 Last time we started the study of relative homotopy groups, see three different definitions of it and then we established one of the fundamental results namely homotopy exact sequence of a topological pair. So today we shall give you one very important application of that namely the homotopy sequence exact sequence of a fibration. So recall that by fibration we mean a map between topological spaces, it is customary to denote such a map by p from e to b, e is a total space, b is the base space which has a homotopy lifting property with respect to every space. The homotopy lifting property I am just recalling means given a data here namely a homotopy from say x cross i to b into the base space and a lift of this map from x cross 0, on the x cross 0 is the initial stage there must be a lift of this to e, lift means what? A map g such that p composite g is f restricted to x cross 0, so this eta is the inclusion map here x, x comma 0 going to x comma 0. So think of this as a copy of x, here you have a function g, p composite g is the initial value of f, then the conclusion is that there is a map g, capital G such that p composite g is little f, g restricted to x cross 0 is little g, so that is the homotopy lifting property. If this is true for every space x and every map f from x cross i to f then p is called a fibration. So the proposition is that any fibration p from e, f, e naught, b, b naught, b naught induces isomorphism of homotopy groups. What is f here? f is again customary to denote the inverse image of the singleton point b which is a base point under p. So this is our basic result used in the fibration. Using this result we will get a very useful statement when we combine it with a homotopy exact sequence of the pair. So let us first prove this statement. So there are two parts here. One is that p is surjective and the other one is p is injective. To prove that p is surjective let us use the definition 6 namely the simplest definition for the homotopy groups. At the homotopy groups we want to show that p is surjective. So start with an element alpha represented by some map alpha of pi n of b b naught. The element alpha is represented by a map alpha from i n boundary of i n to b b naught which means it continues function from i n to b taking the entire of the boundary to a single point. So this is the simplest definition of the fundamental of the nth homotopy group. Now you can think of this alpha as a homotopy on x equal to i n minus 1. So take this as f in this picture here and then we will try to apply this homotopy lifting property. You must have a map from i n minus 1 to e but that is very easy here because this particular f namely alpha take the entire of the boundary to a single point. So I can take i n minus 1 cross 0 which is x cross 0 which is a part of the boundary go to the single point here e naught then p of that would be p naught here. So that diagram this diagram will be commutative therefore I can apply to that situation and get a map capital G here that is the first conclusion that we get ok. So taking x is i n minus 1 and thinking alpha as a homotopy and G as x cross 0 to be the constant map at e naught apply homotopy lifting property to get a G such that the diagram fits. So that means what G of i n minus 1 cross 0 which is little G of i n minus 1 cross 0 because it is a restriction map that is e naught by our choice and p of G is alpha is the definition of G that is how the lift means where p of G is the given alpha and on the boundary it is single point. If p of the whole thing is a single point that means this entire thing is contained in the fiber f G of boundary of i n is content fiber therefore G represent an element of pi n of E f e naught that is the definition that we are using definition 6 is here ok pi n of E f e naught represented by a map from i n to E such that boundary of i n goes into f and of course the base point is going to e naught and p check of G is alpha ok that is how we have taken p composite G is alpha so p check of G this proves the we had p check is surject. Now we have to prove the injectivity what is the meaning of injectivity start with an element alpha now in pi n of E f e naught ok this time and a relative homotopy of this alpha to a single point ok when you pass on to the base base B so H is a homotopy H is i n cross boundary of i n cross i to B B naught inside B such that this H restricted to i n cross 0 is p of alpha and H is i n cross 1 is B naught it just means that p composite alpha is null homotopy in B so in pi n of E B it is trivial element I want to show that it is alpha is a trivial element in pi n of E f itself right so that means we have to find a base point preserving the relative homotopy here namely i n boundary of pi n o to E comma f comma E naught ok such that H restricted to i n cross 0 is alpha and H restricted to i n cross 1 is E naught ok so this is