Kaori Kitao Lecture 2019
Emily Riehl, assistant professor in the department of mathematics at Johns Hopkins University, recently delivered the the 2018 Kitao Lecture at Swarthmore College. Riehl attended Harvard for her undergraduate studies and also completed a post doctorate in category theory. Her dissertation involved geometry in art as she showed that Saint Peters Square’s decoration had to do with interlocking circles, not an ellipse. This interactive talk aimed to prove that the distributivity of multiplication over addition a(b + c)= ab + ac and question what numbers really mean.
“What category theory is, the way that most research mathematicians would know it is as a language for developing some very abstract ideas, but it’s also a set of proof techniques," says Riehl. "And those proof techniques are like magic tricks that you can deploy in all sorts of mathematical contexts. I want to teach you some of these magic tricks today so you’ll hopefully see them in the future.”
The Kitao Lecture, named for and endowed by retired Professor of Art History Kaori Kitao, is in its seventh year of inviting speakers in the field of mathematics and statistics.
Speaker 1: Good afternoon. I'm Aimee Johnson. I'm the chair of the Department of Mathematics and Statistics. I wanna welcome you to the seventh annual Kitao Lecture. And I wanna answer for you the obvious question, what is a Kitao Lecture? It's a special once a year talk. It is supported by a generous request from Kaori Kitao who's a retired professor of art history here at Swarthmore College. A little background about Professor Kitao, she came to the U.S. from Japan after World War II. She did her undergraduate degree in architecture from UC Berkeley. Then she did a PhD in art history from Harvard and eventually came to Swarthmore College. Taught here from the 1960s to 1990s. Upon her retirement she endowed various funds and one of which is for what we now call the Kitao Lecture.
Now, why you might ask? She's an art historian. What's Math got to do with it? In general, she wanted to encourage students to study across the curriculum. The true liberal arts education. Kind of specifically, given some of her own work, like for instance her dissertation, had to do with the geometry involved in art. For instance, in her dissertation she studied [00:01:30 Bernice] Saint Peters Square and showed that part of the decoration had to do with interlocking circles and not an ellipse like was previously thought. So you could see how it came into play. And so she similarly wants to value these intersections across the disciplines. I want to publicly thank Professor Kitao for her support of this lecture and then I will cede the floor to my colleague who will introduce our speaker.
Speaker 2: Today we are delighted to have Emily Riehl giving our talk today. Emily grew up, went to high school in Central Illinois and then went to Harvard for undergrad where in addition to her studies she was captain of the rugby team. Then she went to Cambridge in England on a Churchill Scholarship where in addition to all the other things you can do there she took a course in category theory which was love at first sight. She's been a category theorist ever since. She then returned to Illinois for her graduate work at University of Chicago where in addition to studying category theory she picked up Australian rules football. Became pretty good at it, represented the United States at Australian rules football seven times. Then she returned to Harvard for her Post Doc where in addition to her studies in category theory she played the bass in a all women's rock and roll band. Finally, she came to Johns Hopkins University as an assistant professor of Mathematics where in addition to her teaching and her research in category theory she's a 2018 Kitao Lecturer at Swarthmore College.
Emily Riehl: I feel like I can't top that introduction, so I should just go home right now. If it were me I'd probably feel most comfortable sitting up here in the front so I could, you know you can lean back. You're welcome if you want. No? Okay. We have one. Great. Yeah. I mean. You can sit on the floor.
The goal for this talk is to prove the following theorem, namely ... I'm gonna try to use some color. We're gonna prove that a times b plus c is equal to a times b plus a times c where a, b, and c are natural numbers. So one, two, three, four, five, et cetera. That's the goal for the talk. I think it will take about a full hour. I'm gonna outline the steps we're gonna use. There's a way that you could prove it more quickly, right. But I want to give you a tour of my area of Mathematics which is something called category theory.
What category theory is, the way that most research mathematicians would know it is as a language for developing some very abstract ideas, but it's also a set of proof techniques. And those proof techniques are like magic tricks that you can deploy in all sorts of mathematical contexts. I want to teach you some of these magic tricks today so you'll hopefully see them in the future. These things are gonna have complicated names but I want you to know what to google after the talk so I am gonna give them their complicated name.
