 Welcome to Sperner Lemma. Today we are going to do purely combinatorial result and one single result, okay. As a result in topology, but a combinatorial result, it has a very distinctive role. It is a landmark result which opened the floodgates for many other combinator results inside topology and vice versa. It is very simple to state. It is stated for the simplicial complex, namely the standard n-simplex, okay. So that way it is very simple, simplest object as such. What you have to do is take any subdivision of the standard simplicial complex, simplicial, the cell n-dimensional, simplex delta n, okay. And take a simplicial approximation from delta n prime to delta n. One more condition you have to assume on this simplicial approximation phi. Phi, sorry, see it is just a simplicial map. It is a simplicial approximation restricted to the boundary complex delta n prime. The boundary of delta n, remember, is topologically Sn minus 1. On that Sn minus 1 to Sn minus 1, you have the identity map. That is a continuous function. So that function is simplicially approximated by phi on the boundary. Inside the simplex, it could be any simplicial map, okay. Then the number of n-simplicies of delta n prime of the subdivision which are mapped on to the entire of delta n, this number is odd. In particular, there is at least one, you see. There will not be two. There will be at least three then. If there are two, then there are already three. This number is odd. That is the meaning of this. In particular, there is at least one, and that is very important. Instead of proving that it is at least one, we are going to prove it is odd. So what is the meaning of that? We are only counting everything mod 2 and then show that it is one. So even the counting is done mod 2. We are not bothered about 3,500 and so on. No, no, no. Don't get it. Just mod 2. Go on. So this is what we are going to do. The proof is what in common-to-all method, what one does by counting in different ways, two different ways, maybe three different ways and so on. So I am going to state another version of the same lemma, Sperner lemma which I am going to define and that will be much more elaborate and that will give the proof of this lemma once you prove that. So that is the idea of the proof. So here is an expanded version of the same lemma. So delta n prime is subdivision of delta n and phi is a simple shell map. That is the same thing. Now l I am denoting the original delta n minus 1, the boundary instead of writing every time boundary and so on. Without any its subdivision. So l is the boundary of, sorry l e, one of the simplex is namely delta n minus 1. The last vertex is chakrav. That is sub-complex. For any n simplex f of delta n prime, I will denote alpha f. I am going to define various numbers here now. Alpha f is the number of n minus 1 phases map on to l. You have fixed one phase n minus delta n minus 1 is one of the phases, right. There are in delta n as n plus 1 n minus 1 phases. So I have taken l to be delta n minus. Look at all for any n simplex f inside delta n. Look at alpha f, denote the number of n minus 1 phases which are mapped all to l. Put S1 equal to sum of all these alpha f's where f range is over all the phases of delta n prime, okay. Next, that is S1. S2 denote the number of n simplex is of delta n prime. This time n simplex is which are mapped on to delta n by phi. Then last there is another one S3, the number of n minus 1 simplex of the boundary of delta n prime mapped on to l again by phi. So three different things we are counting. The claim is that all these three are equal if you count mod 2. If any one of them is odd, other two are also odd. If any one of them is even, all the other two are also even. That is the meaning of S1 is congruent to S2 congruent to S3 mod 2, okay. So let us prove this one. Then Sperna lemma will fall away. First note that fix one simplex f. See what is this? f is a n minus 1 simplex, okay. Alpha f will be either 0, nothing is mapped on to l or it could be one, one of the simplex supposed to f or it could be two of them, nothing more than that, okay. Because you see what is f? f is a n simplex, okay. So let us take for example 1 simplex. Then I am taking n minus 1 will have a single point. So these two points, none of them may go to that point, only one of them may go to that point or both of them may go to point. That is very easy. Now we take a triangle for example and take a simplex, one simplex. So there are three edges there. None of them may go there. Only one of them may go there. If