Exercise 1.3.31


Let be the category of finite sets with bijections. We’ll define two functors . First, define . is the set of permutations of . For a bijection , define by .

Now, define . is the set of total orders on – for concreteness, let the elements of be relations in the sense of subsets of . Suppose is a bijection. Define by .

Theorem: There is no natural transformation .

Proof: Assume there were such a natural transformation . Let . And let be the swapping permutation and . Then, we have a component and by naturality . But, observe that: which is some ordering, set . Then, which is clearly a different ordering than , a contradiction.

Thus, since and are the same size as sets, there is a bijection between them and we have in for each , but not naturally. That is to say, and are not naturally isomorphic. This is a counterexample to the incorrect “converse” one might be willing to draw from exercise 1.3.26.