 So, now we are coming to simplicial approximation, given two simplicial complexes K and L, look at all the functions which are simplicial maps from K to L. Naturally, you can think of take the mod phi and think of this as a continuous function from mod K to mod L, that will be a very special kind of continuous function which is much smaller than set of all continuous functions. Question is how big it is? Indeed, what we are going to see is it is not too small, every continuous function can be approximated by a simplicial map that is a simplicial approximation there. It is not all the straightforward like smooth approximations here. The point is that if you keep K and L fixed, you are not going to get simplicial approximation. What you will have to do is you will have to keep dividing the domain according to the function that you want to approximate. And this dividing is precisely what we have introduced last time, subdivisions especially we are going to use barycentric subdivision and we will solve this problem only for K finite. Take K1 and K2 big simplicial complexes and mod F from mod K1 to mod K2 a continuous map. We say phi, a simplicial map, a simplicial approximation to F, the F is there. If the following condition holds namely suppose you have some F alpha inside the closed simplex F2. F2 is a simplex in K2, F alpha happens to be inside the closed simplex K2. Then the corresponding image of alpha under mod phi should be also inside F2 for all alpha in K1, mod K1 and F2 in K2 whenever this happens. So in other words the image of mod phi and image and the mod phi of every point should not go away too far. If it is in a simplex, it should be inside that simplex. So this is the way we want to do the thing. Here the farness and nearness is measured out of simplices inside K2. You do not want any distance and so on here. So here is a picture. Some for some alpha, F alpha has come here in the triangle, inside this triangle. Now phi alpha is also in the triangle, it is on the boundary of the triangle. So this is okay. Now here look at F beta is inside this one simplex in the edge whereas phi beta is inside the larger triangle. So this is where the condition that we started has failed because what should have happened is phi beta should be also inside this F, inside this one simplex. So you have to choose the condition correctly. You can say F beta is in this triangle, so phi beta is here. So that is true but if you choose F beta to be inside one simplex then this condition is not satisfied. So that is the meaning of this condition. For every F2 which contains in mod F which contains F alpha you should have this property. So this F2 and alpha, alpha is taken whatever F2 is also floated here. So all F2 which has this property, whenever this happens it should be true. This is the way. So I hope this is clear, this point is clear. If L1 is a sub-complex of K1 and phi and Fs are all as above, one is a simplex approximation of the other one. That is what I am not debating these conditions here. And if phi is a simplex approximation to F then phi restricted to L1 will be automatically a simplex approximation to F restricted to mod L1 because the conditions are satisfied for the larger simplex complex K1 it will be satisfied for restricted thing also. The beauty is suppose F is already induced by a simplex map on K1 to K2 or from L1 to K1 whatever some sub-complex. Then a simplex approximation will be equal to the map. Equal to this if phi is giving you F, so if phi is giving you F, phi must be equal to F. So this is the meaning. Whenever the map is already simplex here, a simplex approximation to that will be itself. This is quite different from other kind of approximation that we are doing. So how does this one follow? This follows by simple observation that instead of taking arbitrary point alpha, I will take a vertex V1. Then F of V is a vertex because F is simplex. So I can take this vertex F of V to be capital F. Then phi V must be also inside that thing that I will take. So it must be F V. The importance of simplex approximation stands from what happens in the homotopy aspect. So that is what this lemma says. Suppose F is a continuous map and V is a simplex approximation. On a subset, A contains K1. Suppose mod phi is equal to F A. They agree there. This A may be empty. I do not care. Suppose this happens, then mod phi is homotopic to F relative to this set A. F A must be equal to mod phi A if you want relative homotopy first of all. So that is necessary. That is it. Now how does this follow? Because of the condition that F A and mod phi A, F of x or F of alpha, mod phi of alpha always are inside a single closed simplex. A closed simplex has on a structure. Therefore you can join these two them by a line segment and that gives you homotopy. So that is precisely what I am going to give. A H of alpha t phi t times F alpha plus 1 minus t times mod phi of alpha. So this is obviously homotopy. On each simplex containing F alpha, this makes sense. If you take a boundary of that by chance, then if you pass to the other simplex, the two structures will be the same. The line segments are the same. So this defines a very good function first of all and it is linear in each of these. So it is continuous. So all these things now I am sure that you can verify by yourself. So that is why I have written just this homotopy and this will be homotopy. t equal to 0 what happens? It is the mod phi alpha. t equal to 1 it is F. So it is homotopy. So now we go towards how to get a simplex approximation starting with