 In the study of manifolds, we introduced the notion of a topological manifold, gave some examples and then introduced the notion of manifolds with boundary also and we have also shown that every manifold is para-compact including the manifolds in boundary. We also showed that the boundaries of a bounded manifold have what are called as caller neighborhoods. So today we shall study some homotopic aspects and one of the key to that is that every topological manifold is a subspace of some large Euclidean space that is the meaning of embedding. So every manifold can be embedded inside a Euclidean space. So as such it seems that we do not have to go out of R n, we could have taken only subspaces, topological subspace of R n with the locally Euclidean structure like circle or a sphere and so on ok as contrast with the union of two lines and so on that is all. However, in practice what happens is manifolds are not occurring or not displayed as a subspace of R n, some R n they come in different forms and then it is a burden to see them first of all that they are subspace of R n. So the abstract definition has this advantage ok. So let us anyway do this embedding which will itself help in other aspects of the manifolds. The following result tells us that after all we could have just stuck to the study of subsets of Euclidean spaces for studying manifolds. As we shall see this single result has several implications on topological, homotopical and homological properties of a manifold. Being subspace of R n is itself something special. Every n manifold is homomorphic to a closed subset of R 2 n plus 1, very specific about it. Start with n you do not have to go for a very large n but R 2 n plus 1 will do. So this is quite tight, there are examples wherein you may not be able to do it in R 2 n or even lower than that and so on. So we should stick to that ok. For every topological manifold X the set of embeddings of X, if you looked at all the embeddings of X in just I 2 n plus 1 is dense in the space of I 2 n plus 1 power X. What is this one? These are all continuous functions from X 2 I 2 n plus 1 out of which you take only the embeddings. So that would be dense. So what is the topology here? This is the compact upon topology of all continuous maps. This is from X 2 I 2 n plus 1 of all continuous maps with topology of uniform convergence and compact subsets which is the same thing as the topology compact upon topology ok. If you want closed embeddings then you have to take X to be a compact subsets because if I 2 n plus 1 is compact, if a closed subset is a compact ok. So this is the result. What I have chosen is not to prove this result. The proof is quite lengthy and complicated and not very eliminating ok. Therefore we shall only take a very mild form of this theorem only for compact spaces and we shall be liberal with this I 2 n plus 1. We shall not bother about I 2 n, some large n ok. So that is what we are going to do. However here is a reference and so on. The proof of these theorems are somewhat lengthy and hard. The smooth version of 6.3 you know if you have a smooth manifold you can embed smooth and so on. They are under the name easy with the embedding theorems which are likely easier proofs also which you may read from many books such as my own book on differential topology. However for the topological case there are not many references available. You are welcome to see this in an excellent old book of old book by Hurwitz-Bahar and I have given the reference here. This is a wonderful book or you may choose to read a nice proof of embedding 6.3 from Munchres there Munchres book ok. There are two Munchres book I have referred. We shall be satisfied with an easy proof of the following weaker version namely every compact manifold with or without boundary that we have grown further is homomorphic to a closed subset of some Euclidean space. Some Euclidean space ok. So we are not bothered about how large this n has to be taken and so on ok. And this this result this part is very easy. Let us see how X has a chart ok and at last consisting of UIs and homomorphism phi i f i from Ui to where A is either Rn or Hn ok. Since X is compact there will be a finite cover. So I can assume Ui is 1 2 up to k ok. On each Ui I have a homomorphism into Rn or Hn depending upon the point is and the boundary and so on ok. Now pick up a inverse for each of them pick up the inverse eta from Rn to Sn sorry eta from Rn to Sn is the inverse of the stereography projection ok Sn minus a point to Rn you have you have you have what is called as homomorphism which is called stereography projection you have. This eta is the other way around the inverse of the stereography projection and let G i from X to Sn be the extension of eta composite phi which sends entire of X minus Ui to the north pole. See Ui is a homomorphism f i is a homomorphism from Ui to let us say for the