 So on the way, Brauer's invariance of domain, we proved a topological criteria for a point of a subset of Rn to be an interior point of the subset. That was theorem 6.8. Using that theorem, let us complete the proof of Brauer's invariance of domain, which is now very easy. After that, we need to complete the proof of the control homotopy lemma. Then the entire proof of Brauer's invariance of domain will be completed. So let us start proving the final theorem point 6.1, 6.7. So for this, we have two subsets X and Y of Rn and a homomorphism from X to Y. This homomorphism, you have to understand that it is defined from X to Y, not on the whole of Rn to Rn. If there is a whole homomorphism from Rn to Rn, which takes X to Y, then this theorem is obvious. Then open subset image of open subset will be open. The crux of the point is this edge is defined only from X to Y and it may not be extendable to the whole of Rn. But we want to prove that if X is open in Rn, then Y is also open in Rn. So what is the method here? Namely, the boundary, the relative boundary goes to the boundary because the homomorphism, rest of the points namely interior, we will have to go to the rest of the points there namely interior complements. So if you want to say the interior goes to interior, rest of the points, this is the kind of thing that you want to solve. So this we have to do for every closed ball B contained inside X. So if I show that for every closed ball B contained inside X, boundary of H of B is equal to H of boundary of B. Then this means H of the interior of B is interior of HB. From this, what we will follow? We will follow that H is an open mapping from X to Rn itself as a map into Rn. Of course a map is from X to Y but Y subset of Rn. So this open subset in Rn because take any open subset of X which will be the union of interiors of such balls contained inside X. Okay. So union of all this is equal to union of these open sets here. These interior of HBs are open subsets of Rn. Therefore it will follow that this whole mapping is open mapping and in particular Y which is equal to Hx is also open. Because X itself is open, it started, we started it. Okay. Only when you assume X is open, we will get Y equal to Hx is open. So all this open subset is now happening inside Rn. Okay. So we want to show this inequality. H of boundary of B is equal to H of boundary of H of B for every closed ball B contained inside X. Now the closed ball is a compact set. H restricted to B to HB is a homeomorphism. Therefore I can apply the previous theorem for compact subsets. See the previous theorem 6.8 was for compact subsets. So that is how this problem for arbitrary X is reduced to compact subsets. So that is the key here. Okay. So now we can apply 6.8 to both B and HB which are both compact. Because B is compact, HB is also compact. Take a point in the boundary of B. We have to show that Hx belongs to boundary of HB. Right? But the theorem 6.8, it is enough to prove that this Hx has arbitrary small neighbourhoods B and HB such that every continuous map from G from V to Sn minus 1 can be extended to over to whole of HB. This was a criterion for a boundary point. Okay. To start with a neighbourhood V, whatever neighbourhood, V should be neighbourhood of Hx. Take a function defined on V into Sn minus 1. Okay. I must show that it can be extended over to HB. By using H, I will go back to B now. How? Take H inverse of V equal to U. Okay. That will be neighbourhood of X because V was a neighbourhood of Hx. Okay. And this U is now a subset of our X, our subset of B now. Right? So H, so here we have, now we have come to B. Start with Hx inside HB. We have come to B by H inverse of B. Right? So it is a neighbourhood of B. If you take G composite that because H takes U to B and then there is a G. Can we extend it over to B? Okay. To a map such let us call it B hat, G hat from B to Sn minus 1. Okay. It follows that if you go back again, okay, come to H inverse and then take G hat, that will be an extension of G over HB. Okay. So there is a typo here. This will be G composite H. When you take H inverse on this side again, it will be G. So this extension will be extension of G. Okay. Why it works on B? Because we started X as a boundary point of B. Therefore, we use the criteria here of 6.8. It can be extended. So go back so that problem is solved or extension is solved in HB. Therefore, Hx must be the boundary. Okay. So this means started with H of the boundary. Right? It is contained in the boundary of HB. But now the argument is symmetric H from B to HB. But H inverse is from HB to B. Therefore, I can interchange the B and HB. So I get equality. Okay. So this complete is a proof of power simulation of domain. Alright? So it remains to prove the lemma which is purely counter material technique. So we have to work harder here. We have to do what? We have to do simple shell complexes, approximations and various things here. Okay. So I restate this lemma here. K is a finite temperature complex of dimension less than m. A is a close subset of mod k. A map is given from mod k, a to dm s minus 1. We have to show that it is a homotopy mod k cross i to dm of f relative to a such that the net result named H of x1 is a map inside sm minus 1. The whole of mod k now taking values inside sm minus 1. Okay. Instead of dm. So that is what we have to show. Such a result what I want to say is you might have seen in a differential topology course for simply for smooth approximations. There are such results for smooth approximations. Okay. The proof of that and this are not very different. If you have understood that one this will be easier. If you understand this one you go back that will be easier for you. Alright? So I am not assuming that you have read that one. But I say that those things are similar. First we observe that it is enough to prove f is homotopic to a map g relative to a such that the image of g does not contain an interior point q of boundary of q of the disk dm. If one point is