 So, in this module which I have named as NDR pairs, we shall describe the co-fibration in a slightly different way which will enhance the utility of this concept. No doubt you must be having no doubt now that the co-fibrations are very important. Before we actually touch NDR pair, so let us do a little more work on co-fibration. Okay, let me, we have done it already, so let me start with the definition of what is the NDR pair. N corresponds to neighborhood, T corresponds to deformation and R corresponds to retract. An NDR pair is a topological pair XA, okay, satisfying a number of conditions. So it is going to be a long definition. So you have to have a little more patience. So you start with the topological pair X, A, a continuous function from X to 01 and a homotopy from X cross I to X of you know with plot of properties. The first property is that A is the exact zero of the function A is U inverse of zero. For each point of A has gone to zero and nothing else. So that is the meaning of A equal to U inverse of zero. Okay, that is the one property of the continuous function here. The homotopy H is identity to begin with. So it is a deformation. Any homotopy which is identity to begin with is called a deformation. Remember that it is not only deformation, it is identity, it is a constant on A cross I. Each point of A H of XT is X itself for all T, okay. So A never moves, points of A never, A, T will go to A itself. And the fourth property is H1, namely when XT equal to 1 H of X1 is inside A. So H is a deformation namely H is homotopy of the identity map to some function from X to A. So entire of X is folded back into A, okay. Not an entire of X, but only those points wherein U X is less than 1. Remember U is a continuous function. So points wherein U X is less than 1 is an open subset of X and it contains A because on A U X is zero. So this set wherein U X is less than 1 is going to be a neighborhood of this A in X and that neighborhood is pushed into A. The map which pushes is homotopic to identity. Now you can see why all this deformation retracts. The neighborhood deforms into A and A is a retract of that. So A is a deformation retract of the neighborhood, that neighborhood is precisely all points such that U is less than 1. The neighborhood is not given by just some technological condition, but a strong condition namely it is open set given by a continuous function. It is U inverse of open, zero closed open one open, that is an open subset of it, okay. So this looks like too much of a condition, so much, okay. But you will see that this is not a new concept at all. So under these conditions we also say that A is a neighborhood retract in X. The whole A is not a retract of X, but it is a retract of a neighborhood of itself, okay. You can just say that much, that will be weaker than giving the neighborhood by a function like this. At least it looks like, but maybe it is not weaker. So we have the strongest condition, then we will see that this is equivalent to whatever we have been studying all the while. So this notion is due to GW-Vited who developed this notion very thoroughly and he has contributed to homotopy theory a lot, okay. There are two Vited's in homotopy theory. One is JHC Vited, another one is GW Vited. Who has contributed more is difficult to say. So they are quite competitive in that sense. So here is a picture of all that 1, 2, 3, 4 which it says. You start with a X here, okay. Then this line, little bit line segment here is A. It is a closed subset, all right. Then you have a neighborhood, open neighborhood here. How is this given? There is a function from this X to 01. This is a 01 interval. This U is a continuous function. The entire of this line segment here is going to single point 0 under U. This portion outside the shaded region is going to 1. That just means that the shaded region here is the inverse image of 0 closed 1 open under U. So this is the picture which is happening at X cross 0 inside X everything. So V is the inverse image of 0 open, 0 closed 1 open under U. So that is V. Now look at X cross I, there is a homotopy H, X cross I to X. It is identity on this A cross I, okay. It is identity on X cross 0 on the bottom, this H naught and this entire V cross 1 has gone inside A, okay. So under this condition we say that A is a deformation, A is a neighborhood retract in X or XA is a neighborhood retract, neighborhood deformation retract, NDR pair, okay. So this is just the definition, okay. You can visualize it in this picture. Condition 1 implies that namely U inverse of 0 is equal to A implies that A is a closed G delta set because it is inverse image of single point. Automatically it must be closed. In fact it is intersection of open sets 0 to 1 by N upon take the inverse image and then take the intersection over 1 by N all of them. So that is why it is G delta set. That is the meaning of G delta. If you do not know G delta forget about it. At least you can see that A is a closed subset. Put V equal to U inverse of 0, 1 that is what I explained here, okay. Then H defines a relative homotopy, relative to what? Relative to A of a retraction of V onto A. Inside X, you see that is why you have to be careful