 So, in the last module, we discussed what is the meaning of homotopy and the homotopy version of the lifting problem. A lifting data implies the lifting problem means the map has lifting property. If the map has lifting property with every space, such a map is called a Hureviji vibration. This much we had some. Similar to that, some more dual to it is question number 2, which we want to now put in a homotopy theory. Ok. So, the homotopy theoretic version of the same problem number question number 2, extension problem. Remember in the extension problem, very closely associated was the quotient problem, factorizing problem, which was very easy, set theoretically, so we are taking only the extension problem here. So, start with a map from A to X, so this is the map which we are concentrating upon. Now, suppose we have the following data, there is a homotopy from A cross I to Y and the map from X to Y such that this eta composite G or G composite data whatever you want to say is the starting point of this function F, F of A0 for all A and A. Such a data is called a homotopy extension data. What does it mean is that, think of A as a subspace of X, on the space, on the subspace there is a homotopy of a function which is defined already on X. You take the restriction that is G composite eta, so that has a homotopy F, the starting point, ok, this is the data. Now what is the conclusion? We say eta is homotopy extension property with respect to I, if for each such data there exists a homotopy H on the whole of X cross I to Y such that when you restrict it to A cross I, it is the homotopy that has been already given and the homotopy is not of arbitrary map, it is the given function G H of X 0 H 0. So, there is a homotopy of the extended map which extends the original homotopy. So, that is the homotopy extension property. Once again I will denote, I will represent it by schematically so that you will remember how what is the meaning of this one, ok. Before that I have a definition here, if this happens for every homotopy data, extension data then eta will be called a co-fibration, where you call it as a vibration P from E to B, here A to X we are calling it as co-fibration indicating that this property is somewhat dual to the property of being vibration, ok. So, let us look at the figure here. So, A cross 0 to X cross 0 you have the function, function is from A to X, A cross 0 is a copy of A, right, X cross 0 copy of A. So, eta cross 0 you take here. On X cross 0 you have another function G. So, this part is just a function A to X and a function X to Y. On this part you have a homotopy of this function, the restricted function eta, zeta, the G composite eta on A cross I, that is a homotopy. So, this triangle is the given data. This is always given, eta is there. So, eta cross I, A cross I to X cross I and this X cross 0 is contained as X cross I. This part is nothing strange. What is given is this triangle, ok. This is always there. This triangle is given. Now, the conclusion of this one is that there is a map here from X cross I to Y, the dotted arrow, ok. So, this map is when you composite with eta cross I, it is F. So, you can think of this as inclusion and this will be restriction of H and X cross 0 the starting point is the given G. So, whenever you have this diagram, there must be a map here which fits the diagram like this, then it is called homotopy extension property. If eta, this map has homotopy extension property for every data, whatever Y is, whatever G is, whatever F is, then it is called a co-fibration, ok. Once again you will say this is too much to expect. No, the beauty of these things are they are almost always satisfied or what you may say that whatever interesting space is come, they will have this property, ok. So, that is why these both these properties, fibrations as well as co-fibrations are the part and parcel of algebraic property, right. In the beginning you should grasp them what is happening, ok. Having set up these two problems in the homotopy setup, we should now study the homotopy a little deeper, right. So, let us do that one now. What are called as homotopy types. We had already homotopy classes of functions, maps, right. Now we want to rule them on spaces. Start with two topological spaces. You know what is the meaning of their homomorphism. There must be homomorphism between f from x to y, right. Namely, there must be G from y to x which is inverse of f which is both continuous, f is continuous, G is continuous. That is a homomorphism. So, we want to take a weaker equivalence here, namely homotopy equivalence. What is that? So, a map is called a homotopy equivalence. If there is a G from y to x such that instead of G composite f being identity, it is homotopy to identity. Similarly, the other way composition f composite G must be homotopy to identity of y. So, instead of equality, we are replacing it by homotopy, homotopy. Such a thing is called homotopy inverse. G will be homotopy inverse. The point is there may be many G which satisfies this property yet the homotopy class of G is the only one. If at all it exists, there may not be