you can probably imagine that writing out the details is a little tedious! The trick is to at all times carefully keep track of what type of objects you're working with, what the morphisms should be, etc.

Note that naturality of the bijection $Cat(A\times B,C) \cong Cat(A,C^B)$ didn't enter with this argument, since we never had to think about any categories other than $A,B,$ and $C$. So what does that mean? Go back to the thing that took a functor $F:A \times B \rightarrow C$ and spat out a functor $G:A \rightarrow C^B$. Suppose you had a functor $H:A' \rightarrow A$. You can use this to build a functor $H\times 1:A' \times B \rightarrow A \times B$. Suppose you wanted to know what happens when you applied all this stuff to $F(H\times 1):A' \times B \rightarrow C$. You have to go through all that computation over again!

But that's where naturality saves the day. Naturality says that if you've computed one thing, you can compute pre- and post- compositions by just... composing! So the result from applying all this to $F(H \times 1)$ is just $GH: A' \rightarrow A \rightarrow C^B$. As you read the book, you'll notice that naturality is a very powerful condition, since it imposes so many relations on the transformation!

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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