a continuous function. So a little more elaborate analysis of the definition that we have given is involved in this lemma and that will lead you to get the function, the simplex function that we are looking for. So first of all a simplexial approximation whatever should be a vertex map and then it must be a simplexial map and then it must be an approximation that the approximation condition should be satisfied. So start with the rustic function first of all where k1 and k2 are simplexial complexes. Suppose f from mod k1 to mod k2 a continuous function and this phi which is a satiristic function on the vertex set satisfies certain properties. So these a, b, c are all equivalent. So what is this first condition? For every v in k1 namely a vertex f of star of v should be contained inside star of phi. I recall what is the star of a vertex. Star of a vertex is all those alphas such that alpha of v is not equal to 0. It is called star of v. It is an open star of v. It has star shape itself set at the vertex. It is star shape. So f of this open set f of star of v should be in such star of phi v. Phi v is another vertex which is a vertex inside k2. So star of that should contain this one. It should not work of that. For every alpha in mod k1 the support of alpha, support of alpha is a simplex. Phi of that is a simplex. I do not know right because phi was only a vertex map. Phi of support of f must be not only a simplex but it must be contained inside a support of f alpha. Support of any point in mod k1, mod k2 here is a simplex of f2, simplex of k2. Once it is contained inside a simplex this will be also a simplex which will be contained inside that. It may be equal also no problem. The third condition is suddenly phi is a simplexial approximation to f. So these two first two conditions are equivalent to the definition. The definition does not even assume that phi is simplexial. If phi is simplexial it assumes. Here I do not have even simplexial. Simplexialness I am putting here. Only I am putting a vertex map. After all to define a function on the simplexial complex you have to first get the set theoretic function on the vertex sets. So that is only set theory. After that this gives you immediately what that it is a simplexial approximation either condition a or condition b. And they are equivalent. That is the content of this lemma. Let us go through this proof carefully because that understanding that one is important here not only just the statement but how it comes out. That will reveal you what simplexial approximations are about. So let us prove a implies b. The first condition a only says that f of star of v is contained as star of pv. Then I have to show that phi of support of alpha is contained inside support of f alpha. So I start with a point inside support of alpha. I have to show that phi v is inside here right. Support of v is in support of alpha implies this is by definition. It is actually equivalent. We have used that one also. Alpha v is positive which is same thing as again equivalent to alpha is in star v. So this is like duality. v is in support of alpha means alpha is in star v and conversely of course until here it is fine. Now this simple as f alpha is in star of pv by condition a right f of star v is in this one. So if alpha is here alpha f of alpha will be in star of pv. That is condition a way. Now we can go back something is in star of pv means f alpha operating on a pv evaluated pv is positive. But the same thing as phi v is in support of f alpha. So we have completed proof of b. a implies b is over. This step can be exactly reversed to get b implies a. But only you have to use the different hypothesis. The hypothesis is there become conclusion now. So I am rewriting it b implies a start with alpha inside star v. We have to show that f alpha is in star of pv. That is what we have to show. Using condition condition that if something happens that that condition is there now I mean b. So alpha is in star v. This alpha v is positive just like here v is in support of alpha. But by condition b phi v will be in the support of f alpha. Here it was condition a now here we are reversing. Phi v is condition f alpha right. But that is the meaning that f alpha of phi v is positive. The same thing as f alpha is in star of pv. So it is exactly reversing the arrows here. a implies b. a and b are equivalent. Now we want to prove that a and c are equivalent. If we assume a you can assume b also because a and b are the same now. If we assume b then it is same thing as assuming a. So you can take both advantage of both of them and that is what is going to happen so on. So first I will prove c implies a. c gives you that phi is an implicit approximation. So fix a v belonging to k1 and alpha is in star of v. We want to prove a. So what we have to prove? f alpha is in star of pv whatever. We have to show that f alpha of pv is not equal to 0. That will mean that it is in the star of pv. So this is what we have to prove starting with alpha. So let f2 equal to support of f alpha. That means at each point of f2 f alpha is not 0. That is the meaning of support of f alpha. Clearly f alpha is in mod f2. It is actually in the interior. So it is definitely in mod f2 by hypothesis. Simplicial approximation hypothesis now implies that mod phi of alpha is also in f2. But now look at what is the definition of mod phi. Mod phi of alpha at phi v or wasm v prime is sum of all the alpha v's where v comes to this v prime. Phi v equal to that. So in particular this is phi v already. So alpha v will be definitely