missed here from that point I can take the radial projection which is the deformation retract. Okay. On to the boundary dm minus q to sm minus 1. Take this composite with this one. You get a homotopy of the original map to a map into sm minus 1. Okay. So instead of directly showing that there is a map sm minus 1 with homotopic to the given f we take it to two stages. First all that we have to do is f is homotopic to a map g because everything must be controlled homotopy. A should not get disturbed relative homotopy such that the entire image of the new map misses at least one point. Okay. Specifically one point in the interior of the map not on the boundary. All right. Now just to write down the proof so that it becomes writing down the proof becomes easy instead of the round disk dm I take a homomorphism of this pair with the product space j power m into the boundary of j power m. What is j? j is the interval minus 1 to plus 1. You could have taken i power m also no problem but I am taking this one. This is also homomorphic to the round disk. This we know already and the boundary of this one will be under that homomorphism goes to boundary as m1. So take a homomorphism here. Okay. Then you make the whole thing instead of dm prove it for this jm. That is all. So now you have to show that the map g is already taking f is taking values here. Homotopic to a map g such that the g of mod k misses a point in the interior of jm. Okay. Now here is the picture. What I am going to do? Okay. So let me first establish this notation and come back to the picture later on. Cut the cube jm by hyper planes parallel to the coordinate hyper planes at intervals of length 1 by 4, 1 by 4, 1 by 4 and so on. Okay. So into minus 1 to plus 1 length is 2. So you will be cutting it into eight equal parts like this. Okay. Eight equal parts in all the directions, in all the planes that is the meaning of cutting it into this one. In r2, this picture is in r2. You have only two directions this horizontal and vertical. So take all the planes, okay, parallel to the, parallel to this coordinate planes in all the directions, in all the 1, 2, 3 of axial directions and cut it into like this square, smaller square, smaller cubes. Okay. At, at equal interval. This is not very stricter but you have to cut it like this. Okay. So each will be a cube. Okay. I cross, it is like I cross, I cross, I m times. Right. Now you can just cut it into further triangles and so on and make them a simplicial complex. This way they are not simplicial complex but we know that any del dn can be given a triangulation. So you take a consistent triangulation for the whole of this one. Okay. Any triangulation that you like. So that the modulus of that k is isomorphic to this picture. All these lines that are immaterial for me. So I have not drawn them here. Only I should know that this is a simplicial complex. That is all. Underline simplicial complex just plays for the logical part here. Nothing more than that. The picture, the geometry is precisely this much. Okay. So now what are these I am telling you. So that triangulation, triangulated thing that simplicial complex I am denoted by L. Mod L will be equal to jm. Such that each of these little cubes becomes sub complex. Okay. The smallest, the eight of them in each direction. So there will be, there will be, there will be sub complexes. Choose a subdivision k prime of k. Now k is what k is in the domain. Map is from k to, to begin with k to dm. But now then we converted it to k to jm. And now this k I am going to subdivide. Maybe several times by barycentric subdivision. So that you get a simplicial approximation alpha from k prime to L to the function f. Okay. This we know that we can do this one. Now put this little lj equal to minus j by 4 to j by 4 power m for j equal to 1, 2, 3 up to 4. So now you have come back to this picture. L1 is 1 minus 1 by 4 to plus 1 by 4. L2 is minus 2 by 4 is 1 by 2 minus 1 by 2 to plus 1 by 2. L3 is minus 3 by 2 to plus 3 by 2. L4 is the whole thing. So this is what the picture is. L1, L2, L3, L4. L4 is the entire L. Okay. So these will be automatically sub complexes of L one contained in the other. All right. So this is what will help us using the controlled homotopy here. Okay. So these LIs are defined like this. j equal to 1 up to 4. Okay. Then kj is nothing but f inverse of these. I can write it as mod lj if you like. But lj you can think of this as subspace also. f inverse of lj. Okay. lj can be thought of as a simple complex. Underline slash lj, like we have taken l is j m. So don't have to rewrite every time mod is fair. So kj is f inverse of the underlying topological space. f being continuous, these are close subsets. kj should be close subsets. LI contained inside lj, l1 contained, l2 etc. will give you k is contained inside interior of k2 contained inside k2 contained inside interior of k3 and so on. The k4 will be the entire dn. Okay. k4 is entire k whatever to begin with. Whole k may not necessarily dn. Sorry. Let eta from k to i, this is not dn. This is not correct. Eta from mod k to i be a continuous map such that eta is 0 outside the interior of k3. There is a close set and identically one on k2. So this is where I am just using that mod k is normal which we have proved. Okay. On l2 it is identity and outside this l3 it is 0. So here is identity on this part. From here to here it takes the value between 0 and 1. Okay. So such a function continuous function can be cooked out because it is normal. So eta is 0 outside k3 and identically one on k2. Now instead of taking t times f into 1 minus t times mod alpha as usual I take a modified homotopy here. This homotopy is again not very strange because fx and mod alpha fx are there and this is going to be a convex combination. Between f and its simple shell approximation you know that they lie inside the same simplex therefore the convex combination always makes sense. Look at here 1 minus t times eta x is there here and I am adding t times eta x. Some