The first step is something called categorification. The second step is something called the Yoneda lemma. The third step is something called representability. The fourth step, well at that point you just prove it. So the fourth step will finally give you the proof and after we've done all that I'm gonna owe you an epilogue which is what was the point of all that? So if you wanna scowl at me for the first part of the talk I'll try and address your criticism by the very end.
Let's just dive right in. Categorification, maybe one way to explain what categorification is, is by a quote that I've heard attributed to Gian-Carlo Rota but I'm not sure that's right. Anyway, the quote is that, "All equations are lies." The question I wanna ask is ... I mean all equations are lies not in the sense that they're wrong. If you've shown that something is equal to something else you've probably actually done it, but they kind of conceal some sort of deeper structural truth. They flatten reality into just x equals y. And so what I wanna ask, the question that we're gonna start off is what is the deeper meaning of the equation that we're intending to prove? So what is the deeper meaning of the equation a times b plus c is equal to a times b plus a times c? Again, just to emphasize this a, b and c, these are just natural numbers. I'm tryin' to prove the distributivity of multiplication over addition. This is a familiar equation, but what does it really mean?
So we'll get there. Maybe a place to start is what role is played by natural numbers, by the numbers? Not necessarily in this equation, but in life. What are the roles played by natural numbers? There are lots of different ways to answer this question, but what are some thoughts? What is a natural number? What does it do for us as mathematicians or Mathematics students? How would you explain what a number is? This isn't rhetorical. I'm gonna wait until somebody answers. This is probably the hardest of the questions 'cause it is very open ended. There's lots of different roles played by numbers and I only have one in mind so it's a open ended question with only one right answer.
Speaker 4: Counting count?.
Emily Riehl: Counting. Great. Right. We can put numbers in order of you know one, two, three, four, five, six, seven, eight, nine, ten. What does it mean if I were to count the number of people in this room and I got some number 77? What would that mean? What is 77? Maybe I don't actually care what order you're all sitting in which is one thing that you achieve when you count. What is 77 then referring to the people in this room?
Speaker 5: The cardinality of a set.
Emily Riehl: Cardinality of a set. Bingo. Great. One thing that numbers do and this is actually how they're defined ... well it depends on your foundations, but one thing that numbers do is they define sizes of sets. Finite numbers will define sizes of finite sets so number a is the size, that just means the number of elements or the technical term is the one that you used, is cardinality. Or it's called cardinality of a set. Which I'm gonna ... to write the name of a set with a elements I'm going to write a again but in a capital letter and in a different color. So the cardinality of a set a with A elements. This is meant to be grammatical because a is the name of a set and A is the name of a number. It sounds a little funny when I say A is the size of the cardinality, the set a with A elements but, maybe if you can read this the meaning is clear. So 77 in the context of this room would refer to the number of people in this room. Assuming there are 77 people in this room which I really have absolutely no idea.
One of the roles that are played by numbers is sizes and there's a notation for this. If I wanna express this idea that a is the cardinality of a set a with A elements more concisely I could a is equal to and then I put these absolutely value signs around the set a. This is just a notation for cardinality so we're saying that a is a set with A elements. I like this approach.
Let's continue then. If we're focusing on the role played by natural numbers a, b and c as expressing the sizes of sets, so let's say then if A and B are the cardinalities of sets a and b so in other word a is a set with A elements, b is a set with B elements, so what is true of these sets? What is true of the sets a and b if it is the case that the numbers A and B are equal? So what does it mean in terms of sets with A and B elements respectively if those numbers A and B are equal? What information does that give us?
Speaker 6: They have the same number of elements, they have the same cardinality?
Emily Riehl: So they have the same number of elements, they have the same cardinality. I agree. Absolutely.
Speaker 7: You could match the elements of one set to the other to find the size now.
Emily Riehl: Yeah. You could match the number of elements. That way you could pair them up. There's lots of different terms for this. You could put the elements in a one to one correspondence. You could set up a bijection between the number of elements, but I'm a category theorist so I prefer a different term. So I'll say that this is true. I'll say a is equal to be if an only if, so iff for if and only if, so just exactly when the sets a and b are what's called isomorphic. Now I owe you a definition of what exactly that means.
