 Continuing with the topic introduced last time, name is a study of higher homotopy groups. Let us now introduce the so-called relative homotopy groups. This module 16 has been divided into two parts because it is slightly lengthy. The idea is that I would like to take you a little deeper here to the extent that more study of homotopy groups is not possible within our limits unless we use homology and homology theories. So at that door I am going to leave here. So let us do a little more of higher homotopy groups here. Given a topological pair XA with a base point X not belonging to A, sometimes I may forget to write down this base point that is all but it is always there in the mind, okay. Whenever you are discussing homotopy groups, pi 1, pi 2, etc., okay. Let us consider the following subspace of all paths in X, namely continuous functions from closed interval I to X. So this is the function space, it has, remember, it has a compact open topology. So now I am writing a subspace which I am going to denote by omega XA X not. Because omega is usually denoted for loops and pathies in spaces in algebraic topology. So all omega belong to X power I, that means a path from I to X, the starting point of omega is X not and the end point must be inside A, A is a subset, okay. So that is omega XA X not, it is a subspace of X power I. For n greater than or equal to 1, we will have this notation, pi n of XA which is a short form from pi n of XA X not, okay, it is a lazy notation that is all. The base point is always there, is equal to pi n minus 1 of this space omega XA X not, with base point for this one is a constant function at X not, a constant loop at X not. Where C X not is a constant path at X not. Take that, take the n minus 1 thermotomy set, for n equal to 1 this will be just a set if you remember that, for n greater than or equal to 2 this will be pi 1, pi 2 and so on they are all thermotomy groups, so that is just the definition, okay. For n greater than or equal to 2, pi n of XA is a group which is abelian if n is greater than or equal to 3, this is what we have seen last time, these are called the relative thermotomy groups of the pointed pair XA X not. Of course if X not is changed, you know how to relate them and so on, there is some relation but the groups will be different, that is why the base point has to be mentioned, for n equal to 1 in general it is not a group, we treat it as a set but a special kind of set, a set with a base point. The base point of the set itself is now a constant path, okay, it is a set with a base point whereas a group also has a base point and that is the trivial element of the group, okay, that is a pointed set, here is a group structure but pointed set that is there. So for n greater than or equal to 2, there are canonical bijections of the homotopy sets, pi n of XA X not can be thought of as homotopy classes of these triples, I n, I n is I cross I cross I n times boundary of I n and p is standard notation 0 0 0, okay, going into maps going into I n going into X, boundary of I n goes to A, p is always goes to X not and they are all corresponding homotopies, when you take homotopy class when you take I n cross boundary of I n cross p comma p cross I, so I n cross I is inside I, boundary of I n cross I is inside A, p cross whole of I must be a single point, that is the meaning of this notation. So this homotopy set is the same thing as pi n of X X not we have defined in a different way, remember that, pi n is defined as this one, okay, so when you defined like this it is automatically group structure, here you do not know, you have to still define group structures and so on that is a cumbersome method but this description is much simpler, okay, it is just like what you wanted, whatever you wanted, okay, only the second entry here are extra, otherwise it is just I n boundary of I n X comma X not, right, that would have been a definition of pi n, so pi n of X is this one extra, okay, so that is one, okay, there is one relation here that this is a canonical bijection, it is another one, pi n of X X not, it is also equal to complicated looking but this is a very useful thing, pay attention to this, I n boundary of I n and then instead of just single point 0 0 I am taking I n minus 1 cross 0 union boundary of I n minus 1 cross I n, all this is a subset of boundary of I n and it also includes p cross 0 0 0 is already there, this p point is also there, so it is larger set going into X not, so take only maps which take this larger set going into X not, boundary of I n go to A and I n go into X of course, only takes such maps and their homotopes, these two are equivalent and both of them are equivalent to pi n minus 1 of omega X A, that is the definition of this one, so this is a statement of theorem 1.18 which gives you three equivalent way of looking at pi n of X A, okay, let us prove this one, so let us have a temporary notation for this big set, third is I n minus 1 cross 0 boundary of I n minus 1 cross, there is a third subset here, so this I am renoting by z, okay. We know that this z included in the boundary of I n is a co-fibration because it is just a sub complex, okay, remember boundary of I n has this is a subset boundary of I n minus 1 cross I, union I n minus 1 cross 0 and I n minus 1 cross 