A PRIMER ON NATURAL EQUIVALENCES

DMITRY VAGNER

1. Sets 1 Recall that a bijection f : X Y is a map with an inverse map f ´ : X Y 1 1Ñ Ñ satisfying f ´ f 1X and ff´ 1Y ,where1X : X X, x x is the identity map. When a bijection“ X Y exists“ we write X YÑand sayfiÑ that X and Y are isomorphic as sets. We mayÑ also call the bijection– an isomorphism. We know that two finite sets are isomorphic precisely when they have the same cardinality—i.e. number of elements. In this piece, however, we will care not only about the question of whether or not two sets are isomorphic, but rather what kinds of isomorophisms exist between them. Consider, for example, the bijection a, b, c x, y, z a x, b y, c z. t uÑt u fiÑ fiÑ fiÑ But the choice of this bijection was arbitrary! We could have just as well selected a z,b x, c z. fiÑ fiÑ fiÑ Sometimes, though, there’s a bijection that feels—and, as we will see, can be rigor- ously defined to be—not arbitrary. We call these natural or canonical. For example, recall the Cartesian product X Y x, y x X, y Y , and consider ˆ “tp q| P P u X Y Z X Y Z x, y ,z x, y, z . p ˆ qˆ Ñ ˆp ˆ qppq qfiÑ p p qq This bijection works entirely independently of which sets X, Y, Z are involved. This is our first clue to help us identify naturality. We now make an important definition. Definition 1.1. For sets X, Y define the hom-set to be the set of maps hom X, Y f : X Y . p q“t Ñ u Many natural bijections involve hom-sets. Let n 0, 1,...,n 1 and recall the “t ´ u power set X S X —the set of all subsets including the empty set ? and the set X Pitself—ofp q“t a setÄ Xu. We then have the following famous natural bijection. Proposition 1.2. Given a set X, the following is a natural bijection. X hom X, 2 S 1 , Pp qÑ p q fiÑ S where the indicator function 1 : X 2 is given by x 1 i↵ x S. S Ñ fiÑ P We can’t give a proof of this because we haven’t yet defined what it means to be natural. We leave it as an exercise though to check that this indeed instantiates a bijection. Let’s continue with more examples—but first we make another definition. Definition 1.3. The disjoint union X Y of two sets X, Y can be defined as ` X Y X 0 Y 1 a, n a x and n 0 or a y and n 1 ` “ ˆt uY ˆt u“tp q| P “ P “ u 1 2DMITRYVAGNER