what we have to find here we have taken this whole to be 0 0 0 belong to i n ok as a base point for i n it is clear that if we apply homotopy lifting property of p to H directly resulting lift will not specify this requirement namely the base point or the base point the all the time it has to go to the same base point E naught that will not be guaranteed because all that we get is p composite the lift is equal to alpha so it will say it is in the capital F that is all it should be actually just a single point in all that will not happen namely this 0 cross i will not be mapped to a single point so that is a important point here so in order to overcome that we have to work a little harder the key is demanding that H satisfies even more stringent conditions and then appeal to a trade ok we just try to control this H only on O cross i that seems to be more difficult so ask it to be you know more stringent so what is that let me tell you so first have a notation A to be one of the phases of i n plus 1 namely one cross i n then A is an n phase of i n plus 1 that W be the union of all other n phases of i n plus 1 ok note that this W contains i n cross 0 comma 1 i n cross 0 is one of the phases i n cross 1 is another phases ok in particular many other so these two phases are there in W ok as well as the singleton 0 this O O O cross i this will be also there so if I control the map on W automatically this will be controlled in any case I have to control it on i n cross 0 and i n cross 1 on i n cross 0 I have to take it as alpha on i n cross 1 I have to take it as a constant function E naught right and here also I want it to be a constant function E naught so can I do this one is a question so we shall demand that in addition to 8 8 is these two conditions are already there we want H satisfied condition that H of A prime is E naught where A prime is all the n phases A prime other than this i n cross 0 i n cross 1 and 1 cross i n of course is not there ok so that is what W consists of to if you just do not take A and take all other phases W but or i n cross 0 i n cross 1 they are also there so there you take the condition 8 ok and the rest of them you take all of them to a single point in other words on W everything is single point except on i n cross 0 it is alpha so can we get such an H is the question of A so where to use this one homotopy lifting property directly does not give you that so we have to appeal a trick okay so what is that trick I will take now okay define lambda from W to E by taking lambda of i n cross 0 is alpha i n cross n is E naught for all lambda A prime go to E naught for all other phases A prime so that is what I want H to be like that so on W I take this lambda then H is going to be lambda on this one we have to find H on i n plus 1 such that H restricted W is lambda and P composite H is capital H since H of the boundary of y n cross i by by the very choice of this homotopy is singleton B naught it will follow that capital H boundary of i n cross i is inside F that is also one of the requirement but that comes freely for us that will come to the proof okay so how to how to apply homotopy lifting property for this W instead of just i n cross 0 that is the trick so we appeal to the fact that there is a homomorphism tau from A to W A is remember A is one cross i i n one of the phases given by the radial projection from the point 2 comma half half half belong to i n plus 1 okay this radial projection is identity map on the common boundary boundary of A and boundary of tau just consists of the boundary of this A that is what it is full boundary of tau also okay on that this radial projection tau will be identity map okay so let us look at how it is caught in case of when n equal to 1 so this is my picture of i cross i here i n cross i everything and this red thing is i n cross 1 and this is 2 comma half of half of and so on from here you project radial projection each point x on i n cross this one cross i n goes to some point in W this blue part is W so why this is the homomorphism why the bijection all this is very clear first of all this projection so there is no problem about continuity and so on so take a point here okay i am going to instead of defining tau i am going to define tau inverse now take a point here say why how to determine this point it is very easy take this point take this point the line you can write down the line segment t times this point plus 1 minus t times that point that should have the first coordinate 1 so that is 1 cross i n put that condition immediately it gives you a unique solution this is all this is just a linear algebra okay you can write down the formula for tau inverse so that will complete the the claim that this is a homomorphism obviously when you take the boundary point here or here okay namely 1 cross boundary of i n is the boundary of this one okay so there it is in the