 So, we are in the middle of proving Jordan Brouwer theorem and Brouwer's invariance of the main theorem. We prove these two statements, modulo two important lemmas. So, these two lemmas, we need to prove now. The first lemma says that a subset of Sn, homomorphic to ik between 0 to n, k is between 0 to n, the reduced homology group of the complement all vanish. The second one statement says that if we subset homomorphic to sk now, for k less than or equal to n minus 1, then the reduced homology groups of the complement all 0 except at the level n minus k minus 1. At that level, it is infinite cycle. So, we shall prove 4.7 first and then we will prove 4.6. So, while proving 4.7, we will use 4.6 and then once we prove this one, all the four plus the two statements namely Jordan Brouwer's separation theorem and Jordan Brouwer's invariance of the main theorem will be proved. So, this is what we need to prove now. Not this one, this proof lemma. So, we have to use Mary Vitry sequence and apply induction on k. For k equal to 0, Sn has two points and Sn minus these two points is homomorphic to equivalent to Sn minus 1. If you remove one point, it is homomorphic to equivalent to a days. If you remove any two points, it will actually deformation to track to Sn minus 1. Strong deformation to track to Sn minus 1. So, in particular, it is homomorphic to equivalent. And we have already computed the homology of Sn minus 1. So, the statement is true for k equal to 0, right? k equal to 0, h, hn minus k is 0, n minus 1 of Sn minus Sn is infinite cyclic and all other things are 0. Okay, reduced homology. All right? So, k equal to 0 is true. Assume now the statement is true for k minus 1. Now, take a b which is homomorphic to sk. It can be written as union of two open sets, two closed sets a1 and a2, both closed sets. Each homomorphic to ik. ik is two disks, right? It is a disk. So, a1 and a2 are two disks intersecting along the boundary. So, a1 intersection is homomorphic to sk minus 1. Both Sn minus ai, okay, if you throw away the closed subsets, they are open sets in Sn. Hence, we can apply meriture sequence. So, h i twiddle of q plus 1, Sn minus a1 direct sum Sn minus a2, okay, two hq plus 1 twiddle of Sn minus intersection to this is the largest set, Sn minus a1 intersection a2, okay. Sn minus a1 and Sn minus a2 going there. Then hq of Sn minus b, the smallest set, okay, which is the intersection of a1 and a2. Sorry, complement of that. So, Sn minus b will come. So, this is the smallest set. For this open set, I have written like this, two hq again of same thing repeating, okay. Now, 4.6 says we know the complement of disks, right? Sn minus 1 of a, Sn minus a, Sn minus c and this one, these are all 0. The reddish homology groups are 0. If this is 0, this is 0, this arrow, the middle arrow must be an isomorphism, okay. So, isomorphism from a lower dimension, q dimensional thing to q plus 1 dimensional thing. This is Sn minus b, okay. This is Sn minus a1 intersection a2, which is a circle. For this, we have proved the statement. So, from that, we will get a statement for this one. So, first you have S0's, then a circle and then a sphere and so on, 1 by 1 induction. Now, it is k dimensional. So, this will be k minus 1 dimensional. So, you can apply theorem to this one and then write this one. Now, 4.6. Here also we induction on k. Now, I have to prove that Sn minus any disk, which I am assuming i k for convenience for k equal to 0 to n, the complement of that is or has all the reduced homology is trivial, okay. If k equal to 0, what is it? It is just a point, okay. A is a single point and we know that Sn minus a is actually homomorphic to a disk. So, it is contractible. Hence, all this reduced homology looks vanish. Now, assume the result for k minus 1, okay. Now, suppose hq twiddle of Sn minus a is not 0 for some q. So, we are going to prove this one by contradiction. Suppose for some q this, we want to prove for all q it is 0, right. Suppose it is not 0 for some q, then you should have a cycle, okay, q cycle representing a non-zero element in hq twiddle of Sn minus a. A q cycle means what? A q cycle is represented by a q change, such that boundary of that chain is 0. That is the meaning of q cycle, right. A q cycle consists of summation n i from sigma i, where sigma i are q simplexes, singular simplexes. So, how many of them are there? Finally, many of them. So, the support is going to be a compact set. So, this is the technological fact that we are going to use now, okay. So, take h to be a homomorphism from i k to a, okay, because that is what we assume. a is homomorphism to some i k. Write a as a1 and a2, where a1 is, now this, this, this disk is going to be split into half. Along i k minus 1 cross on the last coordinate 0 to half, i k minus 1 cross half