What isomorphic means is, well it's exactly this matching procedure that you described. In other words, sets are isomorphic if there exists a function from a to b, so I'm using some notation to indicate a function. Maybe I'll call it [fe]. So fe is a function that takes an element of a and produces an element of b. Then I also have a function I'll call psi that goes from b to a. So psi takes an element of b to an element of a and then these functions establish this one to one correspondence. So this is saying this element of a corresponds to the particular element of b and this element of b corresponds to the same element of a, which I can express by saying that these are inverses. So that these functions are inverses. This is a structural way to say that we're gonna match up the elements of A with the elements of B. You can match up the elements of A with the elements of B without leaving anything out or without duplicating anything if and only if the cardinalities are the same.
Where are we? We're trying to understand the deeper meaning of an equation. What I've argued is we can think about these numbers in this equation a, b and c as expressing sizes of finite sets. Then we can think about this equal sign to say one number is equal to some other number as asserting that the sets are isomorphic. I wanted to introduce notation for this. In the case where, maybe I'll just summarize this step. In the case where a and b are isomorphic so in other words in the case where the number a is equal to the number b, we write a symbol that looks pretty close to the equality symbol but you put a little squiggle on the top of it. That's the symbol for isomorphism. This just means that I can come up with this one to one correspondence between the elements of A and the elements of B.
We've understood that the numbers in this equation represent sizes of finite sets and the equals sign represents an isomorphism between finite sets. To really understand the deeper meaning of the equation I need to understand what role is being played by the symbols plus and times. By plus and times. When we phrase this question in a less ambiguous way, so in particular let's say if I had my numbers b and c and I'm understanding these to be sizes of sets. So B is the cardinality of a set b and C is the cardinality of a set c. So then what set can I build from my sets b and c that has B plus C many elements? In other words if I had a particular set with B elements and a particular set with C elements, how can I combine these in some way to form a set with B plus C elements? How can I categorify addition? What do you think?
Speaker 8: You can take their union?
Emily Riehl: Yeah. So we wanna just stick 'em together. There's a ... let's think about an example, so let's say b is the set whose elements are flat, sharp and natural. c is the set whose elements are, these are hard to draw, a heart, diamond, spade and, it's a good thing you guys aren't too close to the board, and club. What's meant by union or actually the technical term is disjoint union. So this is a disjoint union. What b plus c is, and I was intending to write sets in pink, so b plus c is then the set whose elements are just these elements so I include flat, natural, sharp and then also these elements. If I had given the elements in the two different sets the same name I should include them twice. Somehow the idea with the disjoint union is these elements remember that they came from c and these elements remember that they came from b.
But anyway, cause there were three things there and four things here and there's exactly one thing here for all the things in here and exactly one thing here for things in here with no overlap, the size of this set is gonna be b plus c. So the cardinality of B plus C is exactly equal to b plus c. Okay. There's one more question to ask and then we're done with this first step. The last question ... I won't write too low on this board but I'm gonna use a little bit of it. The last question is, similarly if I think of the numbers A and B so if A and B are again sizes of sets which I'm also gonna call a and b just to make it challenging to follow along if you're not also reading. So if A and B are the sizes of sets a and b then what set that I can build from a and b has a times b many elements. How do I use elements of a set to define a new set whose cardinality is the product as opposed to the disjoint union? There's a standard operation for this but it's maybe a little less familiar though you've all definitely seen it before.
Speaker 9: The Cartesian product.
Emily Riehl: Cartesian product. Absolutely. I'll give an example again so a is, I'm gonna use the same b as over there so the set was sharp, flat and natural and a has two elements, I'm gonna call them asterisk and star, which I realize look quite similar. What's meant then by this Cartesian product a times b, so the elements of this will be ordered pairs of elements where one coordinate comes from a and one coordinate comes from b. This is like R2 in calculus, multi-variable calculus. So the elements here would be asterisk paired with flat. Asterisk paired with natural. Asterisk paired with sharp and then also star paired with flat. Star paired with natural and star paired with sharp and indeed there were two elements of a three elements of b and there are six elements of the Cartesian product.