This is an awful definition—and this is due to poor conventions that stuck. The problem is the historic primitivity of the regular old Venn-diagram inspired union you’re used to. Consider the following case: X 1, 0 and Y 0, 1 . “t´ u “t u We know that X Y a a X or a Y 1, 0, 1 ,butthisdoesn’tbehave well under cardinalityY since“t | XP Y 3butP u“t´X Y u2 2 4. That’s because, since 0 X and 0 Y we don’t| Y double|“ count| it|`| within|“ the` union.“ P P But we really want an analog of at the level of sets! ` Finding such analogs is called categorification. The disjoint union is precisely what fixes this—by including every element of each summand set—without consid- ering the possibility of equality of elements across di↵erent sets. The disjoint union is more fundamental than the union. One mentally encounters the disjoint union before the union because to form the union one first gathers the elements of both sets—i.e. forms the disjoint union— and then deletes repeats. But one cannot know what it means to be a repeat unless there is some ambient set A such that X, Y A. Then equality of elements across X and Y is determined by the distinctness ofÄ elements imbued by the definition of A. This rant continues in the optional remark below. Remark 1.4. Philosophically, the union of two generic sets X and Y is not even meaningful and relies on the essentially metaphysical assumption that there is some fixed universe of elements from which each set is built. Awareness of this philo- sophical issue is evaded by the fact that, in common practice, elements are denoted with symbols whose shared shape and or familiar linguistic referent (e.g. a letter of some alphabet) implies equality. These pragmatic situations instantiate the condi- tion described above—that there is some ambient set A such that X, Y A—where A just so happens to be implicitly recognized as opposed to explicitlyÄ defined. This leads us to the issue of properly denoting an element—e.g. as we did x, y for X Y —of X Y . The idea is as follows: when we have some map fp : Xq Y ˆA, it acts independently` on X and Y : one simply selects a component ` Ñ map fX : X A that says where the X-elements go and a component map f : Y A thatÑ says where the Y -elements go. The disjointness of elements in Y Ñ X Y evades any constraint on what maps we’re allowed to choose for fX and fY . Hence,` when X Y is the domain of a map, we will simply write ` f : X Yxy f x f y . ` _ fiÑ X p q_ Y p q The symbol ‘ ’ is often used in logic to denote ‘or.’ We denote by an element’s placement relative_ to whether it comes from X or Y . One should read the final _ term as “fX x or fY y ,” where which of the two is true depends on if our input was in X or inp Yq . In contrast,p q when X Y is the codomain of a map A X Y ,we no longer have such a so called universal` property: there is no naturalÑ way` to split such a map into components. Thus, outputs of such a map cannot be interpretable in terms of such a ‘ .’ There is, however, a pair of markedly natural maps whose codomain is X Y ._ In particular, we have the canonical inclusion map ` i : X X Yxx, X Ñ ` fiÑ A PRIMER ON NATURAL EQUIVALENCES 3 and likewise for Y . This allows us even further structure in conceptualizing the component maps fX and fY : they are simply given by respectively precomposing f with iX and iY ; in particular: fX fiX and fY fiY . You may be thinking: why should“ we care about“ this? The answer will be in the form of our most interesting natural bijection yet. Proposition 1.5. Let X, Y, A be sets. We then have the following natural bijection. hom X Y,A hom X, A hom Y,A f fi ,fi . p ` qÑ p qˆ p q fiÑ p X Y q We define the inverse map to the above bijection by sending the pair of maps g : X A and g : Y A to their sum g g g : X Y A,definedin X Ñ Y Ñ “ X ` Y ` Ñ the obvious way: x y gX x gY y . It turns out that the_ disjointfiÑ p unionq_ isp inq some sense dual to the Cartesian product: given a map f : A X Y , we can extract its X and Y component maps by simply forgetting the otherÑ one.ˆ In particular there is a canonical projection map ⇡ : X Y X x, y x, X ˆ Ñ p qfiÑ and likewise for Y . Hence the extraction of the X-coordinate of f : A X Y is Ñ ˆ simply given by postcomposing with ⇡X . This leads us to the following bijection. Proposition 1.6. Let X, Y, A be sets. We then have the following natural bijection. hom A, X Y hom A, X hom A, Y f ⇡ f,⇡ f . p ˆ qÑ p qˆ p q fiÑ p X Y q The story comes together yet further when we reexamine cardinality. To reiter- ate, we have seen, for finite sets, why it is the case that X Y X Y and X Y X Y . | ˆ |“| |ˆ| | | ` |“| |`| | Let’s now investigate the cardinality of hom X, Y . A particularly explicit yet elegant way to approach this is as follows: wep canq conceive of X as the disjoint union of all of its elements, conceived as singleton (single-element) sets. In order to write this down, we need notation to represent the disjoint union of generic collection of sets and not merely a pair of sets. We introduce the notation as follows, via this very example: (1) X x . – x Xt u ∫P Note the use of isomorphism rather than equality. This is because two sets are said to be equal when they consist of identical elements. Just as in the case of defining unions, however, it makes no sense to define equality between sets unless they both sit within a shared ambient set that can imbue equality of elements across them. We now similarly need a notation for generic non-binary products. We introduce such a notation as follows via extending Propositions 1.5 and 1.6.

Proposition 1.7. Let A be a set and Xj j J be a collection of sets indexed by t u P J—i.e. a set J of sets Xj. Then the following bijections are natural bijection

hom Xj,A hom Xj,A f fiX j J . Ñ p q fiÑ p j q P ˜j J ¸ j J ∫P πP

hom A, Xj hom A, Xj f ⇡X f j J . j P ˜ j J ¸ Ñ j J p q fiÑ p q πP πP 4DMITRYVAGNER