to 1, okay. So, cut it in the measure, you see along the last coordinate. Both of them are copies of, again, the disk itself, i k, i k minus 1 cross 0 half is also homomorphism to i k, okay. So, I can apply the hypothesis for a1 and a2, but there is no hypothesis yet now, okay. But we can apply a marivitra sequence and conclude something. This is similar to what you did in, for example, Cauchy-Gurza theorem, okay. So, it is similar, the proof is similar here. So, and easier, apply marivitra sequence to open sets s n minus a and s n minus a1 and a2 to obtain exact sequence, hq plus 1 to the middle of s n minus intersection, hq to the s n minus 1 a is the top thing, then the direct sum of these two things, okay. And then again, s n minus a minus intersection, the union of two things, all right. Now, what we have assumed, we have assumed that there is some element in hq to the middle of s n minus a, which is not 0, okay. So, a1 intersection a2 is homomorphism to i k minus 1. So, here we have induction hypothesis. So, these two, this group and this group are 0. So, there is an isomorphism here. So, this is what we have achieved now. If this is not 0, then one of them must be 0. Not only that, actually what should have happened, look at this, how is this defined, hq of s n to hq of this. So, these are defined by inclusion maps, right. Therefore, what should have happened? If this comma this is not 0 here, it should not be 0 here or here. The image of the same c here, whatever I have taken not 0, either it is not 0 here or it is not 0 here, because it is an isomorphism. If it is 0 on both of them, then the image will be 0, that means c would have been 0, okay. So, this is what a eta i from s n minus a to s n minus a i, a i smaller, a is larger. So, this is the inclusion maps. The middle arrow is nothing but eta 1 star eta 2 here, from here to here, okay. So, eta 1 star eta 2 operating upon c is not 0, means one of the coordinate must be not 0 in the direct sum. So, you choose that as your preferred one, okay. The half, among the two half spaces, you choose the one in which the image of c is not 0. Repeat this process. Now, that is what you do Cauchy-Gerr's theorem. Repeat this process. Now, a 1, you continue it to half, okay, a 1, 1, a 1, 2 if you want and choose the one in which one of them, both of them are, you choose one of them, one of them in which the image of c, further image of c and so on. So, image of c I will keep calling, they are all inclusion maps, inclusion induced maps. The image of c is not 0. Keep doing that. So, what we are going to gain? So, we keep getting a sequence of cells, half cell, a 1 is half of it, a 2 is half of it, a 3 is half of it and so on, okay. Along the same x axis, last coordinate, nth coordinate, okay. So, we keep doing that such that now intersection of a r is nothing but i k minus 1, i k minus 1 cross a single point by Cantor's intersection theorem, okay. So, it will be one single point along the last coordinate. So, i k minus 1 cross a single point. So, this t is the image of c, okay, in h q 2 to the power s n minus 1 is non-zero. For all r greater than or equal to 1. But by induction, h q 2 to the power s n minus t will be 0 because this is i k minus 1, okay. So, what is the meaning of this? When you come here in the image, in the limiting image, it is 0 means what? It is boundary of something, okay. There exists a q plus 1 chain tau such that boundary of tau is c. But this q plus 1 chain has again finite combination of q plus 1 simplicity, right, therefore it is a compact support, okay. So, what are these s n minus a is an open set, contained inside s n minus a 1, contained in s n minus a 2, contained in s n minus t. So, finally s n minus t is covered by the increasing union of these open sets, the complements of a s. And I have a compact subset there. So, it must be contained in finite stage, okay, which means this q plus 1 chain tau must be contained inside some s n minus a r plus 1. But there, we have assumed that by repeated application, the image of c is not 0. That is a contradiction, okay. So, s n minus t is union of this increasing union versus minus a r. a r is closed subset. So, these are open subsets, okay. So, every compact subset must be contained inside some s n minus a r for the mark. And there c is 0 is a contradiction. So, we have chosen a 1, a 2, a 3 so that the image of c is not 0. So, all this because we assumed that h q of s n minus a is not 0. So, therefore h q must be 0. So, that completes the proof of the lemma and all the theorems that we have proposed so far, okay. As I told you, these two lemmas, they give you much more information than the theorems. And they will be quite useful as well also. Okay, now I would like to make a few comments on Jordan Curve theorem itself. This