there plus some more terms which are all coming to phi v under phi. All those alpha of those points are also there. Doesn't matter. This alpha v is there and this alpha v plus something is there. Some non negative terms. So which is bigger than alpha v, bigger than equal to alpha v and alpha v itself is always positive because alpha is in star v. Therefore this phi v is inside f2. By definition for all v whatever you have taken, we have shown that phi v this phi of phi alpha phi alpha phi v is positive. Therefore this phi v must be inside f2. But f2 is the support of f alpha. So this implies f alpha of phi v itself is not zero. So what may happen is you have chosen f2 like this so that f alpha is in the interior. But phi alpha might have gone here. Doesn't matter. It will not get out of this simplex. This is the point because alpha is not equal to zero. Now finally A in price C. So as I have told you, if we assume A, the hypothesis B also comes to your rescue. So you can use whichever one you like. So first to show that phi is a simplex map. Right now what is given? Phi is only a vertex map satisfying A. Now I have to first show that it is a simplex map. What is the meaning of simplex map? Take a simplex. Image must be a simplex in k2. That is what you have to know. Take a simplex v0, v1, vq equal to f1 in k1. Then the bary center f1 to it which is v0 plus v1 plus vq divided by q plus 1. That is an element of mod f1. It is actually in the interior of mod f1. Not only that, it actually belongs to star of vis for each of them because the coefficient, the value of h, vji is a 1 by n plus 1, 1 by q plus 1 for each of them. So this element f1 to it is operating upon a v i is not 0. So it is in star of v for each i. Therefore, this intersection of star of vis is non-empty. A beta twiddle is there. Sorry, f1 twiddle is there. So f of that one will be also non-empty. If we have non-empty set, f of that one will be definitely non-empty. But f of an intersection is contained in the intersections of the f i's, f of star v i's. And each of star v i's is contained inside star of phi v i's. That is our condition a. Therefore, this intersection is non-empty. This intersection is non-empty means what there is a beta, some element which belongs to intersection, which means star of phi v i, sorry, phi v i evaluated at beta. Beta evaluated phi v i is not 0 for each of them. It is intersection. That would mean that phi 1, phi 2, this phi v i, phi v i, phi v i, etc. must be a simplex because every point beta of k2 mod k2 is inside something simplex, close simplex. Therefore, phi v i is in support of beta. These are all in support of beta. Support of any point inside mod k2 is a simplex. So phi of f1 is a simplex in k2. All these phi v i's which are inside this one is a subset of that. Phi v1, phi v2, phi v0, phi v0, vq, they are phi of f1. They are subset of this one. So it is a simplex. So you have proved that phi is a simplex map. Now it must be shown that it is a simplex approximation to f. So that part is here to show that it is simply an approximation start with any alpha in mod k1 such that f alpha is in f2, mod f2. We have to show that what p of alpha, mod phi of alpha is also in f2. This is what you have to show. First of all, f alpha is in f2 means support of f alpha is a subset of f2. Since we have to prove a implies b, we can use condition b also. Sorry, we have proved a already. We can use condition b also from which it follows that support of mod phi alpha which is equal to phi of support of alpha is contained in support of f alpha. This is the condition b. Just now support of f alpha is contained in f2 we have seen. So support of mod phi inside f2. It just means that p alpha is inside mod f2. So now we can harvest this criteria which we have got. You can use that to get another criteria for simplex approximation. Take any map k1 to k2 which is a continuous function mod k1 to mod k2. Admits simplex approximations. You do not know what they are. If and only if this purely condition in terms of f if and only if k1 is finer than this open covering star v varies over where v varies over all the vertices. This is an open covering for mod k2. Take f inverse of that that will be an open covering for mod k1. So that is my u. Members of your f inverse of star v. That covering have the property that mod this k1 is finer than this just means that for each star v is contained in one of the members here. Remember the finer means that. So what is the proof? Let v i denote the vertex set of k i v1 and v2 or vertices of k1 and k2. k1 is finer than u just means that each star v, let us say star u, u is inside v1 is contained inside f inverse of star v for some v. That is the meaning of finer. So what means that? There is a definement function phi from v1 to v2 such that f of star u is contained as star of phi u. What is it? I am writing that as definement function. The same thing is saying if you take f inverse star u is contained as f inverse of star phi u. Whatever v I get I am calling it as phi u. So that is the definement function. This function is a saturated function now. You see the condition in the theorem in this lemma there is no reference to phi. Now I have cooked up the function phi here namely it is a refinement function from one vertex set to another vertex set. So that is my phi. It is a vertex map. But this has another now use the previous lemma. This will tell you that previous lemma phi is a simplicial approximation to f. If this condition is stratified, phi simply has a refinement function