I'm trying to do exercise 1 of II.5 in the Mac Lane's book Category theory for the working mathematicians.

Te exercise asks me to find a natural bijection

$$\operatorname{Hom}(A\times B,C) \cong \operatorname{Hom}(A,C^B) $$ for small categories $A,B$ and $C$ (so the hom-sets are thought in the category of small categories $\mathbf{Cat}$.)

I thought to define the following functors

$$ \begin{array}{rcl} F: \mathbf{Cat}^* \times\mathbf{Cat}^* \times \mathbf{Cat} & \longrightarrow & \mathbf{Set} \\ (A,B,C) & \longmapsto & \operatorname{Hom}(A\times B, C) \end{array} $$


$$ \begin{array}{rcl} G: \mathbf{Cat}^* \times\mathbf{Cat}^* \times \mathbf{Cat} & \longrightarrow & \mathbf{Set} \\ (A,B,C) & \longmapsto & \operatorname{Hom}(A,C^B) . \end{array} $$

But, as functors, I have to define them also for arrows. And here is my problem. If $H:A\times B\times C^* \rightarrow A'\times B'\times {C'}^*$ is a functor (an arrow in $\mathbf{Cat}\times\mathbf{Cat}\times\mathbf{Cat}^*$) I have no idea of how to define $FH$. If I'm not wrong, it is a map (in the usual set) which takes a functor $T:A\times B \rightarrow C$ and gives a functor $FH(T):A'\times B'\rightarrow C'$. So given $T$ and objects $(a',b')$ I have to define

$$ FH(T)(a',b') $$

and also

$$FH(T)(f',g') $$

but I have no clue about it.

Is my procediment fine? In this case, can you help me, please?

  • $\begingroup$ Morphisms are natural here. $\endgroup$ – Randall Jul 5 '18 at 17:06
  • $\begingroup$ @Randall: Sorry, can you develop a little more your comment? What morphisms are natural? $\endgroup$ – Dog_69 Jul 5 '18 at 17:08
  • $\begingroup$ My comment isn't all that helpful, to be honest. It's just that 9 times out of 10 when you're having trouble with something like this it's because you've forgotten that morphisms are natural transformations, which is an incredibly strong assumption. $\endgroup$ – Randall Jul 5 '18 at 17:23
  • $\begingroup$ If $H$ above is an arrow in the category you say it's in, it shouldn't be a functor, but a triple of functors, and its not between products of categories, but triples of categories. It's an important distinction, and it seems like it may be making it harder than it needs to be. $\endgroup$ – Malice Vidrine Jul 6 '18 at 15:08
  • $\begingroup$ @MaliceVidrine Yes, if you see my question, $H$ is defined from $A\times B\times C$ a triple into a triple $A'\times B'\times C'$. and $FH\in\operatorname{Hom}_{\mathbf{Set}}(F(A,B,C),F(A',B',C'))$. $\endgroup$ – Dog_69 Jul 6 '18 at 17:14

Remember that you don't have to define Hom functors from scratch; you should already know how they work. Given $\langle f,g,h\rangle:\langle A,B,C\rangle\to\langle A',B',C'\rangle$, $F(f,g,h)(a:A\times B\to C)$ is just $a\mapsto h\circ a\circ (f\times g)$. (Notice that you have the variance backwards above--the first argument of a Hom functor is contravariant, the second covariant.) $G$ requires a little more thought, but it's easy to sort out with a little time.

$H$ is, I think, where you're getting confused. All you need to show is that for every $A,B,C$ there's a function between sets $H_{A,B,C}:F(A,B,C)\to G(A,B,C)$ that commutes with the $F(f,g,h)$ and the $G(f,g,h)$. That's it. You don't need any extra functors. These functions are given by the currying operation described above.

  • $\begingroup$ For me $H$ was precisely the triple $(f,g,h)$ and it should be a functor, shouldn't it? And to show that the diagram commutes, I needed to know how $F$ and $G$ acts. Thanks to your answer now I know how $F$ does it. I'll try to compute $G$. Thanks. By the way, probably you were right in your previous comment. I mean, even I was thinking of $H$ as functor of triples, I wasn't using this fact actually. $\endgroup$ – Dog_69 Jul 6 '18 at 17:20
  • $\begingroup$ $H$ won't be a functor because its domain isn't a category. Note that two of the components of such a triple aren't even functors; they're what you might call "opfunctors", morphisms $A\to A'$ in $\mathbf{Cat}^{op}$, which are not functors $A\to A'$. This may sound like splitting hairs, but the differences between pairs and products, and between a morphism and its counterpart in the opposite category, are essential to keep straight. $\endgroup$ – Malice Vidrine Jul 6 '18 at 19:09
  • $\begingroup$ You were right, I was mixing triples and products of 3 elements. It is difficult because they are very related. But yes, if $H\in\operatorname{Hom}_{\mathbf{Cat}}((A,B,C),(A',B',C'))$ then $H$ is a triple of functors $(H_A,H_B,H_C)$ and it can′t act as a functor because we are regarding $(A,B,C)$ as an object, not as a category. Am I right now? My problem was that the hom sets in $\mathbf{Cat}$ are sets of functors. Of course they are, but you shouldn't consider them as functors. Finally, would you develop a little bit opfunctors? Because I have read about them but I don't remember where. $\endgroup$ – Dog_69 Jul 6 '18 at 20:07
  • $\begingroup$ You are correct now. And my use of the term "opfunctor" in this case is not standard, it just seemed handy as I was typing it to have a special name for them. In the literature you might find the "Op functor" from Cat to itself that takes every category to its opposite. Or there's the opposite functor of $F:C\to D$, $F^{op}:C^{op}\to D^{op}$. Or there are op-lax functors (a variant of lax functors, which are a 2-categorical notion). $\endgroup$ – Malice Vidrine Jul 6 '18 at 20:58
  • 1
    $\begingroup$ Ok, thanks. Probably I read it for the $op$ category.- By the way, the statement of my question was wrong. the duals in the definitions of $F$ and $G$ are interchanged. $\operatorname{Hom}(-,-)$ is CONTRAVARIANT in the FIRST argument and covariant in the second. This is why I had so many difficulties to define $F$ and $G$. $\endgroup$ – Dog_69 Jul 6 '18 at 21:55

I think the most natural "currying" bijection is

$f(a,b) \mapsto f(a,-)(b)$ where $f(a,-):B \to C$ is a function on $B$ determine by taking $f(a,b)$.

  • $\begingroup$ Yes, I think I have chosen the same natural transformation. However, to show that it is natural, I have to pick categories $A,B,C$ and $A',B',C'$ and study the usual commutative diagram, which involves the term $FH$ and $GH$. $\endgroup$ – Dog_69 Jul 5 '18 at 17:14

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