theorem belongs to a class of problems or results namely s 1 contained inside s 2 or a plane, okay, embeddings of s 1 is a plane. So, more generally, net apologists are interested in embeddings of one sphere in another sphere. These two lemmas tell you a lot about them, but there are many, many more things of interest which lemmas do not tell you. So, some of them are useful also, okay. So, Brauer while proving this, he did not prove just one single result, but he has done a lot of work. But now let us understand in this slide, let us understand these problems in general. Let us consider embeddings of s k in s n, okay. Embedding means just topological embedding, 1 1, okay, continuous 1 1 map onto the image that is all, okay. Being compact, escaping compact in the s n v host or automatically it will be a homeomorphism onto the image, okay. Two such embeddings are said to be equivalent if you can move one embedding to the other embedding. This moving we have already defined homotopy, but here homotopy will not work a homotopy with a restricted condition namely at each stage it must be an embedding through embeddings, okay. These are embedded objects you are you can move it from one to the other, okay. Through embeddings do not collapse and make it overlap each other or cross cross each other and so on. So, then these two are called equivalent. That means I will more specifically you must have a continuous function h from s k cross i to s n such that for each fixed t x goes to h of x t is an embedding for each t, h of x 0 is f x and h of x 1 is g x. Then we can say that f is isotopic to g. Clearly f is always isotopic to f, f is isotopic to g implies g is isotopic to f and you can even prove that just like you have proved homotopy is a transitive action. This is also transitive therefore it is an equivalence relation. So, one is interested in knowing equivalence classes of embeddings, okay. So, this is a general problem. Now, quite a bit is known and quite a bit is unknown. So, I would like to just give you some information here no proofs, okay. Consider the case k equal to 0. What is the meaning of k equal to 0? You have s 0 that means two points embedded in s n, okay. If this n is also s 0 then there are only two possibilities either identity or the transposition and you cannot connect them one with other. So, there are two classes. If it is bigger than 1 bigger than equal to 1 namely s 1, s 2, s 3 and so on. Then we know that all s n are connected and an isotopy of a point is nothing but just a homotopy nothing more than that which is same thing as just a path. So, since s n's are all path connected for n greater than or equal to 1 any two embeddings of s 0 in s n are what isotopic to each other. So, take two points take another two other points you can join them so that the two paths do not intersect each other. That is not very difficult to see, alright. So, let k equal to n, okay. k equal to 0 I have a definition. It is a classical result. What is the meaning of num embedding? Some homeomorphism from s n to s n, okay. By the way this needs a proof. Can you have a homeomorphism in embedding of s n inside s n which is smaller than s n not the whole of it. That needs a proof. I am not going to tell you the proof here. So, when you have k equal to n it will be on to surjective, okay. It follows from whatever we have done also but I am not going to do that one now. You have to think about that, alright. So, it is now question is it is called these are what are the homeomorphisms of s n which are equivalent in this case isotopic to each other. So, this was a classical problem at least in the differentiable case it was solved quite long back by Smale and Hersch and so on. So, there are exactly two classes, okay. So, when you talk about topological classes, topological mapping there are always more difficulties are there. Sometimes these are not true and so on. In the differentiable case namely the defumorphisms there are exactly two classes. This is the standard result, okay. Namely homeomorphisms of s n orientation preserving ones and orientation reversing ones they are distinct classes and there are only two classes, okay. Now, in general lemma 4.7 is very effective tells you that the complement of any embedding as trivial homology except in dimension n minus k minus 1. So, in some sense to distinguish between two embeddings it is useless. So, he is a very good theorem but it is useless as far as distinguishing any two of them complements are homologically same. So, what to do there is no other way. Whereas in some cases namely when k equal to 1 and n equal to 3. So, I will come to that case there are other things like this which are useful. So, before that I will tell you a deep result in PL topology tells you that instead of topological embedding