 So, today's topic is the Big Theorem, Brauer's Invariance of Domain. We have already seen last time, a weaker version of this theorem, namely, for n and not equal to m, the Euclidean space Rn is not homomorphic to Rn. This will be a consequence of the Big Theorem that what we are going to prove today, namely, if X and Y are two subsets of Rn, homomorphic to each other, if one of them is open, then the other one is also open, okay? So, what you can do is, think of the Rn and Rm, n less than m, then you put Rn as a coordinate space of Rm, now you have two subsets, one is Rn cross 000 and another is Rm, both of them are subsets of Rm. If they are homomorphic, since the whole Rm is open, Rn cross 000 should be also open, that is very easily seen, that is not the case. You can easily show that coordinate inclusions are not open sets, they are close sets but definitely not open sets, it is very easy to see. Therefore, Rn and Rm are not homomorphic, so that will be an easy consequence of this Big Theorem. However, how we arrived at that, we arrived at, by showing that Sn and Sm are not even homotopy equivalent to each other, this we would not have got from this theorem, Sn and Sm are not homomorphic we can prove from this theorem, but they are not even homotopy equivalent to each other, you will be able to prove. So, that step, going through Broward-Schwein theorem and proving and going through Spurner Lemma is not achieved by this Big Theorem, so you must know that is why that is justification of proving that one separately. So, let us now prove the Big Theorem. Here again, the key lemma is a strengthening of our homotopy simple shell approximation, namely controlled homotopy and that will give you a topological criterion for a point to be an interior point of a subset, I am talking about a subset of Rn, take a subset of Rn, whether some point is an interior point, it will give you a topological criterion for that one, topological means it will be an invariant of the homomorphism type of X, whereas an interior point is a concept of an embedded object X contained inside Rn, so what is its interior? If you take just a topological space just like that, interior of X is the whole X, there is nothing else, but X as a subset of Rn has a different interior. However, this is independent of how X is contained inside Rn is the crux of this thing, so that is what we are going to achieve, which is actually much more stronger than invariance of domain itself, in your domain we will be catching it as a corollary as a consequence. So this is the general idea and what is ignored in the proof of this one, so let us now go step by step, as I have told you the key lemma is whatever say key step is lemma 6.6, which will lead to another theorem 6.8, I have stated separately because it seems to be much more stronger than in some sense, not exactly, then you are always invariance of the domain itself. So this lemma is the following, ak finite simple shear complex, set us say its dimension is less than some number m, take a subset which is closed inside mod k, given any function, any continuous map f from mod k to dm, dm is the unit disc in Rm, such that the subset A goes inside the boundary sm minus 1. The lemma says that there exists a homotopy h from mod k cross i into dm of this function f, namely h of x, 0 is fx for all x in mod k. This homotopy is relative to A, which means the point is fixed throughout the homotopy and the end result h of x1, that you can call it as g of x, this takes values inside sm minus 1. In short, you can say that a function like this when the dimension of k is smaller than m, such a function can be homotopically pushed into the boundary completely. A homotopy is achieving that, so that is the way to understand this one. The proof of this itself takes some time, so that we will postpone it for next time. However, we shall assume this result and go ahead towards proving Brauer's invariance of the main. So, the first step is a readymade homotopy theoretic corollary. Now, you take a subset, a close subset of sm, again assume that this n is less than m. Then every map alpha from a to sm minus 1, m minus 1 might have become equal to n, but it is n is still less than or equal to m minus 1, that you have to understand. Every map from a to sm minus 1 can be extended to a map from sm to sm minus 1. If you replace this sm minus 1 by a Euclidean space, this is nothing but or dm or im and so on, rm and dmi, this is nothing but T's extension theorem. But with, you know, rm or in etc. instead of that, if you have a round sphere here, this is not an easy thing, but this comes immediately as a corollary to this theorem. So, this is homotopy theory now, you see, this can be extended. Like we told you that the homotopy theory right in the beginning consists of these kind of things, extensions and liftings. So, here is first time you are saying some such thing very clearly, that a map defined on a closed subset, any closed subset can be extended to the whole space provided the extending space itself is of smaller dimension than the sphere. This, without this condition, it is even false. In a way, this resembles T's extension theorem and T's extension theorem will be used. How you have a map from a into sm minus 1, you can include it in dm, sm minus 1 is contained in dm. Once it is dm, this is just like, this is exactly T's extension theorem that you can extend it to the whole of sm because sm is normal, that is all. If you might have seen it only for real valued functions, but then this can be, you can take this as various coordinate functions. First of all, dm can be replaced by i power m. Then there are m coordinate functions, each of them you do, then you get a functional reaction of this, that is all. So, such that, that is a map such that f a equal to alpha a means that this alpha can be extended. Now, you take k to be boundary of delta n plus 1 as a simple shell complex. Then mod k is precisely sm, it is a homomorphic sm. Therefore, I can apply this lemma. I started the map here, here instead of f I have alpha. So, h be a homotopy given by the above lemma. See alpha is given on a, but f is an extension, but it is