 So, now we come to the celebrated results of Brauer, namely Jordan Brauer, Jordan Brauer separation theorem and then using that we will prove the Jordan Brauer's invariance of domain also. I would like to tell you that in part one we had already proved the Jordan, sorry the Brauer's invariance of domain directly using cellular simplicial approximation theorem and Sperner lemma. But we could not prove Jordan Brauer separation theorem there, ok. So, here we are proving Jordan Brauer's separation theorem first and then using that we will prove the Jordan Brauer's invariance of domain. So, this has completely a different approach and historically the Brauer more or less came with these ideas of developing homology just to prove this invariance of domain theorem one might say. So, the idea, basic idea of the proof that we are going to present is more or less as in Brauer's original paper 1911. So, let me state them together because they go hand in hand in some sense though they are quite statements are quite different. So, Brauer Jordan Brauer separation theorem is an extension of Jordan's Jordan-Karu theorem for S1 containing S1 embedded inside in the complex plane R2, ok. So, such a thing is called a Jordan-Karu theorem or a Jordan loop which means 11 mapping of S1 and 11 continuous mapping of S1 into R2, ok. So, this theorem was a what can you say a sensation at that time because after defining what is the meaning of continuous function, lots of weird examples were constructed by many many mathematicians. Continuous functions like Antonis and you know what is called as pianos, pianos curves which were continuous functions from closed interval, but they were filling up the whole of space not only just you know I cross I, but I cross I cross I, I cross I cross I countably many times also and there are many other kinds of weird examples. Nowhere differentiable continuous functions and so on were there is by Brauer-Strauss. So, this came as a present surprise, but his own proof, Jordan's proof was rejected by people you know contemporary people. Later on it was Weblen who gave a proof that was accepted. After that several people have given several so called proofs, ok. Why actually so called? Most of them are wrong, but some of them are right, ok. There are different proofs of this one also. Some of them are right, ok. Here we are not proving that special case separately. We are going to prove directly, we are going to prove the n dimensional version of that namely an n minus 1 sphere embedded in an n sphere separates it into exactly two components. These two components have their common boundary as the embedded n minus 1 sphere. So, this is the statement. Jordan-Brauer that is Jordan-Brauer separation theorem. The Brauer invariance of domain says that if you have u and v are any two subspaces of r2 homomorphic to each other. If u is open, then v is also open. So, that is the meaning of invariance of domain, homomorphic invariance of domain. One is open, the other one must be open. They must be subspaces of the same rn, ok. So, towards the proof of this, we will state two lemmas which are themselves give you better picture of what is going on, ok. Sometimes the lemmas can be quoted and used elsewhere than the theorems. So, these are quite interesting lemmas. So, first lemma 4.6 says that take a subset of Sn which is homomorphic to a closed re-interval cross itself k times. Instead of the sphere, it is a disc. So, I am taking the square model here. So, you can take a disc model no problem. It is homomorphism. Homomorphic to ik where k is between 0 and n. Like you can take a point or you can take an arc k equal to 1 or you can take a square and so on. Any kind of homomorphic picture. So, it need not be in a rectangle even. It need not be twisted and so on, fine. Then the reduced homology groups of Sn minus a all vanish. You know that if you throw away one point from Sn, it is contractible. So, reduced homology vanishes, ok. So, this is a far-reaching generalization of that phenomena. Sn minus a, you do not know whether it is contractible or not. I can say that all the reduced homology groups are trivial. No matter what it is, a is homomorphic to ik is all that I need, ok. So, I would say that this lemma itself is quite useful, ok. So, the next thing is now suppose b is a subset of Sn, homomorphic to sk, ok, for some k between 0 and n minus 1. Then the reduced homology groups of Sn minus b, I mean you are throwing away b and what is left out. These are all 0 except in one single dimension. So, h to the power of n minus k minus 1 of Sn minus b is that, ok. So, this is also of importance in many or not theory problems and so on, all right. So, what we are going to do is first we prove these two lemmas, ok. So, first we assume these two lemmas and complete the proof of the these two theorems. Then we will prove this 4.7 first, ok, which is easier, though this is a little more complicated statement. The Kruskas matter, this one is the, this one will come later, the lemma 4.6 which is a very foundational result one would say, ok. So, that is the, that is the plan, all right. So, granting 4.6 and 4.7, we shall prove this 4.15 and 4.16 theorem, ok. So, 4.15 first, namely Jordan-Brower separation theorem. Take b, be a subset of Sn homeomorphism Sn minus 1. We have to show that Sn minus b has exactly two components to begin with. And then those two components, components means what? Connected