I'm experiencing some confusion regarding the following exercise in Chapter 2, Section 5 from Mac Lane's Categories for the working mathematician:

For small categories $A$, $B$ and $C$ establish a bijection $$ \mathbf{Cat}(A \times B, C) \cong \mathbf{Cat}(A, C^B). $$ and show it (is) natural in $A, B$ and $C$. Hence show that $-\times B: \mathbf{Cat} \to \textbf{Cat}$ has a right adjoint (see Chapter IX).

My attempt: I demonstrated a bijection by creating a function $f: \textbf{Cat}(A \times B, C) \to \textbf{Cat}(A, C^{B})$ such that if $F: A \times B \to C$ is a functor, then $$ f(F) = F(a, -): A \to C^B $$ where $a$ takes on all values in $A$. I then showed injectivity by supposing that if $f(F_1)=f(F_2)$ then $F_1(a, -) = F_2(a,-)$ for all $a \in A$ which implies that $F_1(a, b) = F_2(a, b)$ for all $b \in B$, hence showing that $F_1 = F_2$. I then showed surjectivity by showing that if $G:A \to C^B$, then I can find a corresponding $F$ such that $f(F) = G$. So, $f$ is a bijection.

There are other posts on this question, although I don't think they particularly answer all of my following questions:

  1. Is my attempt correct? My work feels shakey.

  2. is $\mathbf{Cat}(A \times B, C)$ really the "hom-set," or set of all functors from $A\times B$ to $C$? Based on other posts, I assumed yes, and produced my attempt above. But Mac Lane never defines what $\mathbf{Cat}(A,B)$ is for two categories $A, B$. Again, my work feels shakey because I'm working with notation that Mac Lane decided to randomly introduce without any warning. At least, it's not obvious to me.

  3. Can someone explain the last comment? Mac Lane name drops "right adjoint," but only spent a few sentences on a "left adjoint" in a previous section (since he promises to introduce it later). He vaguely refers the reader to Chapter IX, but Chapter IX is on "Special Limits" and I don't see anything about a right adjoint and it includes a lot of advanced language (so I'm not sure why he'd vaguely send the reader there; not even with a page number).

  4. How should one approach naturality? From another post on this topic, it was demonstrated that if one wanted to show naturality in, say the category $A$, one should show that, if $T: A \to A'$ is a functor between categories $A, A'$, then the diagram commutes. $\require{AMScd}$ \begin{CD} \textbf{Cat}(A \times B, C) @>{f}>> \textbf{Cat}(A, C^B)\\ @VVV @VVV\\ \textbf{Cat}(A' \times B, C) @>{f}>> \textbf{Cat}(A', C^B) \end{CD} However, I'm having a hard time figuring out what the downward arrows should be. I suppose I could set the leftmost one to be, say $T_1(F(a, b))= F(T(a), b)$ and the right most one to be $T_2(G(a)) = G(T(a))$ where $F \in \textbf{Cat}(A \times B, C)$ and $G \in \textbf{Cat}(A, C^B)$, but how do I know these are the right choices?

  5. What is the point of this exercise? I assume the point was revealed in his last comment, but given that his last comment is not making sense, maybe someone can enlighten me on what this exercise really achieves.


1 Answer 1


Your proof is fine enough as far as it goes, namely showing bijection but not yet naturality. I don't know if it is intentional or not, but you don't actually show surjectivity, just state what you would need to do. I would, however, recommend instead of proving bijection by showing injectivity and surjectivity to instead simply explicitly write the inverse. That is, define bijection as an isomorphism in $\mathbf{Set}$. This is typically more useful and leads to nicely calculational proofs.

I'm pretty sure Mac Lane introduces the notation $\mathcal{C}(X,Y)$ generically for any category $\mathcal{C}$ to stand for the hom-set. I'm also fairly certain that Mac Lane defines what the arrows of $\mathbf{Cat}$ is, namely functors. For broader context, though I don't think it comes up in Mac Lane, $\mathbf{Cat}$, particularly when defined as all locally small categories and not just small categories, is an archetypal example of a 2-category and so often $\mathbf{Cat}(\mathcal C,\mathcal D)$ will be the hom-category of the 2-category $\mathbf{Cat}$. Again, that's not happening here.

A left adjoint is the dual concept to a right adjoint, but a functor that has a right adjoint is a left adjoint (and vice versa). So a definition of right adjoint is a definition of left adjoint. It is clear that you don't need to worry about it right now.

Mac Lane defined the action of hom-functors on arrows, i.e. $\mathcal C(f,B)(h) = h\circ f$ and $\mathcal C(A,g)(h)=g\circ h$. I'm pretty sure he defined the action of $\times$ as a (bi)functor, though you could likely correctly guess it anyway. Presumably, he defined the action of $C^B$ in each argument (though this is somewhat unnecessary in a way). The relevant functors are just compositions of these. You have mostly gotten it except you should write $T_1(F)(a,b)$ not $T_1(F(a,b))$. $T_1$ (and similarly for $T_2$) is a function which takes a functor $F$ and outputs a functor. $F(a,b)$ is an object so $T_1(F(a,b))$ would imply that $T_1$ takes an object. You know these are the right choices because these are the definition of the action of the relevant functors on arrows.

The point of the exercise is presumably to start getting you familiar with proving things like this and to start spelling out the properties of $C^B$. The point of the theorem is that it shows that $\mathbf{Cat}$ is cartesian closed (well, technically you also need the existence of a terminal object) which is indeed basically that $-\times B$ is a left adjoint for all $B$. Adjoint functors are one of the most important and ubiquitous concepts in category theory and knowing something is a left/right adjoint immediately entails a lot of other nice properties. This will be elaborated on throughout the book as almost all results will implicitly or explicitly related to adjoint functors. One particularly important aspect is that left/right adjoints are determined up to unique isomorphism and more particularly given the action on objects of a functor which is left/right adjoint to a given functor, you can calculate what the action on arrows must be. This means that adjoint functors have a kind of definitional power. What an exponent is, i.e. what the notation $(-)^B$ means in general, is exactly that $(-)^B$ is right adjoint to $(-)\times B$. Indeed, a cartesian closed category is exactly a category equipped with three functors whose only relationships are given by adjoint functors.

Finally, I should say that I don't actually recommend Categories for the Working Mathematician, especially as an introduction. It is not very well organized or systematic, and, as the title suggests, it is not aimed at a "student". It also omits or only sketchily covers content that I think is very important, though admittedly this is a flaw in virtually all introductions currently. There are many freely available or otherwise reasonably readily available introductions. I recommend Awodey's Category Theory and Barr and Wells' ESSLLI notes. But I really recommend just reading multiple sources and getting multiple perspectives. If one book seems not to be working for you (though you should expect a good amount of struggle regardless of the book), then try another and return to the first later. I think it is helpful to read Mac Lane at some point, but it is probably better read after going through some other introduction. The main gap of virtually all of these resources is a discussion of (co)ends which is well handled by Fosco Loregian's notes, but those assume familiarity with basic category theory.

  • $\begingroup$ This is super helpful, thank you so much and for the additional resources! $\endgroup$
    – trujello
    Aug 27, 2019 at 17:52

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