taking values inside dm. So, push it back to sm minus 1 by a homotopy. That homotopy is a relative homotopy, controlled homotopy. So, it does not change the function on a which is alpha. Therefore, the result h of x1 is equal to alpha x on a. So, that is the proof of this. So, we will use this one now that any function from a to a sphere can be extended to the wall of sm provided this n here is less than or equal to m minus 1. So, the homotopy theoretic background is now complete. Now, we have to do purely point set topology. So, I recall a few things here just to make sure that we do not have any confusion. These are elementary things. Take any subset of a topological space. A point z, point x inside z is called relative interior point of x. If there exists an open substitute of z such that x is in u is contained in z x. Of course, x must be a point of x if it is an interior point. Not only that there must be a neighborhood, this neighborhood is not in x, it should be in z. So, that is the meaning of relative interior point of x, okay. A point z, x belongs to z is called relative boundary point of x if it is not a relative interior point of x but not a relative point of the complement also. So, I think it is called a boundary point which is the same thing as saying every open set around x will intersect both x as well as this complement. So, I am just recalling what is the meaning of boundary and interior to make it, you know, the point relative interior emphasis is on that one, okay. We are not looking at manifold theoretic boundary. For example, d and wherever it is, its boundary is Sn minus 1, right. If you include it as a subspace of Rn, then also it is true but there is a different concept of boundary. So, that is not the boundary that we are left into here. This relative interior relative boundary is purely embedded notion here, okay. So, now we come to relative boundary point of x as a property that every neighborhood enters. That is what I told you, okay. A subset of x, a subset x itself is open if all its relative interior points, okay, are all of its points are relative interior points. That is sorry, yeah. The following theorem, it characterizes intrinsically the relative boundary points and hence the relative interior points. Whether you give the characterization for interior points or boundary points, it will be the same. It will be read both ways because its characterization of a subset of a Euclidean space, okay, is that is what I am telling you. It is perhaps the strongest form of Brauer's invariance of domain because once we use this characterization to prove Brauer's of domain, invariance of domain, obviously that will be stronger than this. Let us take the statement of this theorem. Take a compact subset of Rn, okay. Take a point in x. It will be a relative boundary point of x if and only if for every t positive, there exists a R smaller than t, 0 less than R less than t such that every continuous function on the complement of this open bound x minus the R of x to Sn minus 1 as a continuous extension over the whole of x. This n minus 1 is the same as this corresponding to this n. n is the same here. So the statement is about compact subsets but it can be easily extended to non-compact species also. So let us concentrate on compact set, concentrate on other parts of the theorem rather than why x should be compact and so on. That just helps. So here Br of x is the open ball of radius R around x, okay. So there are if and only if parts, right? So I am going to prove the implication here. Take U equal to Br of x intersection x. This is just a short notation every time instead of writing this one for some arbitrary R. Take any function and then I will extend it to the whole of x. That is enough because then given any t, I can choose R to be smaller than that and apply the, apply whatever I have done here, okay. So look at the function restricted to the boundary of Br of x intersection x. That is A. Now this A is a subset of boundary of Br of x, right. And this is taking place inside Rn. So this is a subset of Sn minus 1, close subset, okay. This gives a map F prime from A to Sn minus 1 and A is a close subset of this sphere. Therefore, by previous corollary, there is a continuous function g which extends F prime on the whole of Sn minus 1. So this is the theorem that we are using here, 6.3. That it can be extended to the corollary 6.3, all right. So now take a point P in the complement of Br of x minus x here on x. Why I am saying this? Because x is a relative boundary is what we have assumed. So every open set around x will intersect both x and its complement. So take a point inside here which is not in x, okay. Now let eta from x to the boundary of this is a radial projection from the point P. The radial projections are defined on the entire of Rn minus P. What do you mean by radial projection? Every point of Rn minus P has a unique expression like a polar coordinates as if P is the origin, right? So that eta is the radial projection. After that, all that I do is define hx to be fx on all points not in u on the x I am defining this whole map. Not in u, it is fx. If we see in the comp in the u bar inside u as well as the boundary, u is this set, right? So inside u as well as the boundary define it as g of eta x. When x belongs to the boundary as well as inside x, then these two will coincide because of what? The eta on the boundary of this point is identity if x is a point here itself, then eta of x will be x itself, okay. And this is just, g is just extension of, what is g? g is just extension of this f prime that we had. So f prime is nothing but f restricted. So it is fx only. It is just a notation here. f prime is f itself restricted way. Okay. So on the intersection, they coincide. Therefore, both of them are close subsets. Therefore, h is a continuous function, okay. And obviously, it is an extension of f. So what we have done is every function f defined on the complement of an open bar can be extended to the whole of x, okay, assuming that x is a relative boundary. Notice that relative boundary has been used that p is an outside point is there, all right. Here is a picture. So p is an