definite test case Rn ok first ok. Then outside Ui where R to send them send it to the north pole. So for that you have to take G i as the the composite of f eta composite f i. So f i is from Ui to a eta is from a to Sn ok. It already fills the whole space except whole of Sn except the single point. So the rest of U you send it to that point that is a continuous extension automatically ok. So these are G i's are continuous function to which send X minus Ui to north pole. So put G equal to so I have got some continuous functions. What is the what is the property of this continuous function? Restricted to each Ui they are 1 1 mappings they are embeddings ok. But outside your X minus Ui whole of X minus Ui is going to a single point it is very bad there ok. But look at G equal to G1 cross G2 cross Gk from X to Sn cross Sn cross Sn and k times which is a subspace of anyway Rn plus 1 raise to k. So this sphere is coming in a nice way for us to help we will see that. So finally I have got it a map here. I want to say that this will be an embedding now. All that I have to see that it is already a continuous function because each G i is continuous and I have taken just an inclusion map here ok. All that I have to verify Gc is for 1 1. If it is 1 1 mapping ok then X is compact this is AUSDAR automatically it will be a 1 1 and it is a homomorphism on to the image. The proof will be over ok. Verify that G is a 1 1 mapping ok. This I have left to you but now I will do that in a short minute. So why this is a 1 1 map tell me. So how does this fail to be a 1 1 map? How does this fail to be a 1 1 map? So there must be a point X comma Y such that G1 of ok suppose X comma Y such that G1 cross G2 cross Gk of X equal to G1 cross G2 cross Gk of Y right. If X and Y happen to be in one of this Ui ok then G i of X will be different from G i of Y if X and Y are different because there it is injective. Therefore they are they cannot be in the same Ui. So suppose X is in Ui and Y is in it is not in Ui. As soon as Y is not in Ui what happens to say what happens to this one this definition extension of this one G i of that point will be the north pole. But what happens to X i which is inside Ui it is it is eta composite F i. So it is something which is not the north pole. So the image is different ok. So now we will use this property we will use this theorem itself that every manifold is a subspace of and can be embedded in a Euclidean space ok. Though we prove it only for compact spaces we shall use this one for all manifolds. So in order to derive some homotopical and homological properties of manifold the first thing we prove is the following theorem and perhaps this is the this is the only theorem it is the only result that we can prove today ok. So have patience because this is slightly longest. Let X be a locally contractable closed subset of Rn a closed subset of some Rn and U be an open subset of Rn such that X is inside U. That means this U is a neighbor of our set. What is the assumption on X? It is a closed subset and it is a locally contractable subset. Then there is an open subset V such that X is inside V contained inside U and there is a retraction V to X. So this is what is called as you know local retracts. In fact you can say that it is a deformation retract and so on that is what leads to our co-fibration center. I am not trying to prove such a strong result here. Only retraction every every subset which is locally contractable is a retract of a neighbor road that neighborhood can be chosen as small as you please. This is the meaning of this theorem ok. So let us see how we are going to do that. So recall our notation and lemma for what are these lattice, we had this lattice structure. Remember that which gave you Cw complexes and so on. While proving Cw approximation theorem we have used that one. So let Lk denote the lattice points that of all be X belong to Rn such that Xi is some integer Ri divided by 2 power k. The denominators are only 2 power k ok. Of course some 2 may cancel out it may become smaller than that. So this k is a fixed k is greater than equal to 0 is a positive quantity. So it is these are numbers ok. These Xi's are n of capital N coordinates. I ran into 1, 2, 3, up to n ok. For all of them it must be all the coordinates must be of this form ok. Rational numbers of the form Ri's are integers the denominators are only 2 power k. Put L equal to union of all this Lk's. So L1 for example is all the lines planes you know drawn at length at the interval 1 at all the integer points. L2 will be at half integer points. L3 will be at one fourth integer points and so on ok. So that is the mean you have sharper and sharper lattices are there. At pkd in order to set of all closed n cubes sigma of length side length 1 by 2 power k ok like 1 by 2 or 1 by 4 and 1 by 8 and so on ok and with corners inside one of the Lk's corresponding