projection you see at that point itself it coincides with the point of W okay so it is identity tau is identity on the boundary okay so that is my tau here okay which is identity on the common boundary now take phi to be boundary of i n plus 1 to boundary of i n plus 1 such that phi on the W part is tau inverse phi on a part is tau on the intersection its identity so they have to patch up to define a homomorphism phi from boundary of i n to boundary of i n finally take the linear affine linear transformation from r n plus 1 to r n plus 1 which interchanges the first coordinate and n plus 1 coordinate followed by a reflection in the point half so that is by formula t 1 t 2 t n t n plus 1 going to 1 minus t t n plus 1 t 2 t n as it is in the last coordinate is t 1 the first coordinate has come to last coordinate so this is an affine linear isomorphism obviously if t i's are between 0 and 1 that is a definition of i n then 1 minus t n plus 1 to t 2 t n are also between i n plus 1 and conversely so i n plus 1 goes to i n plus 1 and boundary of i n plus 1 goes to boundary of i n plus 1 okay if boundary of i n plus 1 is defined exactly by taking one of the coordinates at least one of the coordinates is equal to either 0 or 1 okay so boundary goes to itself therefore I can take this t from boundary of i n plus 1 to boundary of i n plus 1 and compose it with phi so I get a map from boundary of i n plus 1 to boundary of i n plus 1 which is a homomorphism after that you can extend it to a homomorphism of the i n plus 1 to i n plus 1 by taking the cone construction so this psi hat is the extension of psi to the i n plus 1 to i n plus 1 okay so this homomorphism has the following properties to sum up all these why I have done is the following property namely on i n both phi and psi hat phi is on the boundary psi hat in the intersection in clear map i n cross 0 homomorphically on to w so now the trick is revealed see we can control we want to control w but we can control i n cross 0 by the very definition of homotopy lifting property so use this homomorphism then we can do whatever you want to so now go back to our homotopy lifting diagram I have projection map p this vibration here I have an h here I want a capital H here such that restricted to w it is lambda and p composite h is h I cannot apply the homotopy lifting here because this w is not just i n cross 0 so what I do I take psi hat compute h s f and take the larger rectangle here on i n plus 1 i n plus 1 i raise to n plus 1 I take this map now namely it is h composite psi hat i n cross 0 just take whatever restricted to i n cross 0 whatever it comes out that happens to be just you know restricted to i n cross 0 it is psi right and psi goes i n cross 0 goes to w that is what we we examine that and there you take composite with lambda this is my g this is my f little f then homotopy lifting property of p gives you capital G here okay now all that I want to take capital H to be come here by psi hat inverse and then take g okay so g company psi hat inverse is capital H okay so that will give you whatever h we wanted okay now let us come to the the final statement of this theorem the proof is very easy namely take any vibration take a base point in the top space base points must be taken such that p of e naught is the base point below and a capital F is p inverse of b naught this is convention then we have a long exact sequence of homotopy groups and homomorphisms what is that pi n of f e naught to pi n of e e naught p check suddenly instead of relative homotopy group here pi n of b b naught and then boundary operator whatever pi n minus 1 of f e naught and so on so pi n minus 1 then again next one pi n minus 1 of e minus e pi n minus b and goes on and finally it will end up with pi n pi 1 of b b naught to pi naught of f to pi naught of e pi naught remember is the set of path components of f okay and pi naught of a is is again path components of f then this is the inclusion map okay the proof is now very clear if i replace this b here with pi n of e comma f comma b naught namely relative homotopy group okay and then this map by the inclusion map then I have an exact sequence the proposition says that the relative homotopy group is isomorphic to pi n of b b naught under p check so what this map I am going to do just p composite I check which is again p check so that is precisely what the proof says start with a start with the homotopy exact sequence of the pair e f okay pi n of e f is sitting here pi n minus 1 of f is the boundary boundary map here and this is a one inclusion map there is another inclusion map from here you go via p check which is an isomorphism to pi n of b b naught to take the composite it is just the restriction of instead of pairs it is just the same map e e e naught so this was a p check only this thing I have to define to be p check inverse followed by daba so delta