any, even any function there may not be any inverses, right. But if the inverse exists, the homotopy class of that is unique, okay. So, you can call that class as the inverse of the class f, homotopy inverse, okay. Whenever f, there exists such an f from x to y, it is homotopy equivalent. We call x is homotopy to y. Like the function f is homotopy to G is now the space x is homotopy to y or x and y are homotopy equivalent or they have the same homotopy type. So, all these terminologies are used, okay. Again, using lemma 1.1, remember that namely compositions of homotopies, you can go on verify. If fx is homotopy to y and y is homotopy to z, then x will be homotopy to z. Every space is homotopy to itself because identity map is homotopy to l and how inverse of itself. If f is homotopy to y, its inverse will give you y is homotopy to x and so on, okay. So, homotopy equivalence is a equivalence relation. Therefore, we can give equivalence classes. Each class is a homotopy type, okay. Once you have that suppose x and y are homotopy equivalent, then for every space A, the set yA and the set xA will be in bijection. So, this is also a consequence of lemma 1.1. Keep using that compositions and compositions are associative. Similarly, all functions from b to x, okay, are in 1-1 correspondence with all functions from b to y when you take homotopy classes. The functions themselves may not be in 1-1 correspondence. When you take homotopy classes, there is a 1-1 correspondence. How is it given? Take a homotopy f from homotopy equivalence f, f to y, composite with suppose you have a map from y to A, composite with f, you get a map from x to A. To go back, you composite in g. You will go back because g composite f is identity. That is the whole, okay. So, these things are all straightforward following of our remark after the lemma. Lemma 1.1, we said 2.1, let us not correct. Now, we make a special definition here. Anything which is homotopy type of a single point, this single point does not have much properties, topological properties. So, we want to single it out. So, such a thing is called contractable space, okay. Up to homotopy type, there is no difference between any contractable space and a single point because they have the same homotopy type. So, all the homotopy theoretic properties will be the same. So, such a thing as contractable space. A map f from x to y, which is homotopy to a constant map, we should also do for maps also. Such a thing is called null homotopy. A constant map will not have much properties, okay. If you compose with a constant map, that will also become constant map, whichever way you compose. Post-compose or pre-compose, once one of them is constant map, the composition will be constant map. So, that kind of properties first, the trivial properties have to be first understood carefully. So, there are some easy way of identifying what is a contractable space. Let us go through that one, okay. This is the theorem. So, this is the first theorem of the course. Following conditions on a space are all equivalent. The first condition is x is homotopy equivalent to a single written space, which is we have named it as x is contractable. That is the first one. The second identity map of x to x, it is null homotopy. The third one, every space y and every map from h to y, h from y to x is null homotopy. All functions taking values in x, they are null homotopy. Every space that, every map from h, from x to z, it is awesome. Everything if this has domain of the function is x, that is awesome. So, both domain and codomain, okay, as a both domain and codomain, x is behaving very nicely. That is precisely what I told you. If you pre-compose or post-compose for the function, it will be constant map and that is reflected here in the spaces. Whether you start from x or you end in x, if x is contractable, such maps are all null homotopy. So, these are the characteristics of point space and up to homotopy, they are the characteristic of contractable space. So, that is the gist of this theorem. All right. So, being a mathematics lecture, we should examine this thing carefully and study the proofs also. All right. The proofs are very easy. Let us go through that. So, let us first prove one implies two. What is the meaning of that? I assume x is homotopy type of a single point. Then I want to show that identity map of x is homotopy to a constant map. That it is homotopy type of a single point means I am taking y as a single point space. f is a map from x to y and g is a map from y to x. What is y? y is a single point. They are homotopy inverses of each other. What does it mean? g composite f is homotopy to identity. And g composite f is what? See, first you start with g single point, sorry, first you start with f from x to single point, then g to x. Where does it go? Single point only. g composite f is homotopy to identity by definition, but g composite f is a constant map. So, identity is homotopy to a constant map. So, that is the conclusion two. So, one implies two follows. Now, let us prove two implies one, reverse. Reversing is also very easy here. Given that identity