What is the summary then? I'll write it up here. What I'm gonna do is I'm gonna summarize step one. We can finally answer the question what is the deeper ... actually I'll write it here first, and we'll rewrite it there. Let's come back to the original. What is the deeper meaning of the equation a times b plus c is equal to a times b plus a times c? If we're thinking about the numbers as representing sets and we've understood that if we replace these numbers by sets we can also replace the other symbols. Can someone put that all together? What's being asserted if a, so in other words if A is the size of a set a, B is the size of a set b, C is the size of a set c, what must be true in order for a times b plus c to equal a times b plus a times c? Yeah.
Speaker 10: [inaudible].
Emily Riehl: Yeah. Absolutely. The answer is this is true if and only if, so this is equivalent to saying that the set which is formed by taking the Cartesian product of the set a with the disjoint union of b plus c is isomorphic to the set that's formed by taking the Cartesian product of a and b and then taking the disjoint union of that set with the Cartesian product of a with c. This equality is categorified is enriched by the presence of this isomorphism. If we think about it, it's not unreasonable that we would have an isomorphism like this. What is this a times b plus c so I formed the disjoint union of sets b plus c so I have this seven element set with flat, sharp, natural and all the card face symbols and then I'm gonna pair each of those elements with a. So this'll be the set ... I'm not gonna write out all 14 elements or we'd be here for a really long time, but these elements would be the elements on that list of seven paired with asterisk and paired with sharp.
Now what's this set? So I take this Cartesian product here and then I form a similar Cartesian product with star and asterisk and the four suit shapes and then I take the disjoint union. Definitely both of those sets have 14 elements. That's good 14 equals 14. That's what we're trying to do. Two times three plus four is equal to two times three plus two times four. Special case of our theorem, but moreover those two sets somehow have the same elements. It's not just that they're isomorphic. They're isomorphic in a very natural way and that's kinda a hint of what's going on.
Let's summary the first step. So again, we're trying to prove a theorem so to prove the equation a times b plus c is equal to a times b plus a times c what I've argued for you is we're going to instead we'll prove that the sets that is formed by taking the disjoint union of b plus c and then the Cartesian product of a is isomorphic to the set that is the Cartesian product of a with b and the Cartesian product of a and c and the disjoint union. Okay? So we've simplified matters clearly.
All right. So that was simple. So what's step two? Oh wait, this one's gonna be the mess. I'll just erase. Step two is something called the Yoneda lemma and it's every category theorists favorite theorem because ... there's this joke that if you wanna prove anything in category theory you just say it follows from the Yoneda lemma and then everybody nods their head. So what is this all about?
I'm gonna state the Yoneda lemma for you in the particular case that we'll use it. It's statement is as follows. I'll state it as a theorem even though it's called the Yoneda lemma. I'll say it like this. Sets a and b are isomorphic so a and b are isomorphic which remember means you can put their elements in a one to one correspondence if and only if the following is true. If and only if for all other sets x, so for all sets x, the set of functions or maybe I'll say the sets of functions ... I'm gonna introduce some notation for this. These are also sets. I'll write them in pink to help us remember that these are sets. So fun ax, and by this I mean the set whose elements are functions like we were thinking about before whose domain is a and whose image is or whose codomain, sorry. I try to always avoid ... the target is x. And also the set of functions from b to x which is defined exactly the same way so an element is just the function from b to x. So the statement is sets a and b are isomorphic if and only if for all sets x the sets of functions from a to x and from b to x are themselves isomorphic and moreover this isomorphism is what's called natural in x, which means something or other that I'm probably not going to explain.
So the universal experience to learning the Yoneda lemma is you just look at the statement and you're like okay maybe that's true but why on earth would anybody care? It's a fun result because it'll take years to understand exactly why this is the best theorem you will ever learn in your life. If you're completely confused, welcome to the club.
Let me give some indication of why this is true. What's going on here. The other thing is this sounds like a crazy idea. I'm just trying to show that a and b are in bijection have the same number of elements and what I'm saying is instead let's prove for any set x, there's not even a sets worth of sets so we're quantifying over something absolutely ginormous. These sets of functions are isomorphic and moreover this isomorphism is natural. Whatever that means. This is a good idea. I promise. Let's give some indication of why this is true.
There's two directions to this proof. I said a and b are isomorphic if and only if these sets are isomorphic. So let me start by ... well in fact maybe I'll leave that first direction as an exercise because it's a more evident. This is the hard one. This is the interesting one. Let's assume that I have proven this crazy statement for every set x there is some sort of isomorphism, some sort of natural bijection between the set of functions from a to x and the set of functions from b to x. I wanna conclude from that that the sets a and b are isomorphic. This is how that's done.