two of them go there, the third one will have to go to a single point. There is no other choice. The same way, if two of the things go to n simplex, to n minus 1 simplex, the rest of the other things will have to be going into a smaller set than n minus 1 simplex. Therefore the first thing to observe is that alpha f is either 0, 1 or 2, okay. What does it mean for alpha f is 1? That means one n minus 1 phase goes to n minus 1 phase. It is only one. That means the rest of them must have gone to n simplex, okay. So that means phase restricted f is a bijection onto the whole of delta f, okay. Thus the collection of all n simplexes of f of delta n prime is divided into three groups, depending upon alpha f is 0, 1 or 2, a naught, a 1 and a 2, okay, according to the value of alpha f. It follows that if you work modulo 2, a naught and a 2 will not contribute anything. Only the class a 1 will contribute 1, because either this is 2 or this is 0, right. So that is what we have. So s 1, which is sum of all alpha fs, okay. Instead of taking all f, you can take only f inside a 1. So each of them will give you 1. Sum total is precisely now equal to s 2, because a 1 corresponds to those f which are mapped onto fully onto delta n. That is the definition of s 2. So we have already proved that s 1 is congruent to here, congruent. This is equal, that is equal. So it is equal to s 2. A congruence is modulo 2 here. Next. Now s 1 also counts the number of n minus 1 phases of g, okay. Now 1 phase is g of delta n prime boundary, which are mapped onto l, except that you see if you look at all n minus 1 phases, each n minus 1 phase will occur twice in a n phase, in a n simplex as boundary of one side and other side. There are exactly two sides inside a delta n. There is no other choice. So it will count twice if and only if it is an interior n minus 1 phase. On the boundary, it will be only one. It will count only once. Therefore, cutting down all these entries which are occurring twice leaves us with only those g which are in the boundary of delta n. The boundary of delta n n minus 1 phases occur as a phase of an n simplex only once, only one simplex. The interior thing will occur exactly in two different places, two different simplex. So when you are counting s 2, those things won't contribute anything. So this last thing tells you that cutting all these entries, okay, the sum total is actually equal to s 2. So s 2, sorry s 2 congruent to s 3, okay. So therefore s 1 and s 2 and s 3 are congruent to each other. If you still have difficulties, what you have to do is do it for n equal to 1, where it is completely obvious, okay. You work it out. You have to read carefully. I have read it in the beginning at least three times and now I have taught it in n times. So that's a different thing. So everybody has to work it out on its one. Do it for a single edge. You have to cut it into number of edges, number of edges, right, because it is subdivision you have to take and then work out what happened, what is happening for s 1, s 2, s 3, understand that. If you still have doubts, do it for a triangle, the next stage, okay, see what happens. Beyond that, you don't have to do, it's not easy either, okay. After that you have to do, by that time you must be very, very confirmed that this argument works. Only logic will remain there after that. No pictures, okay. So now we can prove the Sparer lemma by induction, because we have a passage from delta n to delta n minus 1, right. That is the whole idea. So let c n n greater than 1 be the statement for Sparer lemma. Accordingly, we shall temporarily denote these numbers s 1, s 1 n, s 2 n and s 3 n, where they are located, okay. And we want to show that s 2 n is equal to 1 mod 2. That is the Sparer lemma. The first, the same middle one, okay. For all n we want to show. For n equal to 1, this is very easy, okay. Namely, s 3 is 1. Since phi is the identity map, now you have to use that well. Phi is the identity map on the boundary means say 0 goes to 0 and 1 goes to 1. There is no other choice, because that is already simple here. After that, so b delta 1 is boundary, b delta 1 c 1 e 2, okay. So that's just two vertex, okay. So now, having shown that n equal to 1, this is true. Assume the statement for c n minus 1, n minus 1 and n greater than 1, okay. Since phi is a simple approach to the identity when restricted to the boundary, each n minus 1 phase of phase g prime of g, where g is an n minus 1 phase, okay, of the boundary will be mapped into itself because it's identity map. In other