Lk. Lkk is fixed at this pk is those points ok a closed cube. Let p equal to union of pk. I am just recalling all these things which we have done earlier before. We restate a lemma namely 1.3 for ready use. What was the lemma? That lemma was that take a collection of subsets a collection of points from p the a is a subset of p such that sigma 1 and sigma 2 what are p? p's or cubes remember that ok. Sigma 1 and sigma 2 belong to a implies sigma 1 is not contained in a sigma 2 ok. So if we have one fourth cube and one eighth cube at one eighth cube should not be inside one fourth cube. It should be maybe adjacent to it and so on. Touching is alright. It cannot be contained inside ok. Then if that is the only property then this w a is the union of all the sigma sigma belong to a the underlying topology space is the support of all this. Take union of all the sigma belong to a is a locally finite countable pure n dimensional CW complex with all its n cells being members of it. So come things forthcoming all the n cells will be this one ok. Today we have seen that this is also a actually can be further cut down into a simple shell complex. A cube if you cut it into a simple shell complex then it will extend it to the whole thing that is all. Since you do not want to cut the cube into triangles and tetrahedrons and so on. So it is a CW complex ok. So that is the lemma that is ready meant. Now we directly go to the proof of the theorem. The theorem is what? I have to produce this v as well as r. v is the neighborhood contained inside u neighborhood of a r is a retraction ok. So we vote for that one. So let A1 be the union of all cubes in P0. P0 is what? Interior points right side length 1 now that in P0 are cubes of side length 1 P0 contained in u ok and which do not meet x. There may not be any I do not care. Let A1 be the union of all cubes in P0 contained in u and which do not meet x ok. Inductively for k greater than or equal to 1 let A k be the union of all those cubes in P1 by 2 power k minus 1. Actually P I should just write P k minus 1 in my notation P k minus 1 which do not meet x. So I repeat inductively for k greater than or equal to 1 let A k be the union of all those cubes in P k minus 1 ok which do not meet x and contained in u but r not contained in any member of A k minus 1. If you have already taken some A k minus 1 and things are inside that you should not take them. So for A1 there was no such problem ok which was starting point. So for A2 along with the conditions like A1 ok inductively I should if they have been something is larger box has been taken a smaller box inside that should not be taken ok. So that is the definition of A k minus 1 put A equal to union of all these A ks. Now let W equal to W A be defined as in the lemma by the very definition that but not contained in any member of A k minus 1. It will satisfy that this condition of this lemma namely sigma 1 sigma 2 implies sigma 1 is not contained in the lemma. No once my member is taken a smaller cube inside that will never be taken. So that is why this condition is automatically satisfied. Therefore we can apply the lemma W will be defined as in this one then ok rn minus x ok rn you have you have a substitute x right rn minus x will be the entire of W and W has a CW structure in which all these cubes of various sizes form n dimensional cells. So this part is lemma A by the very choice none of these ok none of these will be inside will intersect x. So they are contained in rn minus x but everything in rn minus x must be inside W because p1 itself covers it p1 p2 all this small small they will cover the whole thing anyway ok. Suppose you have some small set for x if you take large far away integer the whole cubes will be inside x inside W right. So rn minus x is actually W this W will be covered by all of that W has CW structure. The funny thing is we are not doing anything for x compliment of x very nice structure we have got. So W has CW structure all these cubes of various sizes all of them n dimensional cells smaller and smaller maybe smaller and smaller keep coming nearer nearer x but you know we will not intercept it ok. So that is what it is now let us see how we shall construct a sub complex V of W and a function r from x union V to x ok. So x union V will be finally a neighborhood ok such that rx is equal to x on x belongs to x we shall then show that it is x union V contains a neighborhood of x ok and r is a continuous r is continuous on this neighborhood which will complete the proof of the lemma. If it contains one neighborhood then given the open original V original U you intersect further with the U and restrict the whole thing. So that is the idea. So x union V V is now a sub complex portion of W x union V has this property it is what we have to show and then of course we have to show that it contains a neighborhood of x ok and also I will show that