If we assume that I'm given an isomorphism between the set of functions between a to x and the set of functions from b to x for all x, then in particular ... I can take my arbitrary set x to be either a or b. So in particular I have an isomorphism between the set of functions from a to a and the set of functions from a to b so this is the case where I take x to be a and I also have a corresponding isomorphism between the set of functions from a to b and the set of functions from b to b. This is just the special case of the general statement that I've assumed taking x to be either a or b. What's advantageous here is I haven't told you what the sets a and b are. Before we thought about an example with particular elements but now we're just thinking abstractly. I have no idea what a and b are, but nevertheless I can identify an element of this set. I can identify a function from a to a and similarly I can identify a function from b to b without knowing anything at all about a or b. Can anyone think of a function like that? So b is a set. I have given you no additional information. Define a function from b to b. Gotta be a well defined function. Yeah.
Speaker 11: The identity?
Emily Riehl: The identity. Absolutely. That's right. This is, I'll call it idb, it's the function that would take any element of b to that same element again. For computer scientists its lambda x [thought] x. It's that function. Similarly, of course, I have the identity function from a to a and so by the presence of these bijections, if I have a specific element here there is a corresponding element in this set. Under this bijection I have a corresponding element of the set of functions from a to b. I'll call it fe and I have a corresponding element of this set of functions ... I wrote something wrong here. That was supposed to be b and that was supposed to be a. I didn't substitute my variable correctly. So the corresponding element here, this was a function from b to a and then this is the function from a to b. What's great about this is now I have the data that's required for an isomorphism. I'm tryin' to construct an isomorphism between a and b and what I found is a function that takes every element of b to a and a function that takes every element of a to b. These corresponded to the identity function somehow.
We need to argue that these functions phi and psi are inverses because remember what makes a isomorphism between sets is I need a function like this and a function like this but I want them to be the inverse of each other. So if I do phi and then psi it's the same as the identity or psi and then phi it's the same as the identity. This is where the naturality comes in which is the part of the statement of the Yoneda lemma that I didn't explain. I'm not going to either. It's not actually that complicated but it's a little bit obscure. I have to draw a commutative diagram and then we have to stare at that for awhile. What naturality is in category theory is it's a technical term that's very properly named in the sense that if you have written down an isomorphism and it looks natural it will be natural. So if it seems like a natural construction then it's natural in the technical sense. Let's pretend that I finished the proof.
That's actually the entirety of step two is this completely crazy. Forget about that proof if you didn't like it. But what's the summary of step two? Let me actually remind you about step one. Again, our goal is just to prove distributivity of multiplication over addition, a times b plus c is a times b plus a times c. In step one we reduced this result to instead proving an isomorphism involving sets a and b ... a, b and c and the isomorphism is I'm gonna take my sets a, b and c from the disjoint union of b plus c then the Cartesian product with a and argue that that is isomorphic to the set that you get by the disjoint union of the Cartesian products of a with b and a with c. That was step one is I've reduced it to that.
So what step two is ... to prove that I have this isomorphism a times b plus c is isomorphic to a times b plus a times c, what I'm going to do is instead prove that if I consider the set of functions from a times b plus c to an arbitrary set x then that's isomorphic to the set of functions from this disjoint union of Cartesian products of a with b and a with c to x for all x and naturally in x for all x plus naturality. Okay. Again, I promise that this is actually gonna help us prove the theorem. Let's get to step three.
Okay. Again, in steps one and two I've reduced ... firstly I've categorified the question of ... just continue right here. Step three representability. In the first step I've categorified the question of this equality of cardinal arithmetic and instead expressed it as an isomorphism between sets. In step two, I replaced this isomorphism between sets with a naturalized morphism between sets of functions. What step three is gonna be about is we're gonna try and understand what these sets of functions are. I'll phrase this as a question. What is the set of functions from b plus c to x from the disjoint union of b plus c to x? What I mean by asking this question, let's think about an element of this function. What is needed to define a function fe from b plus c to x? So what is needed to define a function from b plus c to x? b plus c remember is the disjoint union of sets. If I asked you to define a function b plus c to x, what are the steps required to define a function like that? What data determines a function from b plus c to x? Again, I'm looking for a pretty general answer because I haven't actually told you what these sets are or anything useful like that. Yeah.