here that it is not V is a different, A is a deformation retract of V. It is not happening, and I think deformation is not happening inside V, okay. So you have to be careful about in these kind of wordings here. That is why you have to stick to the definition finally, alright. So that is what I wanted to drop your attention to, you know. A is a relative homotopy of retraction of V onto A inside X. The homotopy is taking place inside X. A is a deformation retract of one of its neighborhoods is you can say. The two of them are homotopy equivalent inside X, okay. This justifies the name neighborhood deformation retract. The point of introducing this definition is, is that this is not new. Namely, these four conditions are all equivalent. The last condition is the new definition. X is NDR pair. What is the first one? A is a co-vibration. Standing assumption is that A is a closed subset. You start with that one. Then whether A is a co-vibration, X cross 0 to A cross i is a retract of X cross i. This is our proposition 4.1 whatever we have been using this one. X cross 0 union A cross i is actually a deformation retract of X cross i. This part I used in the previous lecture. Maybe I did not prove it, but now we will prove that one also. So you do not have to worry about this one. The last condition is X is actually NDR pair. So you see co-vibration is giving you all such data, so much of data and it is equivalent. So just using co-vibration we have to construct all these things. That is the task here. If you have NDR pair with all these things, proving that it is a co-vibration is not difficult. So we shall prove that these 4 conditions are equivalent. Then you have co-vibration is such a powerful notion. That is what you would come to know. Nevertheless, we know that it is very common. It is occurring in lots of cases. Only very bad cases like comb space etc. You have to take to get counter examples. So proof of this theorem finally. We have already seen in proposition 4.1 whatever A and B are equivalent. This is just a repetition here. So we do not have to worry A and B are equivalent. Now I want to prove you C implies B. That is also easy because deformation retract, it is retract. So C implies B is also easy. Now we will prove D implies C. For this we have to do a little more circus. Namely start with the H that is given by D, H of X, D, S I want to define now capital H. This H is the one which is given by the hypothesis D. H of X, D, S right as H of X, S change T and S here. Take the second coordinate. Remember H of X says is where it is inside X. Okay. This is X cross I cross I to X cross I. This is T times 1 minus S of U X. U X is a function from X to I. So S times I, S times some U X makes sense. 1 minus that makes sense. This number is always less than or equal to 1. Therefore 1 minus that will be also between 0 and 1. You can multiply it by T. Again you get a point in 0 and 1. So second coordinate takes place inside 0 to 1. So this is a map from X cross I cross I to X cross I. Take capital H to be this map. The two coordinate functions I have defined here. Okay. I want to say that this will already imply C. What is C? That what is the X cross 0 even in A cross I, deformation of X cross I. I have given you this deformation. Okay. Let us verify. If S is 0, what is H of X T 0? This is 0. H of X 0 by definition H of X 0 is X. Since S is 0, S of U X is 0. This is just T. So the second coordinate is T. And that is the identity map. So to begin with F is identity map. Okay. When you use condition 2, 3 and 4 together, it will imply that H of X T comma 1. At the end what happens? When you put S equal to 1, this is just T times 1 minus U X. But what is this? It is H of X 1. So it is H of X 1, T times 1 minus U X. Okay. This is a retract of X cross I on to X cross 0 in A cross I is what I want to show. Why? This whole thing takes place either in X cross 0 or inside A cross I. Okay. When it is X cross 0 means what? This is 0. Either T is 0 or 1 minus U X is 0. Then this will be 0. Then this is any point. If it is not 0 means what? T should not be 0. U X should not be 1. That means U X is strictly less than 1. When U X is strictly less than 1, look at the condition 4 here. This one. When U X is strictly less than 1, namely points up here. Okay. H of X 1 is inside A. Okay. This last condition 4. So coming back here, when this is not 0, it is inside A cross I. This is in A. This is A cross I. So this map actually takes values X cross 0 in A cross I. Suppose you take T equal to 0. T equal to 0, this is 0. This will be H of X 1, which is X. So this is identity X cross 0 on X cross 0 inside. Similarly, on A cross I, as soon as something is in A, remember U is identically 0 on A. Therefore, it is H of X 1 comma 1. What is H of X 1 comma 1? Look at this condition. As X is inside A, it is identity for all T. Therefore, this is identity on X cross 0. This map is identity on X cross 0 in an A cross I. Therefore, it is a retraction. This retraction is somewhat appropriate to identity. Therefore, it is a deformation retraction. Thinking of this is something new and I myself don't know how to explain this. See, there is a circus here. The third coordinate here S