is what daba composite p check inverse you define that way that is why there is different notation here automatically this will be an exact sequence so coming from here to here and come down here go that way and that is the precise the statement of this okay so that completes the proof of the big theorem okay homotopy exact sequence of a vibration of course it has several applications we will only mention a few of them which are immediate constant without much trouble we should be able to get one of the thing is that a covering projection is always a vibration it is a very peculiar vibration very very important vibration which has more properties than ordinary vibration in particular the fiber f is a discrete space when it is a discrete space what happens even pi naught will correspond to number of points in it and all other pi 1 pi 2 pi 3 all of them are all 0 so if you use that hypothesis here this pi n of f will keep appearing and it will be 0 this will be 0 so this will be isomorphism so this p check will be isomorphism that means what that means that p check from the top space x bar here instead of e to pi n of x instead of b it is b here this is an isomorphism for all n greater than or equal to 2 why 2 because when n equal to 1 your your problem will become pi pi naught of f naught that is not trivial pi naught of f naught is not a single point there are several components okay that is not a single point so n equal to 1 it will not work obviously for n equal to 1 we know exactly we have studied this one under covering space revolution so here is a case wherein it works even much better look at the look at the action of s 1 on s 2 n plus 1 think of s 2 n plus 1 as a unit sphere in c cross c cross c n times complex numbers taken n times product n times okay so on that s 1 x unit complex numbers the quotient is as we know if the compressor should face c p n this quotient map is a vibration that is not very easy to see and but we we cannot prove that one here it is a standard result and the standard result is the following if we have a lee group acting on at a manifold a smooth manifold so that is what if you lee group is compact and the smooth manifold is also compact then what we get is a vibration the quotient is a vibration okay this is the richest source of vibration by the way but discourse we cannot cover that one okay so this is a vibration and this is it has a name of vibration because when n equal to 1 here this is s 3 2 c p 1 here it is s 2 and this was the important contribution of the P check is an isomorphism for n greater than or equal to 3 so greater than or equal to what a 2 n plus 1 is what I have here okay in particular pi i of c p n is 1 for 1 less than or equal to i less than or equal to n minus 1 and pi 2 n of c p n is isomorphic to pi 2 n plus 1 of s 2 n plus 1 okay hence not trivial so non-triviality of this one is seen in many ways in one of the thing is we have seen it namely the identity map cannot be null homotopy okay this fact was proved while proving Brouwer's point theorem okay how do we get this one apply this apply this one here with e equal to s 2 n plus 1 and b equal to c p n the fiber is equal to s 1 and we know that the homotopy groups of s 1 n greater than or equal to 2 they are all trivial okay that is what you have to do so isomorphism n greater than or equal to 3 because 2 there will be a problem because pi 1 of s 1 is non-trivial it is infinite cycle by 2 onwards is trivial so I think you know 3 you get this okay so that is what I am telling you I repeated pi n of s n is non-trivial is a fact that we have proved while proving Brouwer's point theorem namely the identity map itself is not null homotopy okay indeed what happens is just like pi 1 of s 1 is infinite cycle you can prove that pi n of s n is also infinite cycle and that goes under the name half hoffs theorem okay hoffs degree theorem but we have this is not a part of this course okay usually that that result is proved in differential topology okay so one more remark that fact the pi 3 of s 2 is non-trivial was a great discovery by hoff when he did it in his time okay and it was a landmark result even today so he observed that the fibers of phi from s 3 to s 1 they are all s 1s all fibers are s 1s here they are linked linked in a very nice way namely simple way like this and that is that is precisely what they are called half links okay so this is a non-trivial link that is easy to see but hoff says using non-triviality of this one the map itself should know should be non-trivial non-trivial means now not null homotopy so that part I cannot explain you to more than that thank you okay so here are a few exercises which you can try on your own trying the exercises is a part of the learning process okay the trying is more important than just knowing the solutions okay so keep doing that hope you will learn a little more thank you