0
$\begingroup$

There is an example at the beginning of Section 7.4 of Awodey's Category Theory, where natural transformations are first introduced, and I'm trying to figure out how it relates to the definition of a natural transformation. To give an intuitive feel for the notion, the author refers to functors $F:\mathbf{C} \to \mathbf{D}$ as "constructions" with the intention of showing what a relation between them would look like.

To give a simple example, suppose C has products and consider, for objects $A, B, C \in \mathbf{C}$, $$ (A \times B) \times C \text{ and } A \times (B \times C). $$ Regardless of what objects $A$, $B$, and $C$ are, we have an isomorphism $$ h : (A \times B) \times C \xrightarrow{\sim} A × (B × C). $$ What does it mean that this isomorphism doesn’t really depend on the particular objects $A, B, C$? One way to explain it is this: Given any $f : A \to A′$, we get a commutative square $$ \begin{array}{ccc} (A \times B) \times C & \xrightarrow{h_A} & A \times (B \times C) \\ \downarrow & & \downarrow \\ (A' \times B) \times C & \xrightarrow{h_{A'}} & A' \times (B \times C) \end{array} $$ So what we really have is an isomorphism between the “constructions”, $$ (− \times B) \times C \text{ and } − \times (B \times C) $$ without regard to what’s in the argument-place of these. Now, by a “construction” we of course just mean a functor, and by a “relation between constructors” we mean a morphism of functors (which is what we are about to define). In the example, it’s an isomorphism: $$ (− \times B) \times C \cong − \times (B \times C) $$ For functors $\mathbf{C} \to \mathbf{C}$.

What is the category $\mathbf{D}$ in this example? What's the explicit definition of the functors $F, G$? I'm thinking something bringing us to a space of products of objects in $\mathbf{C}$, but I'm not sure what exactly that would look like. We define natural transformations as having components for each object $C \in \mathbf{C}$. What is $C$ in the example? Is it just $A$?

$\endgroup$
5
  • 2
    $\begingroup$ I think the author is being rather explicit about the fact that they are talking about functors $\mathbf{C}\to \mathbf{C}$, so $\mathbf{D}=\mathbf{C}$. Here $F(A)=(A\times B)\times C$ and $G(A)=A\times (B\times C)$. $\endgroup$ Feb 12 at 17:20
  • $\begingroup$ Ah thank you! Yes I think you're right. For some reason I didn't think the products could be in the same category. $\endgroup$
    – bxw
    Feb 12 at 17:34
  • $\begingroup$ @CaptainLama Do you think you could make this an answer so I could mark it as accepted? $\endgroup$
    – bxw
    Feb 14 at 11:31
  • $\begingroup$ For anyone who encountered similar difficulties, my confusion stemmed from not understanding the definition of a product. I was under the mistaken impression that a product is an object of the product category, but this is not the case. Products of objects in C live in C; ordered pairs of objects in C live in $\mathbf{C} \times \mathbf{C}$. $\endgroup$
    – bxw
    Feb 14 at 11:38
  • 1
    $\begingroup$ I wrote an answer with some additional explanation. $\endgroup$ Feb 14 at 11:47
1
$\begingroup$

The product of two objects of a category $\mathbf{C}$ (when it exists) is another object of $\mathbf{C}$: the product of two sets is a set, the product of two groups is a group, etc.

When the product of two objects exists for any pair of objects, this defines a bifunctor: a functor $\mathbf{C}\times \mathbf{C}\to \mathbf{C}$, sending a pair $(A,B)$ of objects to the product $A\times B\in \mathbf{C}$.

A bifunctor is similar to a bilinear map in linear algebra: just like a bilinear map $f:V\times V\to V$ is such that $f(x,-)$ and $f(-,y)$ are linear functions $V\to V$ when we fiw some $x$ or $y$, a bifunctor $F:\mathbf{C}\times \mathbf{C}\to \mathbf{C}$ defines a functor $F(-,B)$ and a functor $F(A,-)$. In other words, the product is a functor if you fix one of the terms.

Here the author is talking about the functors $\mathbf{C}\to \mathbf{C}$ defined by $A\mapsto (A\times B)\times C$ and $A\mapsto A\times (B\times C)$ (which turn out to be naturally isomorphic by some form of associativity of products).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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