Speaker 12: Function from b to x and a function from c to x?
Emily Riehl: Yeah. That's exactly right. What do you need to do to define a function? For each element of the domain I have to tell you what it's image is. I have to specify fe of each element of b and fe of each element of c. Now the elements of b plus c this disjoint union either come from b or come from c and there's no overlap between those two cases. By case analysis what I need to do is I need to define two functions. So maybe I'll call one of them fe sub b from b to x and another one I'm gonna call fe sub c that goes from c to x.
I'll summarize this step. I'm gonna call this pairing. By pairing I have an isomorphism between the set of functions from b plus c to x and the set of functions from b to x Cartesian product with the set of functions from c to x. What this correspondence is, is again, if I had a function fe with a pair of functions that determines fe is, is it's given by it's component on b so the constituent function on the part b and then the constituent function on the part c. If I had a function like this from b to x and a function from c to x I can pair them together define a function from the disjoint union. This is one of these, I don't know, this feels like a pretty natural construction, so this is one of these naturalized amorphisms as I've characterized this set of functions.
Let's ask a similar question. What is ... how do we characterize a set of functions from a times b to x? In other words what I'm asking is, what is needed to define a function, we're gonna call it psi from a times b to x? What data would you need to define a function? I'm not quite tall enough for that. There's gonna be a little bit of a gap. What data do you need to define a function from a times b to x? This one's a little harder. What's one way to think about a function whose domain is a product of sets a and b? I feel like this is a weird question I think computer scientists actually have better intuition for than mathematicians because they're used to working with functions in this particular way. Yeah.
Speaker 13: [inaudible].
Emily Riehl: Yeah. Right. We can think about a and b as being separate parameters. One way to think about it is ... so the answer is for each element of b, maybe I'll call it y in b, what I need to define is the function, which is psi blank y from a to x. So for each ... if I've agreed that my second component is gonna be y, what I need then is to specify a function from a to x so then I would take it further input of a, I'd put it in there and I'd get an element of x. What is this decomposition called? There's a name for this. I mean, sorry, what's another way to say this is in other words, y is defined by a function that I'm gonna call psi again, which takes as input the set b and then to each element of b produces a function from a to x. In summary, by ... does anyone know the name for this process? It goes one way or the other. I never remember.
Speaker 14: Currying?
Emily Riehl: Absolutely. So by something called currying or maybe uncurrying, but I'm not gonna give a direction to this isomorphism so who cares? We can think about functions from a times b to x as functions from b to functions from a to x. Again, what is the correspondence if I had an element of this set? If I had a function from a times b to x, I can think of this as a functions from a to parameterized by an element of b. This corresponds to a function that I'll give exactly the same name, which evaluates the second variable first and then evaluates the remaining variable. Again, this feels pretty natural to me. I can make this definition without knowing what the sets a, b and c are. We're gonna run with that.
Let's summarize where we are. Again, we're trying to prove a times b plus c is a times b plus a times c. What we've argued using categorification is that instead of suffices to consider sets a, b, and c whose cardinalities are a, b and c respectively and prove that there is an isomorphism between the Cartesian product of a with the disjoint union of b plus c and the Cartesian product of a with b disjoint union Cartesian product of a with c. And to prove that this isomorphism of sets exists I'm going to instead prove that there is an isomorphism of sets of functions from a times b plus c to x and functions from a times b plus a times c to x for any x and naturally an x.
Now, maybe what I'll summarize step three is, by pairing we can describe the set of functions like a plus c to x as a ... I can say that that set is isomorphic, in fact naturally isomorphic to the set of pairs of functions one from b to x and one from c to x. And by a process called currying I have a similar description to say that the set of functions from a times b to x is isomorphic to the set of functions from b to the set of functions from a to x. I'll give you a chance to catch your breath because now we're finally ready for the proof. Now we're on to step four, the proof.
Ordinarily I don't, to be honest I probably wouldn't call the result that we're aiming for a theorem, but we've been working pretty hard so maybe you'll grant me the theorem. What I'm gonna prove is for natural numbers, again natural numbers are just like the counting numbers zero, one, two, three, et cetera. Maybe not zero, I mean zero's fine but I don't wanna think about that. What I'm gonna prove is this distributivity of multiplication over addition. Let's see how this is going to go. We're just gonna follow the outline that I've been forcing down your throat.