words, only some of the n minus 1 phase is l prime of l prime are mapped onto l, okay. Therefore, it follows that s 3 of n is same thing as s 2 of n minus 1. But now we have proved that s 2 of n minus 1 is already odd. Therefore, s 3 of n is odd. The s 3 of n is again s 2 of n, modulo 2, okay. I will repeat this part. So what is happening here? Look at this one. Look at n minus 1 phases of l prime, l is one of the simplex, okay. If it's divided into some, you know, into smaller pieces, okay, n minus, there will be n minus 1 phases, okay. They, because identity map, they could not have gone out of the l. They will have to be inside l. So that is why you can apply induction for this delta n minus 1, okay. Therefore, you can look at this s 2 n minus 1 is already odd, but that is equal to s 2 of n, s 3 of n for this part. So I am computing s 3 of n, but it can be counted by s 2 n minus 1. But s 2 n minus 1 is odd, so s 3 of n is odd. So on the boundary, its identity map is used to get into the induction hypothesis, cutting down the dimension. So Sperner lemma to many results we can apply now, okay. Our aim is to prove the Brouwer's invariance of domain, okay. But first we can prove Brouwer's fixed point theorem for general case. We had proved it for n equal to 2, right. So let us, let us be done with that. For any integer n greater than or equal to 1, each three statements are equivalent. I think I have done it for n equal to 2, but the proof is exactly the same in general case. So I will repeat, what is this? The first statement is Brouwer's fixed point theorem. Every continuous map on the closed unit disk to unit disk has a fixed point. The second statement is the boundary s n minus 1 of d n is not a retract of d n. The third statement is s n minus 1 cannot be contractible. Yeah. So if you might have forgotten maybe we will repeat this proof quickly, okay. So how a implies b? a implies b we are trying to prove what we are trying to prove. If not b then it is a not a. Suppose s n is a retract of d n, okay. If r is a retraction then we will get a contradiction to a. So if r is a retraction, a retraction means what? A continuous map such that on the boundary it is identity map. So take I think take alpha to be the antipodal map. Then inclusion into d n. We get a map from d n to d n. Take a point on the boundary. It goes to the same point here. Under this it goes to minus. Under eta it goes to again minus, right. Take x here on the boundary. It goes to x itself here. This under alpha it goes to minus x. Then again goes to minus x. So x has gone to minus x. What is happening? What is happening? Any point here, okay. Post something goes to minus x goes to minus x. So where alpha x is minus of this. Eta is the inclusion map. Then f has no fixed point contradiction to a, okay. So if there is a retraction then there is no fixed point. If you take any point on the inside the boundary, in the interior it has gone already to some point in the boundary. Again under alpha it goes to a boundary point and retouch it goes to boundary point. But the original point was in the interior. So those two cannot be equal anyway. So chance was only in the boundary. But the boundary points go, x goes to minus x. So that is also no. So there is no fixed point at all. That is the contradiction, okay. Now let us prove beam prices. The proof is exactly the same as that we wrote for n equal to 2, okay. So what we have to do? We have to draw a picture x, fx, draw the line, joining x and fx, extending towards fx, get a point gx. That will be a retraction. So that is what we have done, okay. Now let us prove b and cr equivalent, okay. We have seen that if x is a contractible, if fint only if x is a retract of the cone, this was one of the theorems, okay. The cone over sn minus 1 is dn. So what does be say the boundary is sn minus 1, dn is a cone. The cone is a retract of the x is a retract of its cone. Sn minus 1 is retract of the cone over that. Then what? Then x is contractible. So here is sn minus 1 is not contractible. Sn minus 1 is not a retract. The same thing as sn minus 1 is retract. Sn is contractible. Sn minus 1 is contractible, right. Therefore, since we have proved b, not contract, say sn minus is not contractible and conversely, okay. Now how we are going to use this to prove Brauer 6 point theorem itself. This is our equivalent. I have proved any one of them. We do not know whether any one of them is proved. If you prove any one of them, all the three gets proved, right. So for n not equal to m, yeah. Finally here, to prove that