what you call as piecewise linear embeddings. If the co-dimension which is named n minus k is bigger than equal to 3 then any two embeddings are isotopic. So, this is a fantastic result. So, s1 embedded in s4, s5, s6 or s2 embedded in s5, s6 and so on there is only one embedding essentially. In other words what you can move them around and bring it back to the standard embedding. So, that is the meaning of that any two embeddings are isotopic features. So, the complement must be at least three dimension that must a room must be there then you can move it around and bring it back. This is the result in PL topology. So, the most important thing is now the complement being equal to n minus 2 that means 2 k equal to n minus 2 and k equal to n minus 1 these two things remain to be settled. The case could n minus 2 itself is that means co-dimension to embeddings of spheres this goes under the name North theory which is a quite a deeply studied branch of algebraic topology okay and lot of applications in many, many other sciences like chemistry, physics and all of them use this North theory okay. Even physics very high physics I mean string theory and so on. So, we shall not be able to discuss this problem this particular thing any more detail except I want to tell you the case when n equal to 3 and k equal to 1 is the most interesting one that is where lots of problems are there and classically the only thing that was done was to look at the fundamental group of the complement. Look at Jordan Brower separation theorem all these theorems are two lemmas they do not tell you anything about the fundamental group they tell you only about the homology. The fundamental group okay can be computed by what are called as floating representations and so on using some hybrid when it comes to theorem actually. So, that will give you a lot of information yet up to 1980s lot of problems were unknown the group theory was not able to give you enough thing till CRF Jones came into picture and cracked the whole thing and opened up a new way of thinking completely. So, it is now quite flourishing branch of algebraic topology okay. So, finally k equal to n minus 1 let me give you a little more information on this one for n equal to 2 this is Jordan Kuru theorem okay here we have very strong theorem from complex analysis namely Jordan Schoenfritz theorem which is an extension of Riemann mapping theorem which says that any two embeddings of the circle inside r2 are equivalent in other words if you have any simple curve you can bring it to the standard circle by this isotope okay with a little more extra assumptions the same thing can be done in all other dimensions the extra assumption from it was first it was major and then the assumption was weakened by Morse and then further by Brown but still there is some assumption I do not want to go into the technicalities of that. So, you can read it in Bradon's book or some of one of the thing Brown's paper I have given you Brown's paper is very readable paper so you can look into that one the last thing that I wanted to tell you is that what is suppose you remove the extra assumption namely any topological embedding of s2 inside s3 that is the first thing right so that was cracked by Alexander Longworth in the negative. So, it is called Alexander's horned sphere obviously the embedding is somewhat wild okay wild embedding okay so he is going to put horns to these spheres and what happens is it is the actually the embedding of the disc d3 the boundary is obviously a sphere two sphere but the complement is a complicated complicated usually if you standard sphere d3 what happens the complement is also d3 so complement we are able to show that is not simply connected therefore it cannot be a disc okay so for lack of time I will not be able to present that one but I will just give you the picture the original picture of this Alexander's horned sphere okay so this is supposed to be the disc okay if you cut it away here it is just a twisted disc with a two horns the horns here this is a big horn this is another horn they each of them grow two horns again like stags okay two horns okay and then those two horns all the four of them they each of them grow two horns again so they are all trying to get interlinked as they keep growing they are coming closer and closer and at the end they do come closer in the limiting case so there will be a counter set of points wherein they meet finally okay so one can show that the complement the inside thing will be still a disc complement inside means inside this one this is a this was a disc we have started growing horns okay so this inside thing will be a disc and so boundary will be a sphere but the complement is complicated okay so that's all I can tell you for this one so let us stop here thank you