Remark 1.8. The above notation—of form xj j J —denotes a generic element within a Cartesian product of an arbitrary collectionp q P of sets indexed by J.This generalizes the ordered pair x, y . Note, however, that our notation was order- free—and this is the case unlessp theq indexing set J is itself imbued with an order. This should imply that the use of order for pairs is not inherent; rather, it is a pragmatic notation that uses position within the order to imply the set to which a given component belongs. We need one final piece to complete our puzzle of dissecting hom X, Y towards computing its cardinality. Consider the singleton set 1 0 .Thereisaunique—p q and hence, by virtue of avoiding arbitrary choice—natural“tbijectionu between 1 and any other one element set—say A a —simply given by 0 a and vice versa. Furthermore, and perhaps more deeply,“t u we have the followingfiÑ natural bijection. Proposition 1.9. Let X be a set. The following is a natural bijection. X hom 1,X x 0 x Ñ p q fiÑ r fiÑ s Remark 1.10. When the type of a map is evident from context, and we are primarily concerned with the rule that defines this map, we will take the convention of simply denoting the map by its rule via placing the rule in brackets, just as above in Proposition 1.9, where we wrote 0 x for a map 1 X. r fiÑ s Ñ Combining the above isomorphisms yields a proof for the cardinality of hom-sets. Proposition 1.11. Let X and Y be finite sets. Then X hom X, Y Y | |. | p q| “ | | Proof. We proceed via a sequence of natural isomorphisms:

hom X, Y hom x ,Y p q– ˜x Xt u ¸ ∫P hom x ,Y – pt u q x X πP hom 1,Y – x X p q πP Y – x X P πX Y | |. – The final isomorphism simply denotes the fact that we are taking the Cartesian product of X copies of Y . Applying cardinality then yields the result: | | X X hom X, Y Y | | Y | |. ⇤ | p q| “ “| | ˇ ˇ Xˇ ˇ This result motivates the notation Y ˇ for theˇ set hom X, Y of maps from X to Y . We even call this the exponential, and write p q X X Y Y | |. | |“| | It will turn out that all laws of elementary algebra involving addition, multiplica- tion, and exponentiation are categorified to natural isomorphisms of their respective A PRIMER ON NATURAL EQUIVALENCES 5 categorified operations disjoint union, Cartesian product, and exponential. As an example, we restate Propositions 1.5, 1.6, and 1.9 in our new notation: X Y X Y A A A 1 A ˆ A A X Y X Y X X. – ˆ p ˆ q – ˆ – Perhaps the most important among these is the following. Proposition 1.12. Let X, Y, Z be sets. Then the following is a natural bijection. hom X Y,Z hom X, hom Y,Z x, y f x, y x y f x, y p ˆ qÑ p p qq rp qfiÑ p qs fiÑ r fiÑ r fiÑ p qss which, in our new notation, evidently categorifies another law of exponents: X Y Y X Z ˆ Z . –p q This bijection demands some further explanation. The idea is that we begin with a function f : X Y Z which takes X-Y pairs as arguments and returns Z values—and we send thisˆ toÑ a function f : X hom Y,Z , that takes X arguments and returns hom Y,Z values, i.e. functionsÑ from Yp to Zq. This procedure is given by freezing a chosenp xq X, thus rendering the two-argument function f , P p˚X ˚Y q into a one-argument function f x, Y . This isomorphism is called currying, and the property of possessing it is calledp ˚ qCartesian closed. We complete our discussion by considering the empty set ?, which in our schema will be sensible to denote by 0. How should we think of maps X 0 and 0 X? It is simply to argue that the former cannot exist: a map must takeÑ each domainÑ element to precisely one codomain element—but when we lack a codomain element to assign to domain elements, it must be the case that we cannot form a map. The latter is a little trickier to conceptualize since there are no domain elements! It is perhaps helpful to think about this in terms of Proposition 1.7: a hom-set with empty domain set is equivalent to an empty product of hom-sets. But what’s an empty product? It should be an inert product, i.e. one that, were it to be multiplied by some other product, would leave it fixed. In short: it should play the role of multiplying by 1; i.e. should be isomorphic to the singleton set. In symbols: 0X 0 X0 1. – – We close this section by aggregating all the natural isomorphisms which categorify laws of elementary algebra. The reader is encouraged to convince themselves of the isomorphisms—(2) and (7)—not explicated in this section.

(1) X Y Z X Y Z (6) 0X 0 p ˆ qˆ – ˆp ˆ q – (2) X Y Z X Y Z (7) 1X 1 p ` q` – `p ` q (3) X Y Z XZ Y Z (8) X0 – 1 p Y ˆZ q –Y ˆZ 1 – (4) X ` X X (9) X X Y Z – YˆZ – (5) X ˆ X –p q