1, that is the full boundary, so what is missing, only I n minus 1 cross 1, one of the phase is missing here, you take a cube, okay, take one of the phases and look at all the other phases, so that is what is happening here and that third is a included boundary of I n is a co-fibration, this we have proved in part one actually, not only that this third is contractible, so that is also easy to see, it is like a, on the sphere it is like a half half disk, half sphere, it is contractible. Consider the quotient map cube from boundary of I n to boundary of I n by z, we are z to boundary of I n is a co-fibration and it is contractible you are collapsing it, okay, therefore this is an equivalence, this is the amount of equivalence which we identify, okay, where z is to a single point z, okay, so that is the meaning of this quotient map, only all the points of z are equivalent to every other point of z, so that is the equivalence relation here, okay, so that is what we keep saying collapse that to a single point, from theorem 0.6, this is reference you can have from our part one course, okay, so I will just give you this reference here and show you, so this is what it is, so a to x is a co-fibration, it is contractible, then the quotient map is a homotopic equivalence, okay, so this was the theorem, all right, let us go back now, okay, so this is a homotopic equivalence, this quotient map, take lambda from boundary of I n by z to boundary of I n as a homotopic inverse of q, okay, so what is the meaning of that lambda composite q and q composite lambda are homotopic identity of the respective spaces, by composing it with a rotation of boundary of I n, boundary of I n is nothing but Sn minus 1, and Sn minus 1, you can have a rotation which takes any point to any other point, therefore you can assume that lambda of the single point z is the point p which is 0, 0, 0, lambda is a fixed homotopic inverse, you do not know where it z goes, if z goes here fine, otherwise you just rotate it and then you put lambda equal to this one, boundary of I n that, now you take put eta equal to lambda composite, okay, so lambda is, look at this one, lambda is boundary of z to boundary of I n and q is from boundary of I n to this one, so what you get, lambda composite q is from boundary of I n to boundary of I n itself and we know that it is homotopic equivalent to identity map, therefore it is homotopic equivalent, okay, so eta lambda composite q is a homotopic to the identity map and since the entire of z, z is a set under lambda it is going to a single point, under q the z goes to the single point z here, the whole set goes to the whole set, so under composite, so this eta sends the entire z to a single point and yet it is a homotopic equivalence from the boundary of I n to boundary of I n, this what is the point you have to understand. Now whenever you have a map from s n to s n, s n minus 1 to s n minus 1, you can extend it by cone construction to d n to d n, so that is what we do, that eta hat be the extension by taking cone construction I n boundary of I n to I n boundary of I n, automatically this will be also a homotopic equivalence, it is actually homotopic identity map, okay, it is stronger than saying homotopic equivalence and since the entire of z to a single point, now you already why I am doing all this, now you automatically have appreciated why I am doing this, this whole thing can be treated as a single point like this, that is the idea, okay, therefore eta hat induces a bijection eta hat prime, eta hat star from I n boundary of Y n p to x A x naught, suppose you have a map here into x A x naught, okay, then by composing with eta, eta hat, you will get a map from I n boundary of Y n as it is, this whole thing going to a single point p, okay, so omega goes to omega composite eta hat, that is the map from the first set to second set, these are all function spaces, oh, that is a bijection, automatically this composition is under, you know, under function space topologies, it is, if eta is continuous, composing this will be continuous, so this is a, this is not just a bijection, it is a homeomorphism, right now we are going to use it only as a bijection, under the exponential correspondence, the latter space is actually equal to this what you have to see, why we have taken this one, this is the same thing as maps from I n minus 1 boundary of I n to omega x A x naught, remember what is omega x A x naught, these are all paths in x, starting at x naught with end point in A, right, look at a function from I n minus 1 to here, under the exponential correspondence, it corresponds to map from I n to x, okay, so fixing a point x, you have a path x cross I, the starting point is x naught, but that is I n minus 1 cross 0 goes to x naught, the end point is inside A, so that is already, you know, whatever path cross I, that is going to be inside A, okay, so that is not a problem, boundary of I n also goes inside A, okay, more important is its entire thing going into boundary of I n, okay, boundary of, boundary of I n, on boundary of I n, the maps are here going to see x naught, it is a constant map, okay, therefore, this boundary of I n cross I goes into see x naught, okay, by the very definite boundary of I n here goes to single point, constant