r is continuous on this neighborhood which will complete the lemma. So this is all visual thinking it takes some time to prove this one let us go ahead with this. First the constructions of V and r will be done simultaneously and inductively ok. Later we shall check the continuity of r so defined right now do not bother about the continuity of r. So we have it in mind that it will have to be continuous that is how we are trying to do that ok. Of course we start with rx equal to r because there is no choice because it has to be a retraction right on to on to x. So rx is x take the 0 skeleton of V skeleton of V I am I am going to define V define V most we will have to be a CW complex right to be W 0 W is a CW complex right. So take the 0 cells of W all these lattice for some of the lattice points will come ok that is all vertices are W 0 that is W 0 given sigma belonging to V 0. So V 0 is constructed now V 0 is W 0 choose any point r sigma. So sigma is a vertex ok but I am writing it as if it is a simplex it is just a V 0 it is a vertex but each point choose another point inside x such that it is distance between sigma and rx sigma and r sigma. See sigma is given r sigma I am choosing such that it is less than distance between x and sigma. Remember this sigma does not intersect x they are inside W. So there is a positive distance x is a closed subset there is a positive distance twice this distance ok you take that that must be bigger than distance between sigma and sigma and rx this rx must be a point of x all right. So there is nothing very great about this one. Here d denotes the distance between namely a infimum of the Euclidean distance dxy the distance between a point and a and a space means that ok. So the construction of V 0 and r on x union V 0 is over the 0 skeleton is over ok. Now we want to do the 1 skeleton, 2 skeleton and so on ok. Actually here you can easily see that it has to be continuous function because x on x is identity and V 0 which is discrete you could have taken any function. So but I will not discuss continuity right now ok let us go ahead. Having defined V k minus 1 and r from x union V k minus 1 to x ok. Let us take sigma to be any k cell in W such that boundary of sigma which is you know sigma is a box it is a very regular box. The boundary of this will also consist of various you know union of boxes actually of all over dimension. The boundary must be contained inside V k minus 1 ok. Let take sigma to be any k cell of W such that boundary of sigma is contained in V k minus 1. Suppose you for V 1 suppose 2 points are already there then look at one of the cells. So take such a cell one cell a line segment ok V 0 is just what is this. Now the line segments which belong to these lattice points only, lattice is only. It may be of length 1 by 2 or 1 by 3 or sorry 1 by 4 or 1 by 8 and so on whatever ok. So take such a sigma such a boundary of sigma is V k minus 1. If r from boundary of sigma to x has continuous extension see r is defined on boundary of sigma already right because boundary of sigma is in V k minus 1. If it has a continuous I do not know if it has continuous extension then let us put this sigma inside V k. Something is defined on boundary if it has continuous extension on the whole of sigma then that is an ideal thing you take that that that sigma inside V k not otherwise. If you cannot extend it do not take those cells ok. Having put this sigma inside V k take r from sigma to x to be any continuous extension now which satisfies the property among the all continuous extensions you must take one there may be several you must take one such that distance between sigma and r sigma is less than twice the distance between sigma and r prime sigma. This is the mini-mull take the one which is mini-mull in some sense what exactly I cannot do that so but I have doubled it and take one of them for all extensions r prime from sigma to x ok. I would like to have mini-mull but I do not know so but take double of that definitely this will be less than that there will be one because given any r sigma that r sigma it will satisfy this property so among the stack therefore there must be one ok which satisfies this one ok this completes the construction of V k and r from x union V k to x by induction V and r are the definition of V and r complete ok it remains to prove a number of things what we have to do this V x union V contains a neighborhood of x inside r n r k then I have to prove that this r is continuous on that neighborhood once you have done that the proof is over alright so let us do that not that by very choice r is continuous on V can you see that what we have done we introduced a first first V naught on V naught everything is continuous on V 1 we will took only those sigmas on which we could have extended the already defined sigma on the boundary to a continuous function then we took one of those