components, path connector whatever, u and v, call them boundary of u must be equal to boundary of v must be equal to b, this is what we have to show, ok. Now, look at the above lemma. What is this lemma? Lemma is what? b is homeomorphic to sk, where k is equal to n minus 1, right. So, put that h twiddle of n minus n minus 1 minus 1, ok. So, that will be h naught. H naught twiddle is isomorphic to z. H naught twiddle is isomorphic to z means h naught is isomorphic to z direction z, which is the same thing as saying, h naught, what is h naught is that I will use means what? That the space has exactly two connected components, ok. So, very easy proof. So, h naught twiddle of Sn minus b is z and this implies Sn minus b has precisely two components. Let us call them u and v. Now, elaborately, now you have to show that both u and v are opens of sets by definition boundary of u which is u bar minus interior of u, ok. But interior of u is u itself u bar minus u. Similarly, boundary of v is v bar minus v, we have to show that boundary of v itself is boundary of u could be boundary of v. This is what we have to show, ok. So, v is open and disjoint from u, ok. u union v union b will be the whole space that is very clear. So, no point of v because v is open is in the closure of u because u is disjoint from v, ok. So, it take a point of v, v itself is a neighborhood, it does not intersect u. Therefore, it is not a boundary point. Therefore, boundary of u is contained inside v, ok. It is not inside u because by definition boundary of u is u bar minus u. So, there is a chance that it may be inside v. If it is not there also, the rest of thing is just boundary b because u, v and b together all the three things form our space s n, ok. It is a pure set theory. So, boundary of u is inside b, one way of it. So, now the point is we have to show that b is inside boundary of u. Every point of b is a closure point of u is what I would show. Then I am done. Once I show it is closure point, ok. It is not a point of u by the very definition. So, it must be inside b, inside a boundary, ok. So, b is contained inside boundary with the proof if you show that every point of b is a closure point of u. By symmetry, once I show it for u, it will be true for true for v also, ok. This is the same as showing that any arbitrary small neighborhood n of b, take a point b inside the boundary of b, ok. Intersects u bar. If it inter, every neighborhood intersects u bar, then you are done, ok. So, inside a neighborhood, whenever you have a neighborhood of a point, we can find a neighborhood n 1 of b in b, ok. What is b? b is homeomorphic to Sn minus 1, remember that. So, you can take a neighborhood and use the homeomorphism and come to Sn minus 1. So, there will be a neighborhood of Sn minus 1 going inside that neighborhood by continuity, right. So, you can find a neighborhood n 1 of b such that b is actually union of n 1 and n 2, ok. And both n i's are homeomorphic to i n minus 1. In other words, what I am doing is I am taking n 1 to be a small disc around this point, ok. The complement will be also a disc. Inside a sphere, you should take a small disc, the complement will be also homeomorphic to a disc, ok. So, b is a union of n 1 and n 2 and both n i's are homeomorphic to one dimensional over, sorry, is the same dimension is there Sn minus 1. So, it is i n minus, ok. i n minus 1 or a disc, d n minus 1 does not matter, ok. I am using the square model here, i n minus 1, ok. Here is a picture. So, this green thing could be the, so this picture is after all n equal to 1, n equal to 2. And b is a circle, b is the curve and we are working inside a sphere, ok. Same thing as working inside r 2, ok. This picture is for a curve inside r 2. So, this is the curve, all this along with this green part. And I have taken a point here, b on the, where? On the curve, on the b. This is b, right. On the curve, I want to show that every neighborhood of that will intersect u, intersect u bar. Of course, in picture it is obvious. So, that is the problem. You should not use a picture to conclude something, only what are the hypothesis so far. Keep, keep track of that, that is all. So, there is a neighborhood, there is a point b. I have to show that this neighborhood intersects u bar, ok. So, what I have chosen is, this n 1 is homeomorphic to i k minus 1. Its complement is also homeomorphic to i k minus 1, i n minus 1, k is n here, ok. So, now the previous lemma 4.6 says that s n minus n 2 is portion which is homeomorphic to a disc, ok. This is contractable. This has reduced homology or 0, ok. In particular, H naught of that is z or H twiddle, H naught twiddle is 0. So, that is connected. So, if you throw away only the, this part n 2 from s, from s n that will be still connected. That is the first lemma, ok. So, s n minus 2 is path connected because H naught twiddle is 0, ok. So, we take p, two points p belong to u and q belong to v, ok. One is in here, one is there, alright. Point is u and v and join them by a path. This path lying in s n minus n 2 because this p q are in s n minus n 2. What is n 2? n 2 is a part of v. So, I have taken points u and v outside this path and two as well as v itself, ok. They are inside u and v. Therefore, I can join them by p by s n minus n 2. So, that is path connected. But u and v are in different components of s n minus v. Therefore, you cannot join them in s n minus b. But this path is there, where