is homotopy to a constant map, let us call that constant map c to c. Let us name it as, it is a map from x to x because it is homotopy to identity. So, let us call it as c to c. The image of c is a single point. So, let us call that as y. Then we can view c as this function as a map from x to single point because the image is single point after all. This is subspace of y, subspace of x, but I will call it as y. So, now let g from this one be the inclusion map. Take the inclusion map. This is subspace of x. Then what happens? If you compose with g, g, c compose g, g is a constant map, c is a constant map, that will be the identity of the constant map, identity of the single point. Start from single point, you come back single point. Moreover, we are given that g composite c the other way around is homotopy to identity. But g composite c is c. Thus we have what? c is a homotopy equivalence from x to single point. So, which means x is contracted. So, we have got a map x to c is a single point that itself is a homotopy equivalence is what we have. So, x is contracted. So, 1 implies 2, 2 implies 1 is okay. The 3 and 4 as I have told you already, they are built in automatically. There is no problem. You go to single point and then keep composing. Okay, 2 and 3 are an equivalence of 2, 3 or 2, 4. They are very straightforward. Let us have some examples now. Euclidean space r, r square, r q, they are all contracted. In fact, they are all vector spaces. So, what happens? You can join lines. The joining lines is joining any two points by straight lines. That is the way to get homotopies. More generally, inside a vector space, you can take a convex subset. What is a convex subspace? Given any two points x and y inside a the line segment between x and y. So, x and y are vectors. So, t times x plus 1 minus t times y will give you the line segment. That line segment could be entirely inside a. Then you can take that as the homotopy. That is the whole idea. So, all convex subsets are contractible. In fact, we can generalize it to what are called as star-shaped subsets. Namely, take a subset S. It is called star-shaped. If there is one point S0, such that every other point S can be joined to this S0 inside S, which is the same thing as saying that the line segment S0 S is contained in S. In this case, S0 is called the apex of S. This is definition for star-shaped sets. Then all that I do is I will define a homotopy S cross I to S. In the beginning, in the starting, it is identity map. At the end, it is the single point S0, the constant function S0. So, I have S0 here and the identity here. S goes to S0 is the first one. S goes to identity of S itself is a second map. I will just join them. T times S0 plus 1 minus T times S. When T is 0, it is S. Identity, S goes to S. When T is 1, all the, no matter what S is, it is S0, the single point. So, obviously, this is a continuous function. This is just a linear combination of, you know, scalar multiples and adding, subtracting and so on inside a vector space. So, all these things are I am talking as a Rn. You can have Rn or Cn or any vector space. This will be true. So, this will give you homotopy of the identity map, the constant map. If identity map is homotopy the constant map, our theorem 1.1 will tell you that it is a space is contracted. Okay. So, we have a picture of what is the star shaped set is here. This is this, the three lines meeting a sixth point. That point is S0. You take any point inside this figure. You can join it to that point by a line. The line segment completely inside the picture. So, you can take a point here. This line segment is here up to here. To take a point here, for example, of course, the point itself is not there. So, there is no question of the point. The two points must be there. So, this is S3's where any other point must be inside this one in the whole line segment. So, it is called a star shaped set. Okay. This is the star shaped set inside R2. Okay. We will have one very important example. This will give you a set of a sequence of spaces, pairs of spaces. They are not homomorphic to each other but they are homotopic to each other. This time I am not giving you things which are homotopic to constant map, constant homotopic to a single point. They are not contractable but the two spaces are homotopic type same. Okay. So, what are they? For each n greater than or equal to 0, you take the unit sphere S n in Rn plus 1. For example, when n is 0, what is this? This is R. This is S0. S0 is what? What are the unit vectors inside R? Plus 1 and minus 1. Okay. When you take R equal to 1, n equal to 1, we have in R2. In R2, what is the unit sphere? Unit S1, S1, circle. Set of all unit vectors inside R2. Okay. So, what we are doing is throw away the 0 from Rn plus 1. S n is inside, included inside. This is a subspace of that. The unit vectors are never 0. So, this is a subspace. So, this eta is an inclusion map. We are going to show that this eta itself is a homotopic equivalence which is homotopy inverse mu which is very nicely given namely take any vector divided by its norm. So, that is a map mu. This mu is