itself becomes a simplicial approximation. There may be many. All of them will be simplicial approximation to f. Now we know how to get a simplicial approximation to a given function. What we must do is the k1 that you choose here which is going to which is going to you know give k1 is homomorphic to this space whatever. That must be finer than the open covering here. So once you have chosen k1 how to make this one we know. What namely go on taking barycentric subdivision divide divide divide till it becomes very fine that is the meaning of finer. So it must be finer than the covering that is the last thing. If k1 and k2 are simplicial complexes f from mod k1 to k2 is a continuous map k1 is finite k2 could be anything. Then there exists an integer n such that for all and bigger than n there are simplicial approximations f from where from not from k1 from the iterated barycentric subdivision of k1 to k2. So these maps are defined not on k1 but on a finer one in the subdivision and there will be approximation to f. Once your n is sufficiently large namely we know how to choose this capital N. q divided by 1 plus q sorry q divided by q plus 1 raised to n should go to keeps going to 0. So it should be less than your epsilon for epsilon is what whatever the labeling number and so on. So that part we have seen last time. Remember that so that is the meaning of combinail m of 6.4 and 6.2. 6.2 about finer subdivision 6.4 says once it is finer like this then you are done. So simplicial approximation theorem is proved. This has similar applications similar to what? Similar to smooth approximations in differential topology. You must be very calculus that you may have learned. But it obviously has wider applicability because here we are not taking manifolds at all our things could be highly you know corner things may be there non smooth. Okay triangle you know tetrahedrons and so on. Okay and things built up of that very crooked path is and so on could be there. Okay but there are other types of even larger things namely CW complexes and they have also this kind of CW approximations. In some sense they are they have a wider applicability than simplicial approximations. Okay but simplicial approximations have the great advantage you know if you k1 and k2 are some kind of simplicial complexes they have to be given and if you specify a few properties of continuous function F even the computer can find a simplicial approximation to it. Just some a few more you have to give some effort what is F we have to give some information about how what is happening to F some geometry approximation it will do that. This kind of thing is not possible CW approximations and so on. Okay so there are a lot of advantages also as compared to other simply other approximations. So some immediate corollaries some of these things can be done by smooth approximations also so it is not that these are the advantages no. So this corollaries immediate namely take any function from one sphere to another sphere of higher dimension SN to SN plus k where k is at least one any such map is null homotopy. The proof is very simple you have to think of SN given by you know the underlying topological space or geometric realization of the boundary of delta n plus 1. This we have seen already take k1 equal to boundary of delta n plus 1 mod k1 will be SN. Similarly mod k2 will be boundary of delta n plus k plus 1 one dimension higher you have to take each time. Okay then your F is a map from mod k1 to mod k2. Therefore after dividing k1 sufficiently many times I do not want to write it what it is my S t of n some for some we said nothing to do with that n. Okay so it is k1 prime is some subdivision you have a simple shell map which approximates F. Okay we know that F is homotopic to mod phi. Mod phi is again from mod k1 to mod k2. So F is also mod k1 to mod k2 right. These two are homotopic to each other that is what we have seen that is the importance of simplicity approximation. Okay so to study the homotopic behavior of F I can pass on to now mod phi because these two are homotopic. I want to show that F is homotopic to constant map. I should show that mod phi is homotopic to constant map but mod phi has better properties namely because simple shell map its image is contained inside maximum nth skeleton. Okay image of any n simplex here that means the maximal simplex is here cannot be bigger than dimension n. It will be n simplex or some smaller dimension simplex because that is a linear map. Okay this part we have seen so it is contained in the nth skeleton of k2. But k2 you know is n plus k dimensional thing there are n plus k dimensional cells also. So all those things are not covered which just means that mod phi is not a surjective function. Now this is one of the simple exercise I had given you have to better work it out namely any function into a sphere which is not surjective that means this is one point just one point is enough. Okay one point is missed then it is homotopic to a constant function. Okay you have seen this proof must be you have seen this proof by now. So simple shell approximation theorem is valid over arbitrary simple shell complex is also instead of just finite we have as you domain is k1 is finite right that is not a necessity only this proof it was needed. Okay but to prove it in general case you cannot just use bare syntax subdivision you have to you have to go for arbitrary subdivisions also. Okay so let us stop here next time we will derive another common to result out of this simple shell approximation. Thank you.