continuous functions extension ok therefore the function r is continuous on each sigma that is enough for a function to be continuous on the on C w complex a function is continuous on C w complex if the intersection with restricted to each simplex is each cell is continuous so that is what we know already ok so we have to verify continuity the points of x you see only for x it is identity there it is continuous that does not mean that as a function of x union V on the oxygen V why that is continuous so many people make this mistake ok now it is already because I on exit identity restricted to exit is continuous but on the space x union V it must be continuous and that is the crux of a matter here ok so we must we have to verify a continuous at the points of x belong to x continuity must be verified on x union V remember that so given x belong to x and epsilon positive I have put epsilon 2 n equal to epsilon maybe I can do it with epsilon n we will check it will matter and go on choosing the or the later ones in the following order once you have chosen epsilon 2 n the next one will be chosen such that 3 times epsilon n i is less than epsilon n i plus 1 or this one is smaller than one-third of the earlier choices keep doing it till you hit epsilon naught ok smaller and smaller epsilon so that now if B i is x intersection B epsilon of x B epsilon i of x is a ball inside where in R n intersect with x that is B i then the inclusion map B i to B i plus 1 is null homotopy so how do you ensure this this is where local contractability of x is used this is the meaning of local contractability given any neighborhood must have smaller neighborhood ok such that the inclusion map of that neighborhood in the larger one must be null homotopy so that is the meaning of this local contract this is local contractability of x ok so I do not say that this epsilon i is exactly one-third of epsilon plus 1 no smaller than that ok so that is why you have to choose such a epsilon i let U t be B t of x where 0 is less than t less than epsilon naught by 4 after choosing the last epsilon naught I am going to a further divided by 4 maybe 3 is enough again and take t smaller than that that is my U t ok we first claim that if sigma is any cell of W you know how the cells of W are taken contained in this U t there may be very smaller smaller cells then sigma will be automatically in V and the distance between x and r sigma remember that ok r sigma was some point if V was single single point and so on this should be less than epsilon no this picture the bigger this epsilon looks very big now because we have come so so small here ok so this is this claim is what what makes it work like this so if all these sigmas are inside inside V and you know there are sigma is a cell of W contained in U t ok so that is already in V so all portions of U t intersection V there will be inside inside V so that is how we can be able to prove that V is open ok suppose A is a 0 cell and A belongs to U t so we have to prove this one so I start with 0 cell suppose A is a 0 cell and A belongs to U t then distance between x and r A is less than the distance between x and A plus distance between A and r A ok x and A is by definition U t so it is t but this is by definition twice distance between A and x that was our first choice ok so this is t and this is twice t so the whole thing is less than is 3 t ok and 3 t is less than epsilon 0 because t is less than epsilon by 4 so I said that is why this may be you can assume epsilon by 3 t less than that then it is ok does not matter ok so it is less than epsilon 0 now suppose for all k minus 1 cells tau of V contained in U t we have proved that distance between x and r tau is less than epsilon 2 k minus 2 ok epsilon 2 k minus 2 for k equal to 1 we have proved this one ok k minus 1 cell is 0 we have proved it now inductively suppose we have we have assumed this one by the way finally we want to prove epsilon but all these these are less than less than epsilon right so epsilon 0 was very small coming all the way down so once we have this is epsilon 0 then so we are proving inductively if we just prove epsilon here then we will not be able to prove the next step so you have to prove inductively that is less than epsilon 2 k minus 2 let sigma be a k cell of W contained in U t now then boundary of sigma will be contained inside B epsilon 2 k minus 2 ok because we have proved this one and the inclusion B epsilon 2 k minus 2 is contained in B epsilon 2 k minus 1 by choice because this is one third of less than one third of this one ok and the inclusion map we have Bi to Bi plus 1 is null homotopy that is important null homotopy since there exist extensions are from boundary of sigma to Bi ok inside Bi plus 1 sorry once it is a null homotopy a map if the map is null homotopy then you can always extend are from boundary of sigma to Bi ok so Bi to Bi plus 1 null