total is 1 and eta x takes values between 0 and 1. t also takes value between 0 and 1. So this is also taking value between 0 and 1. Therefore this is just a convex combination makes sense and is continuous etc. So this entire thing will be inside l. Okay. So g of xt is a homotopy. What happens when t is 0 and t is 0 this is just 1 and t is 0 this goes away. This is fx. This is starting function is f. What happens when t equal to 1 it is this map g 1 minus eta times f plus eta times mod alpha. Okay. So this is our function g whatever it is. We want to say that this function misses one of the interior points of l1 in its image. Then we are done. All right. Now let us analyze how it is done. This is purely point sector policy. Outside k3 what happens to eta? eta is 0. It is this part 0. This is 0. So it is just fx. Okay. But a is contained inside f inverse of boundary of jm. Remember that. Okay. a is contained inside f inverse of boundary of jm or f of a is going inside the boundary to begin with. The function is from mod k comma a to jm comma boundary of jm. Right. Therefore this a is contained inside outside k3 I mean complement of k3. All right. What is k3? k3 is the inverse image of l3. l3 contains the boundary of l after all. Okay. So look at this picture. So l3, l3 is this part. Okay. So if something is outside this one, so it must be coming from here. But this whole of boundary is there. Right. So that is all. I am asking this set theoretic conclusion here. So we are here since a is outside k3 this whatever we have derived this applies here. Therefore g of xt will be x for all x inside a which means g is a homotopy of f with the map relative to a. So homotopy we have seen so relative to a is important. Okay. So what we are left out now we have to show that this g misses a point in the interior of l1. Okay. So we claim that g does not contain some point in the interior of l1. First of all since alpha is a simplicial map a simplicial complex k prime which is a subdivision of whatever we have taken of k which is again dimension n is again a dimension n. Okay. It follows that the image of alpha okay is contained in the nth skeleton of l. No simplex here will be of dimension more than n because the same thing applies to our k prime. So the image cannot be of more dimension that is a simple thing about simplicial maps because simplicial maps are defined on points right. So nth skeleton of l in particular it follows that when you pass to mod alpha mod alpha of k2. k2 is a sub complex of subset of k prime the whole of k prime this itself will be contained inside ln. ln is the l skeleton okay. But then what happens is g is identically alpha on k2. So go back here on k2 eta is 1 right. So this is 1 minus t times fx plus t times eta is 1 t times mod alpha x okay. And on k2 this is what is this one so what I have to what I am talking yeah. So all that I want is g is mod alpha on k2 okay. On k2 eta is identically 1 right. So look at this one eta is identically 1 so this is 1 minus f this eta is 1 right this is 0 this alpha eta is 1. So g is actually mod alpha okay on k2 right mod alpha of k2 is inside ln which just means g of k2 is inside. Finally we wanted to conclude something about g. So this is one part so I have marked it as 17. Next part is suppose x is not in k2 x is in k2 we have taken here. Then a intersection k2 is empty this we have seen because a is inside k3 k4 outside k3 and k4 outside k3 inside so instead of a is inside a is inside k2 is empty fx and alpha fx are contained in a simple x modulus of they say sigma simple as simple of l which does not intersect l1 at all. So I want you to say whatever x is there is a picture whatever x is if it is x is not inside this f of x is not inside this one we mean fx is either here or here as soon as fx is here mod alpha fx will be also inside the corresponding simplex. Maybe simplex is here or simplex is here no it is in one side simplex which is outside this part that is all I want both alpha both alpha and mod alpha x and fx will be in this part okay in one simplex here it may be one simplex here may be one simplex here may be something simple here and so on okay one of them in one of the simplex is in particularly it will be one of these cubes all right that is all I wanted to know so geometry is very much used here okay so what we have we have what here is that fx and alpha fx are contained in a single simplex of l which does not intersect l1 at all no problem okay therefore the line segment fx and alpha fx the line segment is entire thing does not intersect l1 okay it is contained inside some simplex which does not intersect this whole simplex is inside this whole line is inside this simplex and this does not intersect l1 therefore this entire thing does not intersect l1 which just means that gx is one of the points here finally because g is the end point of this right the gx belongs to this line segment it follows that gx is not in l1 as soon as x is not in k2 gx is not in l1 therefore g of mod k minus k2 this is inside l minus l1 okay so this is purely set theory is it clear now put 17 and 18 together what do I get g of k either point is in k2 or is not in k2 there are two things so it will contain is a ln or in l minus l1 okay ln will not cover all of l1 anyway this part never covers anything from l1 so not all of l1 will be contained inside this one therefore l1 is not contained inside gk because l1 is contained as gk would have l1 is contained here l1 is not here because this is complement l1 is the full simplex okay of dimension n more than n cannot be inside the inside a n skeleton all right therefore l1 is not in gk it just means that the image of gk misses at least some point in l1 and l1 is a all of l1 are in the interior of model so this completes the proof of 6.8 hence theorem 6.7 is also completed okay so the the proofs here are much more educative than the the statement of theorems if you read read carefully so you should learn many many subtle points here okay so thank you