has come second coordinate here X H of X. This is something by S into 1 minus S kind of thing. But you are not multiplying by 1 minus S. You are multiplying by S and then subtracting. So these are all quite somewhat weird way of trying. Of course, why it doesn't explain this one, even this much whatever I have explained. So I am explaining to the extent I understand this. So this is quite tricky thing. But more or less, you know, forewarn you to, if you are not given this one by just looking at this and trying to prove what you want to, you will have to come to this map. So what we have done is T implies C implies B implies A, A implies B. This much we have done. What we have to show is actually we will directly show that B implies D. Then the cycle is complete. We don't need this one. If we use this one, it may be easier. It is not. So we will just assume X cross 0 union A cross I is a retract and show directly that it is NDR pair. So that is a harder part to prove B implies D. So let us denote the retraction from X cross I to X cross 0 and A cross I by R. Take the first projection X cross I to X P1 put H equal to P1 composite R. So take R and then follow it by first projection. This whole space is contained inside X cross I anyway. So you can take P1, P1 composite R. This H satisfies condition 2, 3 of the above definition. What is condition 2? H of X0 is X. Wherever X0 R itself is an identity. Then when you project it to the first coordinate, it is just X. What is 3? H of XT is X for every extension A cross I. So when you take R, R also has the same property. When you project it to first coordinate is X, the second coordinate is T. So that is what we are doing here. So it satisfies, this satisfies 2 and 3. So what you have to do is, it remains to prove that this continuous function U from X to I, which is identically 0 on A and satisfies condition 4. Namely, the set of points wherein it is not 0 is going inside A by H. By our already H is already defined here. H is defined. So we are adjusting U to suit this H so that condition 4 is satisfied. Even to have a continuous function like this, it looks like T's extension theorem. You are given the 0 function on a closed subset. A entire A goes to 0. You have to extend it. It is stronger than that. Namely, I do not want U to have any other 0s. So I was a bit skeptical about this till I read the proof. This is something which I did not prove myself. I have to read it. But it is interesting. It is not so difficult finally. So let us take the second projection X cross I to I. This is second projection. Now for n greater than or equal to 0 integer, let us define a function phi n to be minimum of 1 by 2 power n and the value of P2 composite r of X comma 1 by 2 power n. Remember r is a function from X cross I into X cross 0 union A cross I. So I can talk about the second projection of that map. That will be an element inside I. So this is some real number between 0 and 1. This is 1 by 2 power n. You take the minimum of that. This is a continuous function. This is a constant. So phi n will be a continuous function. Minimum of two continuous functions is continuous. So I have got a sequence of continuous functions taking values between 0 and 1 by 2 power n because if something is bigger than 1 by 2 power n, it is 1 by 2 power n here. So it is maximum is 1 by 2 power n. It is minimum of these two. So it is taking values in 1 by 2 power n. Therefore if you take the series summation phi n X summed over n, this will be convergent. Not only it is convergent, it is uniformly convergent for all X. Why? It is dominated by summation of my 2 power n, the geometric series. So why stress-dominant convergence theorem tells you that it is uniformly convergent. When you have uniformly convergent series of continuous functions, the limit function namely summation will be itself a continuous function. So most of our work is done. Now you put u X to be 1 minus the summation of phi n X. To make sure something works, I should multiply it by phi naught of X. So phi naught of X can be multiplied outside this summation also because this is common thing. It can take it outside also if you write. This series is series of positive non-negative terms. So it is a absorptive convergent. So you can write this one on this side here also. Or you can rewrite the whole thing because 1 is equal to summation 1 by 2 power n, 1 to infinity, 1 by 2 plus 1 by 4 etcetera equal to 1. So you write this 1 by 2 power n minus phi naught of phi naught of X into phi n of X. And then take the summation. So this u can be either defined by this way or defined this way. So it has two formulas. So this formula defines our function u. Why it is continuous? Because summation phi n is continuous. Multiplying by phi naught of X is continuous. 