homotopy inverse of the inclusion. One way is clear. Namely, you take a unit vector, think of this as a vector here, non-zero vector here. If you divide by the norm, it will get same x, same vector because norm is already 1. So, that means mu composite eta is actually identity. Okay. What you have to show is eta composite mu is homotopic to the identity of Rn plus 1 minus 0. So, this is what you have to show. Right? But that is this homotopy. See, whenever you can actually write the homotopy, it is nothing but joining the two points. So, this is identity x, this x by norm x. So, I am joining them 1 minus t times x plus t times x by norm x. The point here is that the right hand side will never be 0. Therefore, its entire thing is taking place inside Rn plus 1 minus 0. Why? Can you tell me why? Because take any non-zero vector, first we start with x is non-zero vector. Then x by norm x is also a vector in the same direction. They are on the same ray emanating from 0. Open 0. 0 is not there. Open the way emanating from x. So, when you join this to the whole line side, it will be away from 0. Okay? So, this is the homotopy inverse of the inclusion map. All right? This example you have to understand very clearly because there will be modifications of this one. This will keep coming again and again. A homeomorphism is definitely a homotopic balance because f composite g is actually equal to identity. G composite equals to the corresponding identity of the corresponding spaces. So, if you have some maps such that, you know, x and y are not, some spaces such that x and y are not same homotopy type, then they cannot be homeomorphic. Okay? So, this is one of the effective way how algebraic topology, how homotopy theory is employed. Okay? So, how to determine whether two spaces have same homotopy type or not? This is our fundamental question. This is answered in various ways by cooking up different algebraic or algebraic topologism, homotopy invariants. Okay? Just like we did in point set topology, if something is a T1 space, other than not a T1 space, they cannot be homeomorphic. If something is a T2 space and other than it is not a T2 space, they cannot be homeomorphic. So, similarly we have to cook up various invariants, homotopy type invariants. Every homotopy invariant is also topological invariant. There is no problem. But there are many, many homotop, topological invariants which are not homotopy invariants. We have already seen one here, namely Sn and Rn plus 1 minus 0. They have the same homotopy type. Okay? They are not homeomorphic. It's very easy to see how Sn is compared. Rn plus 1 minus 0 is not compared. Very easy to see, right? So, they are not homomorphic to each other. But they are the same homotopy type. As an example, I will just discuss a few historical things here. There is a celebrated century-old three-dimensional Poincaré conjecture which states that something which looks like S3 up to homotopy type. Okay? Almost like the three-sphere up to homotopy type is actually homotopy, actually homomorphic to S3. That was the Poincaré conjecture, dimension 3. You can ask the same question in higher dimension also. You can ask in lower dimension also. Take something which is homotopy type of a circle. Will it be actually a circle? Of course, we have to be careful here, namely what is the meaning of something like, so that they say are called manifolds. Take a one-dimensional manifold which is homotopy type of a circle. It is not hard to show that it is actually a circle. This can be studied in point set topology also, but you might not have studied. So, if time permits, we will try to give you a proof of this one. Same question you can ask for S2. And the answer is again, yes. But that is already a little bit difficult. But we will try to answer it in the second part of this course. As soon as you come to three-dimension, this was a problem posed by Poincaré, which was answered only very recently, namely 18 years back. For n bigger than 3, this was already answered by many other people, 50, 60, 70 years back. In 2002, G. Perelman, a Russian mathematician, solved this conjecture using very typical, very strange kind of differential geometry, namely heat equations and so on, unexpected or somewhat. He was awarded Fields Medal in 2006 and a million in prize 2010, okay, both of which he has declined, he hasn't accepted them. His fellow is much more stranger than our Ramanujan. He is a weird guy. Subsequently, the problem for n greater than or equal to 5, not 4, was solved by Smale in a differential topology case, Stallings in the piecewise linear case. There are different versions when you go to higher dimension. G. Manin topology case. N equal to 4 was a another big problem, okay. So that was also solved by 1980 by Friedman and he was awarded Fields Medal for it. So this is all I can say about this problem. We cannot even go nearer to this problem in this course, okay. We will be quite far away from that. However, next time I will tell you there are some very, very great results which we can prove in this course, okay. So this I will tell you next time. Thank you.