each of the statement is true, we shall prove b. That is there is no retraction from dn to sn minus 1. If there is 1, okay, take a simple shell approximation to it. Call that as phi, okay. Apply Sparner lemma. Sparner lemma says that the number of simplices delta n, mapped on to the whole delta n is at least 1. It is actually odd. But if it is a retraction, the whole dn is going inside sn minus 1. Therefore, no n simplex would have been mapped on to n simplex, okay. On the boundary, no matter how many times you divide, the boundary there will be at the most n minus 1 simplices or smaller, right. Because to begin with, there are only n minus 1 simplices and you are going to going on dividing them, that is all. So there are no n simplices left. All the n simplices are inside, they are all going on to the boundary now. The boundary has only n minus 1 simplices, nothing comes into the whole of delta n, all right. So Sparner lemma gives you immediately that there is no retraction of the entire disk on to the boundary. Therefore, Brause-Wiffen theorem is also proved, okay. Now we will prove the simpler version of Brause's invariance of the main. Many books call this itself as the Brause's invariance of the main. What is it? It says that Rn and Rm, n naught equal to m cannot be homeomorphic, okay. I have stated it next to that. Rn and Rm are not homeomorphic. How do we prove that? Suppose there is a homeomorphism, then you know that you can take one point compactifications of this, they will be also homeomorphic to each other. What are one point compactification of Rn? It is Sn, Rm, it is Sn, n and n are different, okay. So we have got a homeomorphism between them. In particular, Sn and Sn will be homotopy equivalent, right. You see the theorem that says for n naught equal to m, Sn is not homotopy equivalent to Sn. So that will complete the proof of Brause's invariance of the main, the next theorem. How does this one is proved? Suppose you have a homotopy equivalent from Sn to Sn. We can assume n is less than m by interchanging the n and m, okay. Now we have already proved that if n is less than m, any map is not homotopy. So this f is not homotopy. By pre-composing with the homotopy inverse, homotopy inverse is from G to Sn, right. So G from Sn to Sn, if you compose it with f, composing with any f, homotopy, if f is homotopic to G, G composite f is homotopic to G composite the, sorry, if f is homotopic to constant map, G composite f will be homotopic to G composite constant map, it's a constant map, right. Therefore, what you get is Sn to Sn, see Sn to Sn is G, then f Sn to Sn back, Sn to Sn, okay. G composite f is homotop, null homotopic is what you get. But G composite f by definition is homotopic to identity map. So we have got Sn to Sn, identity map is null homotopic, but we have seen earlier, any space, if the identity map is null homotopic, then it's contractable, right. Right. Now part C of the previous lemma is you that there is E2 added conditions, Sn minus 1, none of the spheres, if N is floating here, this is not fixed for all N, this is not contractable. So I repeat, because of this theorem, what we have proved is two different spheres of different dimension cannot be homotopic type, same homotopic type, in particular they cannot be homomorphic. From that it follows that the corresponding RN and RM cannot be homomorphic, okay. So when this was proved, it was a very great result, landmark result. How Brauer proved it more or less through his invention of homology, he had different proofs of this also by the way. It's a proof by some dimension theory and so on, which are much, very much more complicated. So nowadays, homology theory gives you the standard proof of this one. So what I have got is, I have got you this one by just simple shape approximation and sperm outlet, okay. There is a stronger version, I obtained you this weaker version, what is that version that says that if you take an open subset of RN, non-empty open subset of RN, suppose it's homomorphic to another subset of RN, just a subset, not the whole entire, there is no homomorphism of the whole space, just a subset, okay. Suppose this subset for example, A is homomorphic to an open disk, just then A itself will be open inside RN. So that is called actually invariance of domain. Remember, domain was the word used for open subsets, open and connected subsets inside analysis. So that is how this came to be. So that will be our next task, proving the full version of of invariance of domain. So we will take it up next time. Thank you.