1 minus that is continuous. So you have got a continuous function. Each of these phi n X summation is between 0 and 1 because they are dominated by 1 by 2 power n. Summation of 1 by 2 power n is equal to 1. So these are smaller than that. So this entire thing takes place between 0 and 1. So I have got a continuous function X on X namely this u. So you have got a continuous function. Now X is inside A. Then you get P2 of R times R of X 1 by 2 power n, X 1 by 2 power n, R of that is identity X 1 by 2 power n. P2 of that is just 1 by 2 power n. Go back here in the definition of phi n. Phi n will be now minimum of this, but this is equal to 1 by 2 power n. So phi n of X will be equal to 1 by 2 power n for all points X inside A. Therefore, u X will be 0. If X is inside A, then P2 itself is 1 by 2 power n for each n, its summation will be equal to 1 and phi naught is just 1. So therefore, this summation and here this is equal to 1. Therefore, u X is 0. So A goes to 0. So u X is 0 if X is inside A. Now I have to prove that if X is outside A, then u X is not 0 because I want to prove that A is exactly u inverse 0. Take X in the complement. Then R of X 0, which is X, because R is an identity map on X cross 0. But X is inside now complement of, because X is complement of C. I take it as cross 0. If you take it as cross 0, it is not necessary here. X cross X comma 0, if I write, then it is A cross 0. R of X 0 is X comma 0, then it is A cross 0. Which is, this is an open subset of X cross 0. By continuity of R, we can choose epsilon positive and a neighborhood V of X inside X such that R of V cross 0 epsilon is contained inside. See some point is here and it is an open set. Therefore, some neighborhood of this must go inside this set. That is all I am telling. How does the neighborhood of X cross 0 look like? It is a neighborhood V inside X and 0 comma epsilon. This whole thing must be inside A C cross 0. A C is complement of A. But if this happens, if you take N very large, then X comma 1 by 2 power N would be inside here. V is a neighborhood of X. This X is fixed here. X is belonging to X minus A. So, X comma 1 by 2 power N is inside here. That means R of that takes values in A C cross 0. What does that mean? P 2 of this point is 0. Therefore, phi N itself will be 0 because phi N is a minimum of 1 by 2 power N and that value. Phi N is 0 if N is sufficiently large. So, look at this expression here. After such a stage, all these terms will be 0. Even if you take all of them to the maximum, it will be 1. So, most of them are 0 and only for some few of them are left. Some total, each of them less than 1 by 2 power N anyway, less than equal to 1 by 2 power N. Therefore, some total will never be equal to 1. That means 1 minus some smaller number. So, it is never 0. It is positive. Therefore, U X is positive. This proves that 1 namely A is precise 0 set. So, major thing is done. One more thing is there now. When U X is less than 1, we have to show that the homotopy H X 1 takes value inside A. Suppose U X is less than 1. That is, U X is not equal to 1. That is all here. By the first formula for U, this part, it is not 1 means what? Whatever you have subtracted is not 0. Suppose U X is sorry, U X is less than 1. It is not equal to 1 means this is not 0. So, none of this will be something. If phi naught is 0, then the whole thing would have been 0. So, in particular, phi naught is not 0. So, that is what I wanted here. So, if less than 1 by formula, this implies that phi naught is not equal to 1. Other things may be 0 or phi naught is 0, then the whole thing you would have. So, phi naught is not equal to 0. The phi naught is not equal to 0, which would not be p 2 R of X 1 is positive. Look at this definition of phi naught. When n equal to 0, it is minimum of 1 and p 2 of X comma 1. The minimum is not 0, this is not 0. So, this implies the second coordinate p 2 of R X 1 is positive. Remember, R, where does it take value? X cross 0 or A cross I. So, it cannot be X cross 0. This is second coordinate is positive. Therefore, it must be in A cross I. So, this implies that H of X 1 is p 1, which is the first coordinate must be inside A. p 1 of R X, this is the definition of H because I have taken H to be p 1 composite R. So, this is inside A. So, that is a beautiful proof. So, this implies this. So, in summary, a simple looking co-fibration implies all this stuff. So, somehow the kind of thing that is happening inside R n or any Euclidean spaces has come back to us by just by a co-fibration. The fact that it is a g delta set A, you said that just you are using A is a closed subset. You are not ready to even assume that, but you assume A is a closed subset then it is a g delta set blah blah blah and it is a small neighborhood can be retracted, deformation retracted to A. All these things happen in nice neighborhoods inside R, inside R 2, inside R 3 and so on, nice spaces. Inside R n itself or R 2 itself there are weird spaces like cone spaces. If you look at one of the exercises Zebra, every point is a bad point of that space and it is subspace of R 2. Every point is a bad. Bad point means what? It is not a co-fibration. Inclusion map is not a co-fibration. So, such things can happen. So, this is more or less for us complete whatever we have planned of homotopy theory. From now onwards we will study special kinds of things namely we will we will specialize to spaces which are built up of straight lines, segments, triangles, tetrahedrons and so on. So that we have plenty of you know homotopy, plenty of line segments to join two points and so on when they are nearby. Okay. Thank you.