homotopy so it can be can extend so inside Bi plus 1 it can extend from 37 what is 37 this choice ok follows that distance between sigma and r sigma will be twice 2 k plus 1 epsilon 2 k plus 1 this epsilon 2 k plus 1 ok so twice comes because of our choice is like that here ok therefore once again the same same thing as you were here d of x r sigma instead of r a this is less than d of x sigma plus d of sigma r sigma and this is always t but this is now twice epsilon 2 k plus this is also less than epsilon 2 k plus 1 some total is 3 times epsilon 2 k plus 1 which is less than the next one epsilon 2 k ok therefore by induction we have r of ut intersection v is contained inside u epsilon ok so each time you are going to express 2 k right so for each k we have done so all of them are less than epsilon anyway so this is contained inside u epsilon ok note that this however does not complete the proof of continuity of x at r nor does this prove that v contains a neighborhood of x in x but there is a hope that is all we have to prove one more step for for we have not yet proved that ut itself is contained inside v since ut will necessarily have many points belonging to the cell sigma in v where sigma itself is not completely inside ut ok so we claim that there exists s equal to some s depends upon t positive such that if any cell sigma of w intersects us then the entire cell sigma is contained inside ut this is what we have to prove the moment it intersect it must be whole thing must be contained combined with the earlier claim this will then complete the both the proofs ok this is the step which is missing in both of them ok so what we have to prove there is some small s sufficiently small but positive such that any cell sigma of w intersects the small neighborhood of us then the entire sigma is contained inside ut ok we have fixed at t what is this t we have fixed in all these namely 0 less than t less than epsilon by 4 any t we have proved this one but that itself does not prove the continuity now fixing at t we will claim that there is a smaller s ok such that if an element of a sigma come beyond w intersects us then it must be contained in ut in a larger one if it comes so close smaller then it must be inside that one that is the whole idea so if this is not the case what happens just take the negative of this one what happens that means that there exists a sequence n i converging to infinity such that even if i divide 1 by n i it will not satisfy this property as any small number it does not satisfy that is the meaning ok there is a sequence n i converging to infinity such that cells sigma i in w sigma i intersection u 1 by n is non-empty yet sigma intersection the complement of this is also non-empty it is not contained inside ut take the complement it is still non-empty in particular this implies that the diameter of sigma i is bigger than t by 2 ok for large for all large n i because this sigma intersection is where this more it has come but the ut is still there so I can check n i to be less than less than t i by 2 then 1 by n i less than t i by 2 then this will happen ok in particular this implies the diameter sigma i is bigger than t by 2 for large n i if m is an integer such that t by 2 is bigger than 1 by m this means that all sigma i are inside w m which has a positive distance from x say delta if i is such that 1 by n i is less than delta this will contradict the choice of sigma intersection 1 by n is ok I repeat this part ok so what we have what we have done suppose for each sigma i sigma intersection this one is non-empty ok you choose n i large enough ok then sigma i is the the diameter of sigma i will be bigger than t by 2 for large n i ok now take if m is an integer such that this t by 2 is bigger than 1 by m ok this means that all sigma j I should say this is 1 by 2 power m ok this this must be 2 power m because these w m's are belonging to our lattice points then inside w m which has a positive distance from x say delta ok this means that all the sigma i's are inside w m ok which has a positive distance from x say delta ok remember all these u i's are u and u i t etc i am writing they are balls of centered at at x one single point at a point that point we are we are we are doing all this our rotation is this ut what is ut ut is bt of x ok u 1 by n is b 1 by nx that is that is the meaning of that ok if i is such that 1 by n i is less than delta but then sigma intersection this will not be so that is the contradiction next time we shall prove one of the very important things that we had raised earlier from using this one namely that the fundamental group of any topological manifold is countable ok so that is the homotopical aspect of this one that is what we get retraction is already some kind of homotopy though we have not proved that it is deformation retraction ok that is not needed ok so let us stop here thank you