 So, in this module, we will try to sum up a number of things that we have done about relative homotopy. In particular, I will try to illustrate how this theory helps you to see that two given spaces are homotopy whenever they are of course, in many somewhat easy situations without writing actual homotopy equivalence, okay. To begin with, you take any point in Rn, it is an Ndr, namely deformation, neighborhood deformation retract. The same thing applies to SN also, once it is for Rn. What is the meaning of neighborhood retract? Neighborhood deformation retract, we have all those big list of things, function u, function h and so on, it essentially says that there is a neighborhood of the given subset that neighborhood actually deformation retracts to the subset inside slightly or larger neighborhood. In fact, this is the case with many other examples of subsets of Rn. For example, any smooth arc will be a deformation Ndr space, Ndr subsets, okay. So the tubular neighborhood theorem in differential topology, if you have learned, will tell you that a manifold and a submanifold, let us say cos-submanifolds or compact, whatever. In general, manifold and submanifold is good enough, that forms an Ndr pair. Later on, when you study simple shell complexes, you will have many examples of Ndr pairs. In particular, when you have Ndr, namely inclusion may have a co-fibration, that is the easier way to say a whole thing, right. If A contains an Sn, is an arc, arc is contractible. I can use the theorem that the arc which is contractible can be collapsed to a single point to get a space namely Sn by A, homotopy equivalent to Sn, you take any closed convex subset of Sn, you collapse it to a single point. Then what you get is Sn, okay. If you are asked to write down homotopy equivalence every time, you will see how foreign does it is. But the theory is easy to remember and easy helps us. This is the point I wanted to tell you. Let us consider a little more complicated example. Let us take the sphere along with one of the diameters, okay. So that is my SpaceX. By collapsing one of the great arcs which joins the endpoints, endpoints of a diameter can be joined inside the sphere through a great arc, right. To join them, okay, there is an arc, okay. Now you collapse that arc to a single point. Earlier I have seen that Sn, S2 here, S2 modulo that arc is again homotopy type of S2. But now you had an arc there, the endpoints of those two arcs come to a single point. Therefore, there will be an S1 now, a circle. And this circle and the sphere will have only one point in common. So such a thing is called one point union, it is denoted by S2 union S1, okay. One point here, one point here are identified. So this Y is a one point union of two spaces, S2 and S1. So our original space S2 along with this diameter is homotopy type of S1 union S2, it is one point union, not disjoint union, it is only one common, it is like a balloon construct, okay. So if you are asked to write down a homotopy equivalent, you will have to go through the whole order of things or write down formulas and so on. There is no need, use the theory, okay. Many topologists have this habit of not explaining this to you at all. They will just say, oh, this is obvious. What they say? If at all if you ask, they say, oh, these two points can be moved on S2 so that they come together, that's all. We can move one end point, end points of diameter slowly to coincide with its other end point along the arc, weight arc. But this is not a proof. But if you do anything in between, they are all of the same homotopy type. In fact, in between stages are all homomorphic to each other. The end result is not homomorphic because the two points of the diameter end points have come together, right. So our angular space and this S2 union S1 are not homomorphic to each other, but they are homotopy type of same thing. So to a beginner or an outsider, all these things are hand waving and so on, okay. But an expert topologist knows exactly what is the proof also, all right. He will give you heuristic arguments like this. So here is a picture. So I have taken the sphere and this diameter, one of the diameter, so this is my X. So I start moving it along this arc all the way here, great arc. So this is the middle stage here, okay. The diameter has become like this, okay. So finally it will go and all the way coincide with this one. So this will also become a loop or circle and along with the sphere, this point is on the sphere, okay. See this, this diameter has come like that, okay. So this is like a, you can take North Pole. So at this point, there is a circle attached to the sphere, so this is S2 and S1. So this picture and this picture are of the same homotopy type. These and these have actually homeomorphic picture. So you can have a homeomorphism mapping any point of the circle, any point of the sphere to any other point, keeping this point fixed, such homeomorphisms are there of the sphere. So you can describe this one one way I told you, namely you have collapsed this arc. But then you have to know what happened to the sphere, but that space X by A, we are