path, therefore this whole thing going to x naught here, okay, so this is the meaning of this set, this set is under, and the same thing as this set, okay, so this implies over seven, so why we have what this one, that is the seven, directly as sets themselves they are same, when you take homotopy classes, okay, if you have two spaces which are homeomorphic, then you take path components, we said nothing but homotopy classes are path components, path components are in one one correspondence, so it gives you seven, okay, now six is one step further, namely since eta hat is a homotopic to identity of you, upon taking path component we get six, okay, the functoriality of these bijections follows from the fact that everything is happening in the domain and has nothing to do with the topology pair x x naught, what is the meaning of this, suppose you have y comma b comma x, okay, a map from here to here composed with that map x a to y b will give you another map there, okay, now if you have another space, so z comma c comma that one, some function f and g and so on, all these things, composition is taking on one side, the map from here to here, this composition are changing of this from here to here etc, it tries taking on the domain, therefore they have because of associativity of the composition, the functoriality of these isomorphisms follows, okay, so this will be explained completely, not exactly this one, when you study the functors and categories and so on, okay, after that you come back and see whether it makes better sense for you, okay, so six gives us an alternative definition of pi n of x x naught, which is easy to remember and useful, however with this definition define the group structure on it will be somewhat difficult as compared to definition, the first definition 1.21, wherein the pi n's are automatically defined as groups, right, pi n minus 1 of some other space, so 7 is just an elaborate description of what is happening the function space x power i, but that is also useful, so you can keep all the three in your mind, okay, so that is just the definition so far, okay. Now take a special case when a is x naught, it follows that from this definition now 6 1, if a is x naught boundary of i n goes to single point, then p going to single point is redundant, okay, so it as shall write i n boundary of i n x comma x naught, right, so that is same thing as pi n of x x naught x naught, which is what which is just pi n of x naught, x x naught, okay, so that is the conclusion here, okay, so pi n of x a x naught is equal to pi n of x x naught x naught, it is pi n of x x naught, okay. We can now state and prove one of the most fundamental results in homotopy theory, given a pointed topological pair x a, we shall use the notation a included in x, x naught x naught is you know it is a pairs that is all, x x naught included in x a, so this is by j and this is by i, these in turn induce respective inclusion maps when you pass on to the loop spaces, the function spaces, omega a x naught instead of all path is sin a with certain property, right, starting at x naught, right, so you compose, you know if path is here compose with i, think of this inclusion map, it will it can think of this as a path in x, so it will give you omega x x naught, this inclusion map, similarly omega x x naught can be thought of as omega x a x naught, okay, so these are loop spaces, this is path and point could be any point inside a, okay, so these are inclusion maps which we shall denote again by i and j, all right, consider the evaluation map, you see x is a path, this x could x raised to i is a path from, this path space is from, right, from i to x, all the starting point is x naught, right, you can take that one, end point is important for us, look at the end point here, e 1 of omega, this evaluation omega 1, okay, so that is the map from x power i to x, now given alpha from i n minus 1 to omega x a, you take even composite alpha of i n minus 1 will be inside a, because that is the definition of omega x a, end points are inside a, therefore we have a homomorphism boundary I am denoting this one, it is induced by even composite alpha, okay, we are taking a simplified notation, this is classical notation anyway, pi n of x a, okay, remember this is homotopy classes of maps like alpha, okay, 2, you get a map from i n minus 1 to a, so that will represent pi n minus 1 of i n, so that definition of boundary of alpha is nothing but the class of even composite alpha, okay, using definition says it is clear that boundary of alpha is nothing but just alpha restricted to boundary of i n, the last coordinate restricted to boundary of i n on the boundary, okay, if you think of this as maps from here, maps from i n minus 1 cross i, that is the i n, so boundary of i n is fine, okay. For any pointed topological pair, there exists a long exact sequence of groups and homomorphism induced by this topological pair using the i, j and daba, so all these three will be occurring again and again and they will be forming a chain, infinite chain of homotopy groups and this is the homotopy groups. Pi n of a, the first one is inclusion map, i check, inclusion induced, then j check, inclusion induced, then boundary, here for the sake of accommodating the entire thing in a single line, I have