outside point. This is the ball Br of x. x is a point here. Well, I have drawn it on the boundary. So you may ask, oh, you are already boundary. It is a boundary point in what sense I am using only point set topology here. This set open ball will intersect both x as well as complement of x, okay. No matter what Br of x is, what r is, x is a relative boundary. That is all I am using. So look at this projection map, okay. And whenever this ball intersects this part, for example, is both in x as well as in the boundary. Where does this point go? It will go to this point only. If the point is here, where will it come? It will come to a point here. That is a real projection. Every point here will come here, okay. So I am taking this relative position which is defined on the whole of rn minus p restricted to x, okay. So this will map like this into this one. Points here will be mapped here and so on. In any case, the entire thing will be contained inside s n minus 1. The projection onto this one, okay. This is the picture of the radial projection, all right. So one way we have proved, we have to prove the other way around now. Suppose this criterion is true, then I want to prove that x is the relative boundary. Instead of that, what I am doing is, if it is not a relative boundary, it is a point of x, not a relative boundary means it is a relative interior point. Then the criterion is false. Namely, I must find a t such that no matter what r I take, smaller than t, okay, there will be some function which cannot be extended, right. So you have to read the negative of that statement correctly, okay. So take x to be a relative interior point. Let t be positive any, okay. If I show for some t, okay, there is something like that, okay. If we are, for each t with some property, you can choose something smaller than that and that t has something smaller than that. So if I prove one such thing, then the criterion is gone. That is why I can assume that u which is this notation bt of x, okay. This u was different earlier, right. So this is now it is different. That is why I am putting a definition here each time. It is contained in the boundary, it is contained in the whole of x. This is possible because x is a relative interior. For every t, there must be an r, blah, blah, blah, blah, right. To say the negative, I can choose my t. So I have chosen this t. Now for every r, I have to say something is wrong. So let f from x minus x to Sn minus 1. See I just throw away the center of this ball, which is a point of x after all, okay. To Sn minus 1, b, this map which is defined on the whole of rn minus singleton point x. This is again the radial projection, y minus x to the power by norm of y minus x. If x was origin, you would have been y divided by norm y. We have studied that one, rn minus 0 to Sn minus 1, okay. So now because x is treated as an origin, so you have to say y minus x by norm of n. This vector is always, this makes sense because y is not equal to x. See I am defining it on rn minus 1. Then I am restricting it to x. This x is subset of that, okay. So y minus x by norm of n minus x is always a norm 1. So it is a map from x to Sn minus 1, okay. It is from x minus x to Sn minus 1. Suppose you can extend this to the whole of x, okay, which will be an extension of f restrict to x minus u also, okay. Suppose there is a map g, which is an extension of f restrict to x minus u only. U whole u is contained at x, right. Suppose there is such an extension. Then what I do is I look at this new map phi of v, which is g of t times v plus x. Take any v, okay, in dn, which is norm of v is less than equal to 1. Multiply it by t. It will become a norm equal to, at the most equal to t, right, from dn. You add it, add the point x. It will be inside this ball, the closed ball. You got it? I have just translated, first multiplied by t and translated the vector. So that vector will be inside d, inside here. So it will be inside x. We take g of that, makes sense. Call that as phi. That is a map from dn to sn minus 1, right. dn is boundary of this one. Start with any dn. When it is on the boundary, it will go to the boundary of this one, okay. So take this map. This is really continuous, translations, scalar multiplication and applying g, so composites of various constants. If norm of v is 1, then mainly it is on the boundary of this, norm of t will be equal to t. So tx minus v will be of norm that, this tx plus x minus v. So that means this plus this is precisely on the boundary of this one, of vt of n, right. So if norm of t is 1, then this is in the boundary. Therefore phi of this which is defined like this one, which is the extension of g is an extension of f, which is f of t v plus x, okay. And that is equal to v. f of t, what is f of anything? f of anything is y minus x divided by norm y, right. So look at that map f, t v of x plus x minus x is t v divided by its norm which is t. Okay. So phi v is identity, phi identity on the boundary, all right. Now you see this means that phi is a retraction of dn on to its boundary. It is a contradiction to our theorem 6.5 that we have used to prove what? Brouwer's fixed point theorem. So we are using Spurner lemma also indirectly here. So we could not have proved this theorem without that, okay. So we have proved, we have used that one. So this is a contradiction. Thus we have proved that f restricted to x minus t cannot be extended to r x. Extension, if there is such an extension, what we have proved is that there is a contradiction that dn retracts to its boundary, okay. Now you take any r less than t, okay. Then take v equal to Br of x, then f restricted to v, the same function cannot be extended because even f restricted to the boundary in the smaller thing cannot be extended. If v is Br of x, then v is a smaller subset than u, x minus u f restricted to even that cannot be extended. So this cannot be extended, obviously. If there is an extension of this one, then there will be an extension of f of x minus u also. But just now we proved that f of f restricted x minus u cannot be extended. So this completes the proof of the theorem. Let us stop here and we will complete the proof next time.