collapsing in arc, it is same homotopy type is assured by that theorem. Alternatively, what you can do is, you can think of this as an adjunction space, S2 is Y, your Z is the interval I, the diameter I. What is X? X is the end points. What is F? F takes one end point to the south pole, other end point to the north pole. When you take, perform the adjunction space, what you get is this one, okay. So your X is two point space, end points of an interval. From that, you can have different map, namely in this picture, one point always goes to this point, but the other point goes to different points here. Think of this as a map from S0 to S2, the two maps are homotopic to each other. The final picture is both the points I have gone to the same point. That is also homotopic to the original map inside S2. The adjunction space, homotopy adjunction, homotopy invariance of adjunction space says that all these spaces, whatever adjunction space you have got, they are all of same homotopy type, okay. So that is what I have explained. So you can think of this as boundary of minus 1 plus 1 with S0. The maps from S0 to S2, which keep one of the points fixed, they are all homotopic to each other because S2 is connected, that is all, path connected, okay. So one can give many such examples of whatever we have done. So I come to slightly deeper question here now. Above result about co-fibration, co-fibration was essential here by the way. So adjunction space performed on a subspace which is inclusion maps co-fibration allows you to do all this, remember that. So the above result of co-fibration may encourage one to ask the bold question such that for namely suppose you have an XA, so for example, I compare when A contains an XC co-fibration, okay. Now the two maps F and G from X to X be such that on A they agree, FA equal to GA, F of any singleton A equal to G of singleton A for every A inside A, that is the meaning of this, F restricted to A, G restricted to A. Suppose F is homotopic to G, then is F homotopic to G relative to A? This question was motivated by weird deformation retracts or deformation retracts and then if singleton point is a co-fibration, then any deformation retracts or strong deformation retract and so on, remember that theorem. So would you like to have just arbitrary homotopy, will it be relative homotopy? Of course without A being a co-fibration, we know that this is not possible, alright. So under the assumption that A to X is co-fibration, will it be true? Answer is again negative, okay, that is why I said this is a bold question, but the answer is negative. So this is where an expert and a hand waver will be distinguished. If a person has learnt only hand waving, he will go and do this kind of mistakes, okay. This is only an example. There are lots of people who have fallen into this kind of traps while doing algebraic demolition. So it is important to learn where your theorems come from, how they are originating, the fundamentals things should be very clear, okay. So here is a counter example which is, which cannot be completed at this stage because you have to do some computations. I have given those exercises, if you have not done those exercises, then you will not be able to understand it completely, but modulo that I will explain it to you, okay. So what is this? This counter example is also simple. You start taking the cylinder S1 cross i, X is S1 cross i, A is S1 cross 0 union S1 cross 1, namely the two brims, the boundaries of S1 cross i, okay, the two circles, okay. A to X is a co-fibration. So this we have seen before, okay. Double points minus 1 plus 1 or 0, 1 contained inside i, that is a co-fibration, then you take the product of S1. So this one way, there are several ways of seeing this one. The first thing is that the boundary included inside S1 cross i, that includes in my opinion co-fibration, okay, that is the first exercise. Now you take the function f of Z t equal to e power 2 pi i t Z times t. So I am multiplying the first coordinate Z, I am thinking this as a complex number of unit length, okay, first coordinate is complex number, second coordinate is real number, e power 2 pi i t times Z will be another complex number of the, of unit t. So multiplication will be again a complex number of unit length. So f of Z t is e power 2 pi i t Z, it just means that at time t, I have rotated Z through an angle t, 2 pi t, that is the meaning of this. At t equal to 0, equal to pi i t is 1, so this is Z, no rotation. At t equal to 1, what will happen to this one? It would have rotated 2 pi i, right, through 2 pi, so it comes back to Z. So this map t equal to 0 and t equal to 1 is the identity map, f of Z t, f of Z 0, f of Z 1, both of Z comma t, okay, Z comma 0, Z comma 1 whatever, okay. So take this and g to be the identity map of S 1 cross i to S 1 cross i, all right. So this is f, f is a map from x to x. So I have two maps here, let us take H of Z t as namely now it is a homotopy, S 1 cross i cross i, e power 2 pi i t S times Z t. If S is 0, oh t plus S, sir, should I write it as t plus S, yeah, it is not t S, it is t plus S, that is the type of here. If S is 0, I