omitted mentioning the base points, the base points are there, okay, always. These are all homotopy groups for n greater than or equal to 2, so these are groups and homomorphisms, what is the meaning of exact sequence, any two compositions are 0, that is funny but it is stronger than that, kernel of j check is equal to image of i check, kernel of daba is equal to image of j check, kernel of i check is equal to image of daba check, this should happen at every group here, goes on from for n greater than or equal to 2, they are homomorphisms and keep coming down, down n minus 1, n minus 2 and so on, last thing will be pi n of x, pi n of x a, pi not of a, pi not of x, this is a group, okay, but this is not a group, this is not a group, this is not a group, so what is the meaning of exactness here, remember that is why, remember these sets are pointed sets, so there is a constant function here which is the base point, okay, so constant function here again base point here, these also base point constant function, so constant function works fine, everything which goes to constant function as kernel must be the entire image of daba, similarly everything which goes to the constant function under daba must be entire image of j check, that is the meaning of the, this sequence is exact, okay, kernel and image can be defined when you have pointed sets, so let us prove this, obviously the proof will be divided into three parts, namely exactness set here, exactness here, exactness here, okay, so one by one, the proof is broken into three steps, the first step is kernel of j check is equal to image of i check, kernel of j check here, the i check will come from here, this one, j check and i check, so pi 1 of x, okay, so start with an element alpha in pi n of x which is by definition pi n minus 1 of omega x naught represented by a map i a alpha from i n minus 1 to omega x x naught, where this alpha itself is i composite beta for some beta from i n minus 1 to omega x a, that is you are taking an element here which is the image of i check, alpha is i check of beta, I have to show that, j check of that is 0, that will show that the image is contained in the kernel, okay, that is one part, the second part will be that kernel is contained in the image, okay, so let us prove that image is contained in the kernel, so I have picked up this alpha, so set alpha is i composite beta, we have to show that alpha is in kernel of j, just define a homotopy h of x t s equal to beta s t s x s, s is a s is a real number between 0 and 1, okay, s and t, x is an element of this i n minus 1 or i n wherever, okay, the beta of s x makes sense, okay, beta of s x is a path operated upon t, so this h is a map from i n minus 1, the first slot is x cross i second slot is t, then it is a path in x, so it operates upon again an element 0 to 1, okay, so gives you a path in omega x a, why it is a path in omega x a, put s equal to 0, what is this, it is beta 0 of t, the beta 0 of t is always a constant function, so that is a point x naught, put s equal to 1, it is beta x of t, beta is taking values inside a, therefore the end points are inside a, so that is the meaning of this, so this is the way you have to verify, okay, so verify that h from i n minus 1 cross i to omega x a is a homotopy of the constant loop at x naught with alpha, okay, so if alpha is homotopy of the constant loop and what is happening, only when you go to omega x a, therefore alpha is in the kernel of j check, okay, when you go to omega x a 0, so one part is over, the second and the converse part given alpha and pi n of x a, so put j check of alpha is 1, this gives us a homotopy, j check of alpha is 1 means this alpha when you treat it as an element of omega x a is an homotopy, so take the homotopy h from i n minus 1 cross i to omega x a from alpha is a constant, okay. Now take g as i n minus 1 cross i cross i to x given by, I am reversing the rules of what we did in the previous step, g of x t of s is but it will carefully have to h of x t s 1 minus t times s plus t, okay, s as a general factor here and the constant function 1, constant the point 1 has been taken 1 minus t plus 1 minus t of this plus t times that is converse combination of that, you have to verify that g defines homotopy of alpha with a map beta given by n point namely beta x of s is nothing but h of x of h x 1 of 1, okay, when this whole thing is 1 that is the homotopy of the n point, then when it operates on 0 that will be alpha that is what you have to verify, this means that beta is an element of pi n minus 1 of omega x a omega a, okay. Therefore, if alpha is homotopy to i composite beta it is same thing as i check of beta itself, so we have to prove two more steps here, kernel of dab i equal to image of j check, okay, so now we are coming here this part, kernel here is equal to image of j check, elements are inside pi n of x a, okay, so if alpha is j composite beta for some beta so i n minus 1 of i n minus 1 into omega f, okay, then even composite alpha remember this was different this was used in the definition of the dabba the boundary operator, even composite alpha which is the end point map is constant map c x naught because beta is a loop at x naught, this means dabba of alpha is just a 1 that means the trivial element is c x