The first step is we're going to ... maybe let's remind ourselves the connection between these numbers and sets. I'm gonna let a, b and c, I'm gonna think of these numbers as being cardinalities of sets. Maybe I'm letting A, B and C, capital A, B and C be sets, whose cardinalities are the numbers that I'm interested in. So if A is five I need to find a set with five elements and so forth. Let A, B and C be sets whose cardinalities are these numbers and we'll instead show ... we'll show that the sets that's formed by the Cartesian product of a with the disjoint union of b plus c is isomorphic to the set that's formed by taking the disjoint union of the Cartesian product of a with b and the Cartesian product of a with c. As I've argued, if these sets are isomorphic then I can take their cardinalities ... I can just count the number of elements here and then count the number of elements here and because I know how cardinalities are affected by the operations of disjoint union and Cartesian product, if these sets are isomorphic then certainly a times b plus c is equal to a times b plus a times c. That was the first step, the first piece categorification.
Now to prove this we'll instead show that a set of functions from this first set Cartesian product of a with the disjoint union of b plus c to x is isomorphic to the set of functions from the disjoint union of the Cartesian product of a with b and a with c into x naturally in x. So I need to construct some sort of isomorphism like this for all x naturally, whatever that means in the set x. This is where I need your help to finish the proof. I'm trying to show that his is isomorphic to that. I'll write at the left hand side again so we can see it. A little more space. If I'm looking at the set of functions from the Cartesian product of a with the disjoint union of b plus c to x, what is this set isomorphic to? Naturally isomorphic to. How could I ... what's another way to think about a function from a times b plus c? Eventually we're gonna get here but not in the first step. I wanna take a baby step towards that direction. What's another way to think about a function from the Cartesian product of a the disjoint union with b plus c into x? Yeah.
Speaker 15: It's isomorphic to the functions from a to the set of functions from b plus c to x?
Emily Riehl: That is true, but is I think gonna be a little harder to use than the other way you could do that. What's the other way to curry? I agree with what you said. You're suggesting the currying property. Yes.
Speaker 16: The function from b plus c to the function you have [inaudible].
Emily Riehl: From b plus c to the set of functions from a to x. Absolutely what you said is true and you could use that to get through to the proof but I think it's a little harder to see so I'm gonna go this route. This is by ... sorry that's [inaudible] by currying. I have a natural isomorphism between this set of functions and this set of functions. Great. What do I know about the set of functions from b plus c, the disjoint union of b plus c, to the set of functions from a to x? How can I can I re-express that set of functions? Yeah.
Speaker 17: We can break it up by pairing into the set of functions from b to the set of functions from a to x and then [inaudible].
Emily Riehl: Absolutely. Fun ax is a funny set, but it is a set and by the pairing property if I have the functions from b plus c into any other set I can decompose them in this way. Absolutely. So this is by what we call pairing. All right. What can I do with these sets? So now I have the Cartesian product of two sets of functions, one from b to functions from a to x and the other from c to functions from a to x. How can I re-express this set?
Speaker 16: Currying?
Emily Riehl: Yeah.
Speaker 18: Can you inverse curry?
Emily Riehl: Yeah. Let's uncurry or curry which ever currying we need to do. I'll just call it currying again. A function from b to functions from a to x is the same as the function from a times b to x. I'm gonna use the similar isomorphism in the second component. Similarly, a function from c to functions from a to x is function from a times c to x. This is by currying. Great. Now what? What is this isomorphic to?
Speaker 19: These numbers of pairing?
Emily Riehl: Yeah. I can pair these functions back together and express this pair of functions as a single function whose domain is the disjoint union of the Cartesian product of a with b and a with c and whose target is x again by pairing. That's exactly what we wanted to show.
I definitely owe you the epilogue which was what was the point? What was the point of all of that? The first epilogue, why bother? The first thing I want to observe is, let's note, when I stated this theorem I said that a, b and c are natural numbers but then I started arguing about properties of sets and I didn't actually require those sets to be finite. Note in the argument that I just gave to you, the sets a, b and c don't have to be finite. This means that the exact same argument that I've just proven, generalizes to maybe a slightly more unfamiliar context. What's the name for cardinalities of infinite sets? If natural numbers are cardinalities of finite sets, what's a cardinality of an infinite set? What sort of number is that?