want your identity, if S is 1, t plus S times 2 pi i, the bracket should be there, it is also 1, okay, so it will be identity, gives a homotopy between the two maps. I want when t equal to 0, I want it is 2 pi i t Z t, okay, okay, it is a multiplication only, sorry, okay, it is multiplied t equal to 0, it is t raise to 0 is identity Z t, this identity map g, t equal to 1 it is this map, sorry, this is correct, okay. So the first map is t equal to 0, it is identity map, then this is f when t equal to 1, S equal to 1. So this gives a homotopy between two maps, g to f. I want to say that f and g are not homotopy relative to a, the points of a here in this thing are more, they are related also, okay, so relative to a, they are not homotopy, if you want to keep them fixed, they are not homotopy, when you want to say they are not homotopy, it is not that you cannot construct one, I cannot construct one and so on, we should show that there cannot be a homotopy relative to the end points, the S1, S1 cross 1 and S1 cross 0, should all the points of that should be kept intact while the homotopy takes place. How to do that? So that takes a little more effort. So let us go to the quotient space, namely identify S1 cross 0 with S1 cross 1, Z comma 0 be identified with Z comma 1, then what you get? You will get S1 cross S1, so that is a quotient map, okay, by identifying Z cross 0 with Z cross 1 for each point in S1. Then what happens to this f and g, g will give you identity map, okay, whatever f, f will be also give you because the points corresponding Z 0 and Z 1 are identified, under this map they will also get identified, correct. So capital G, capital F are induced maps from S1 cross S1 cross S1 cross S1, okay, g will be identity, f will be some other map given by capital F, that is a map like this is what you have to see, okay, because Z 0 and Z 1 are not moved by is equal to pi i t, right, that is why it makes sense here, alright. But what is the difference between f and g? f fixes the, there is a image circle here, image of the boundaries of this one, two boundaries of that, on that one both f and g are identities, but along the circle here, the other circle here, f is twisting, f is twisting the other circle exactly once, whereas g is identity, so somehow you have to distinguish these two phenomena and this can be distinguished by looking at the fundamental group, the application of fundamental group to show that there cannot be a homotopy between f and g, if there is a relative homotopy between little f and little g, namely identity, that would have boiled down to give you a homotopy of capital F and capital g, right, therefore by showing that this capital F and capital g are not homotopic here, you would have proved that f and g cannot be homotopic relative to the boundaries. So finally you have to do this one, for this you have to compute the corresponding homomorphisms on the pi 1, you have computed the pi 1 of s1 cross s1 is z cross z, the abelian group of rank 2, okay, because pi 1 of s1 is z, okay, using that you can compute the fundamental group of homomorphisms induced by f, induced by g it is an identity map because g itself is identity, so what happens to f that is what you have to show, okay, so that is what you have to do, okay, so I have left it at this stage, using those two exercises you can see that this is absurd because at the fundamental group level, okay, the first generator goes to identity, second generator goes to this generator plus that generator, not just identity, first generator goes to identity, second generator goes to identity is for g, for f it will be a comma br generator, then f check of a will be a, but f check of b will be a plus b, whereas g check of a is a, but g check of b is just b because g is identity, so this is what you have to write, so finally I conclude one of the theorems that I have read from Hatcher's book, okay, which says that though arbitrary homotopies are not possible, homotopy equivalence is possible, which is a very remarkable result because it is just a borderline thing and the proof is quite tricky proof, therefore I want to say that since I have nothing to add, okay, I have given you a reference here, if you want you can, if you want to learn you can read it from Hatcher's book, okay, what is that? I will state the theorem, the theorem is x, a, y, a satisfy homotopy extension property, okay, they are NDR pairs, suppose f is a homotopy equivalence, then f is a homotopy equivalence relative to a as a pair, so this is a wonderful result, okay, but the proof does not follow from anything that you have done so far, you have to cook it up by using some tricks, so I am not going to give this proof because I am not going to use this result either on here, but this was one of the questions how to determine relative pairs, if they are just homotopy equivalence x and y then they will be homotopy equivalent to each other relative to a, what is this a, a could be anything but inclusion map, okay, a is a common subspace, inclusion map is a co-fibration, a is an s, that is all, all right, so this is the end of this session, thank you.