naught, okay, so one way is very easy, now we have to converse, suppose dabba alpha is 1 for some alpha from i n minus 1 to omega x, this means that we have a null homotopy of even composite alpha, okay, you see even composite alpha may not be a constant function but it is homotopic to constant function is what we get, okay, it is relative to the boundary, okay, that is the meaning of that this is a trivial element in this group, okay, i n minus 1 to a boundary of i n minus 1 going to single point all the time that is the meaning of this group, alright, so what we have we have a null homotopy of even composite alpha from i n minus 1 to a, okay, let h be such a homotopy, for each fixed x in i n minus 1 let us define a path omega x from 0 1 to a, okay, omega x is a path operating upon s is h of x, this homotopy, then omega x is a path from starting from even of alpha x to alpha x of 1, okay, which is 2 x naught inside a, okay, so even of alpha x is not exactly all the time x naught, right, so this is alpha x of 1 by definition and it is homotopic to constant power so 2 x naught you get, each of them will be end point would be just x naught, so alpha x 1 is some point and from there I am joining it to x naught that is the meaning of this one, omega x is what that is what is doing, so now put omega x t of s equal to this omega x t s, okay, so I am defining another path here, okay, tempered path, now you define h from i n minus 1 cross i to omega x by h of x t equal to first part is alpha x, remember this is a path it ends in alpha x 1, so take that point and take omega x t from there, so omega x t starts at alpha 1, alpha x 1 and ends at some other point, okay, so take this path this, okay, I want to say that they are all loops now, what is the end point of this is always x naught, so starting point is x naught, end point is x naught, so they are loops, so we have h of x 0 is alpha x star c x, okay, alpha x star c x naught, what is the point is x is always alpha x, so I have to put 0 here, so alpha x 0 is just c x naught that is the same thing as if you take the constant loop and take alpha x constant loop, constant path it is always alpha x, right, homotopypt alpha x and h of x 1 is alpha x star omega x t equal to 1 it is omega x, remember what is omega x t of this one, omega x 0 will be omega x of 0 comma s whatever s is it is really 0, omega x 1 will be omega x of just s it is omega x, okay, so alpha x star, alpha x star omega x is h of x 1, okay, this we know is an element of omega x that is why we started with this one, okay, this is how we started with this omega x is this one, okay, they are all loops, so this we get a homotopy of alpha with a map beta which is h of x comma 1 at end point, that map is i n minus 1 omega x, this alpha end point was arbitrary point inside a from there we took path is going into x naught, so all of them are now loops and these are homotopy, therefore this alpha is in the image of J check now, so this alpha being homotopypt beta it is image of J check, so this is a clever proof it needed some cleverness here, okay, the third step kernel of i check is equal to image of Davao, this last thing here kernel of i check is image of Davao, so now elements are inside pi n minus 1 of a, okay, let now beta from i n minus 1 s n minus 1 to a x naught be a map such that beta is in the image of Davao then I have to show that i check of that is 0, i check of that is constant function that is what I have shown, right, so this implies that there is a map alpha from i n to omega x a such that this beta x is alpha x 1 that is the definition of that beta belongs to image of image of Davao, okay, putting h of x t equal to alpha of x t remember alpha x is a function, so h of x t becomes homotopy, we have h of x naught is equal to alpha x at 0 x x naught h of x 1 is alpha x 1 is beta x, hence beta is null homotopy the starting point is a constant function and this beta, so beta is null homotopy in x, okay, so that means beta is in the kernel of i check, when you go to i check i is from a to x, right, okay, so one way we have proved conversely suppose you have a homotopy from i n cross i to x, okay, to a constant loop at x naught with the map beta because beta is null homotopy in x that is the meaning, then look at alpha from i n to omega x a given by alpha x of t equal to h of x t the same homotopy, so that gives you boundary of alpha is equal to beta because boundary of alpha is evaluation at the end point, the end point is beta starting point is x naught, so that will be a looping that is really path in omega x a that those are functions of omega x, that is a function from i n to omega x a, beta has property boundary of alpha is beta, so that proves beta is integer, okay, so to be precise 3 and 1, step 1 were easier, step 2 was somewhat difficult but in nature they are all slightly different that is why I have taken pain to write down all these proves, okay, so go through them carefully it is not that if you prove that one other things are similar this is not the case here, okay, so go through them carefully that is the only way to understand the various interrelations here and the definitions and so on, you know properly, okay, so we will take a rest here and start the second part of the module 16 a little later, thank you.