Speaker 20: Transfinite numbers?
Emily Riehl: I don't actually remember what those are so let me not comment on that. Possibly.
Speaker 21: Aleph-naught?
Emily Riehl: Aleph-naught is an example of one of these. I guess this goes back to the question what are numbers for? One of the suggestions of what numbers are for is for counting things or another way of to say that is putting things in order. The first thing, the second, the third thing, the fourth thing. If you have a countably infinite sequence of people and then you have the person who's one after that, that number gets called omega. That's the first infinite ordinal. And then we have omega plus one and omega plus two and omega plus three so these are the numbers that indicate orderings of infinite collections of things. People for instance because infinitely many people totally makes sense. But you use a different sort of numbers if you're counting infinite sets. You might have heard about Hilbert's hotel or got in an argument with a sibling about infinity versus infinity plus one. As ordinals those would be different, though it depends on which side you add the one, confusingly. For a different sort of numbers it's a different story.
Speaker 22: I think you have cardinals?
Emily Riehl: Cardinals. That's exactly right. Cardinals are the numbers that means the sizes of sets. In fact this is exactly how they're defined. A cardinal is a sort of isomorphism class of sets. For cardinals ... think like really big numbers. Cardinals are really big numbers that are for sizes. If your cardinals alpha beta gamma ... I have exactly the same properties. So alpha times ... I can develop arithmetic for cardinals. That I guess was the title of this talk, Categorifying Cardinal Arithmetic. I have exactly the same property. The alpha times beta plus gamma is alpha times beta plus alpha times gamma. How do we prove this? It's okay if you're unfamiliar with these sorts of numbers because the proof is exactly the same one. You take these cardinals. You represent them as sizes of sets. They're not finite anymore, but who cares? You need to argue that there's an isomorphism like this which you can construct in this way via that sequence of naturalized [crosstalk]. Maybe you don't really care about cardinal arithmetic which is fair enough. But another thing to note is that this a, b and c don't actually have to be sets at all. So don't have to be sets at all.
What did I mean by this? If we're thinking of sets as like the thing whose number of elements defines a number, then I guess we're thinking about sets. But for the meat of this argument, for this Yodena lemma step and for this representability step, those principles hold for more sophisticated types of mathematical objects. I'm gonna mention a couple theorems. One thing I could say is that for vector spaces, it'd be like R2, R3, those are examples of vector spaces but this works over any field and they don't have to be finitely dimensional. For vector spaces u, v, w, I can form the tensor products of the vector space u with the disjoint union of vector spaces v and w or I can form the tensor product of u with v and then take the disjoint union of that with the tensor product of u with w. I'm gonna argue that these vector spaces are maybe linearly isomorphic so there's some sort of relationship there.
If you aren't so fond of vector spaces I could say for nice topological spaces, for nice spaces, topological spaces, I'll call them x, y, z, I have some analog of a plus and times operation on topological spaces. I can take what's called the Cartesian product with the product apology of x with the space that's the disjoint union of y and z. I [crosstalk] again in a particular way and that is homeomorphic, so isomorphic to the space that you get by taking the Cartesian product of x with y and x with z and then the disjoint union of those spaces. This is a continuous isomorphism, and it's inverse is also continuous which is another property which is a fun fact of topological spaces.
Or if you like algebra, I could say for abelian groups a, b, c, I can take the tensor product over z of a with the direct sum of b plus c. Or I could take the tensor products of a with b and a with c and then take the direct sum. Again, I have an isomorphism like this. This is a homomorphism in an isomorphism. In each case the proofs are all the same. Exactly the same as above. The point of these category theoretic arguments is you don't have to reinvent the wheel. It's exactly as ... it's proving something in ... doing sort of context free mathematics. It doesn't really matter what sort of mathematical objects I'm arguing about. I can argue about them anyway.
The real point of this is if you continue in mathematics categorification, Yoneda lemma, representability, things called limits and colimits, things called the junctions which appeared secretly in this proof. They're just all over the place so now you've seen them. Thanks very much.