it is? It is in s n minus n 2. Therefore, it must intersect n 1. The path whatever you have taken, omega must intersect n 1. So, let us say I have parameterized from p to q, ok. Omega 0 is p, omega 1 is this one. There will be a first time, the first t for which w t will be on this n 1, on this green part, ok. Which is the least element, ok, or whatever, t between 0 and 1 for which omega t will belong to n 1. That omega t, whatever the point is obviously in the closure of u. You will say that this must be in the closure of u, ok. What does that mean? My whatever n I have chosen, it will be inside n 1 by it, the whole n 1 is inside n, ok. This n intersects u bar. Every point, every point b has a neighborhood n which intersects u bar. That is what we have shown. So, that completes the proof of Jordan Brouwer separation theorem. Now, we come to Jordan Brouwer separation theorem, Brouwer's invariance of domain theorem. So, u and v are subspecies of R n. By taking a one point computation, we can work inside s n. There will be still open subsets of s n also or there subsets of s n. If one of them is open, I have to show the other one is open. The only assumption we have is that u and v are homeomorphism. So, let us fix a homeomorphism fee from u to v, ok. Now, we are assuming that they are inside s n, no problem, ok. Now, I have to show that v is open, right. So, take a point inside v which I can always write it as y equal to phi x where x itself is inside u, ok. What I have to do? I must produce an open subset in R n or c s n actually which is contained inside v and contains the point phi x. If this is true for every x, then why I have completed the proof that v is open, ok. We should produce an open subset of s n containing y and contained in v, ok. So, now you see the Jordan Brouwer separation theorem comes to help here. In a miraculous way, this great theorem will prove now, ok. So, let a be a neighborhood of x in s n because u is open, that is the assumption. So, take a neighborhood around x namely which is homeomorphic to a closed disk, ok and contained inside u, ok. That you know that that is possible for every open subset of s n as well as every open subset of R n whatever. Then its boundary, boundary of a is homeomorphic to s n minus 1. If we apply fee which is a homeomorphism from whole of u into v, ok, fee of boundary of a, ok, that will be write it as b, that will be a copy of s n minus 1 because it is homeomorphic to s n minus 1. And that is contained where? Contained inside s n. Of course, it is contained inside v also now, ok. By Jordan Brouwer separation theorem s n minus b has precisely two components and both components are obviously open subsets, ok. Look at a minus boundary of a, that is the inside of the disk. I have taken a to be a homeomorphic to a closed disk, right. a minus boundary of a, we say open disk, that is connected and that will not intersect fee of fee of this boundary of a, right, when you take fee of that. Therefore, fee minus a minus boundary of a, when we add this interior of that, ok, that must be a connected subset, ok, which is fee a minus b, fee a minus v b is connected. By lemma 4.6 s n minus fee a is connected because fee a is what? a is a disk. So, fee a is homeomorphic to a disk. The other lemma, the previous lemma 4.6 says s n minus fee a is connected, ok. So, far you have got the picture. On the other hand, s n minus b is s n minus fee a, b, remember b is fee of boundary of a, s n minus fee a union fee a minus b is a disjoint union because I have, this fee a I have thrown away here. So, here fee a is subset of fee a here. So, this s n minus b, I can write it like this, ok. So, disjoint union like this, ok. Fee a is closed, s n minus a is open, ok. This b is closed, of course, fee a minus b is semi closed. It is a, it is an open subset of a closed set, does not matter. It is disjoint, ok. Hence, these two sets must be the components of s n minus b. There are only two components here. This whole set and this is connected, this is connected, this is connected. The connected subsets when you have separate are contained in this one and there is a separation s n minus b, they must go to one of them and that is the whole thing, it is the whole of it. So, they must be the components, ok. So, this is an elementary topological result that I am proving. You have, you have subset which is disjoint, this is union of two connected components, ok. You have written it in terms in another way namely as disjoint union of two connected sets, ok. Then those two connected sets must be the components. There is nothing more than that here, ok. This is ordinary topology, ok. Hence, these two sets must be the components of s n minus b. But now components of s n minus b, we know that they are open subsets. In particular, it follows that this fee a minus b is an open subset of s n. So, this should have been obvious. If you think Jordan-Karuz theorem is obvious, but we have to prove it with all this elaborate way, ok. Fee a minus b is an open subset of s n. Also, it contains the point y because it contains fee x and it is a subset of fee, ok. Because fee whole fee is from u to v, ok. Thus, we have succeeded in producing an open subset, ok, around fee x actually containing contained in v. So, this proves that v is open. So, the two lemmas we will prove next time. So, using the two lemmas, we have completed the proof of two big theorems, ok. Thank you.