 Last time, remember this picture in which we showed that the two segments here on the boundary of the square is a deformation, it has a strong diffusion rate of the entire square I cross I. Just reflect this picture along the Y axis. What do you get? You will get minus 1 cross I here, minus 1 2 plus 1 the interval here and 1 cross I here as a subspace of minus 1 plus 1 cross I and it will be a strong distribution rate trap. There is no need to write down another proof here, right? Now imagine you are inside this, this is X axis and this is Y axis and take another Z axis as horizontally in R3. Then along the Y axis, instead of reflecting, now you rotate the picture. What do you get? You will get this disk here of radius 1 cross I, strongly deformation retracting to the tub namely S1 cross I union the bottom disk. We follow this geometric argument. Once I have written down this one, I don't have to write formulas for them. It's obvious. So today we are going to make use of all these and do a lot of more generalizations of constructing deformation traps. So that is the topic today. Generalize construction. Think of the interval, close interval 0 to 1 as a cone over 1. What is a cone over 1? What is a cone over anything? By definition, it is X cross I and then X cross 0 being identified as single point. So if X is singleton 1, what is X cross I? It is an interval. There is no further identification because endpoint is just singleton 1 only. So think of this 1 as a cone over, I interval as a cone over 1. Then the construction above can be generalized to any space X instead of the singleton 1. So what you do is you use polar coordinates for the cone over X. So thus what you have is, let us look at this map X cross I cross I cross I to X cross I, I cross I. See in the earlier case when X was singleton, we had this homotopy capital S from I cross I cross I to I cross I, which is a strong deformation trap. So that you do not write any singleton X here. But now you want to write the whole space X, so you have to write X cross that. So the X factor is a dummy. What you are interested in is the homotopy is capital S which was defined last time from I cross I cross I to I cross I. Okay. So therefore let us define H as H of X comma t, t prime, t double prime is X as it is take S of the other three variables, S of t, t prime, t double prime, which is obtained by you know pushing stereography projection from 0 comma 2, projection from 0 comma 2. So this was the map. Okay. So what will be the image namely when t double prime is equal to 1, all these you have to see. Let us use the notation X comma t bracket, round bracket, X comma a comma b, round bracket, etc. To denote the image of X t, X a, b, etc. inside the product space under the quotient maps where quotient maps are cone construction. There is one cone construct Q from X cross I to C X. Okay. Then if I take one more cross I, C X cross I, cross I, C X cross I cross I, etc. That has one uniform notation instead of too many notations. Whenever I write round brackets, they are in the product space. Whenever I write square brackets, they are corresponding C X cross I cross. Okay. Starting with a map from X cross I, cross I, cross I. The last two factors are not affected at all. But X cross I, this factor, there will be a quotient map to C X wherein X cross 0 has been identified as a single point. The point of this one is when the first coordinate here, this t is 0, then this entire map as of 0, t, t prime, t double prime is a single point. So therefore, you know, this makes sense and it will go factored down from the quotient space to the quotient space here C X cross prime. This is the quotient space here C X cross I. Okay. So that is the whole idea here. Okay. So under the quotient map Q cross I and G, etc., the function H respects equivalence relations defined by the cone because the t coordinate of S 0, t prime, t double prime is always 0. So it will be something, it may be X, actually X comma something, cross sometimes something, but that whole thing X can be collapsed to a single point. Okay. The t coordinate of this part is always 0. So here the t coordinate going to 0, so it is fine. Hence there is a well-defined map, quotient map, I mean the induced map, co-induced map H bar from C X cross I cross I to C X cross I. Okay. What is the formula? H bar of this equivalence class X t, t prime, t double prime is the first coordinate is X and S of t, t double prime, t, t prime, t double prime. Okay. So if you are just right like this, you need to verify why this is well-defined because when t is 0, there will be different X is here. Okay. But when t is 0, this look at the X comma some number. Right. But that is also 0. Therefore, this is a well-defined function in C X cross I. This is two coordinates here, right. S of t, t prime, t double prime may look like some alpha comma beta, where alpha beta are elements of I cross I. When t is 0, alpha will be always 0. Therefore, this is well-defined function is what I wanted to say. Instead of writing S of this one in two different formula and so on, I am just writing this thing. So you have to just check this S is the function that we have worked last time. Okay. So I will look at this one. So you see, when t is 0, what happens to the first coordinate here? That is why I have written down this formula in terms of, fully in terms of t, t prime, the t double prime. If I write like this, it is not clear what happens. Okay. T is 0. This is the formula which is valid. This will not come at all. Never t prime is bearing equal to 2 t. 2. Then t is 0. So it is this one. With t prime less than 2, it is always possible. So look at this formula. t is 0. This part is 0. Right? t here, there is a t here. This part is also 0. The first coordinate here is 0. This will 1 minus t double prime t prime is independent of t anyway. Doesn't matter. This part is 0 is all that I need. Okay. So we have a map like this. Now you can verify that it is actually identity, all those things. Identity when you know this last quantity of prime is 1. This is 0. And then at t equal to 1, it is, takes values inside, blah, blah, blah. What is the image? All that you can verify. Exactly same way as in for just s prime. The x part has no role to play here. Okay. So it can directly verify that. It is a continuous map also. But you don't need to do that because of the quotient space theory. The continuity at points of this one, you can tell because we have formula there. Okay. Alternative we can use theorem 3.4 which is just the quotient, quotient of product. Product by the, that's true. Product of the quotient, the second factor is our identity. Okay. Why it is a quotient map? Because i is compact. So in particular it is locally compact. We have proved that for locally compact space, x to y is a quotient map. If z is a locally compact, then x cross z to y cross z, q cross identity is also a quotient map. If you use that theorem in continuity of h bar is obvious. You don't have to verify. You don't know that theorem. You can directly verify this by looking at the formula. At each part you just say it is given by a certain formula, projection map, x comma something. That something is continuous. So that's what it is. To show that c x cross i is a quotient topology. You can use this theorem and then you are done. Okay. So the conclusion is that for any topological space x, c x cross 0 union x cross i is a strong deformation retract of c x cross i where we identify x with the image of c x with the image in c x of the space x cross 1. Okay. x is identified in that. If you put x equal to 1 for single point, c x cross 0 will be just i. Okay. i cross 0. Union 0 cross i sorry 0 not x 1 cross i at the other end. Right. This is the starting point. The bottom line and the one side bar along the parallel to y axis is a strong differential tract of i cross i. That was the starting point. But instead of that i now I am replacing it by c x. Okay. I do such examples i give. One is by reflecting it again it was whole cone over s naught. The interval can be thought of as cone over s naught. Cone over s 1 is got by rotating the whole picture along the y axis. Why bother? That is all geometrically you can do. But you can do this by just simple argument. Just c x wherever it is. It is a cone over x. Okay. So the bottom c x cross 0 and c x cross i and x cross 1 x cross i itself. Okay. That is the subspace of c x cross i. Right. It is a strong differential tract of c x cross i. So this figure I have already told you. Now I am drawing those two figures. This is the reflection here of the earlier picture and this is got by rotating. So what you got here is now it is a bathtub. The solid thing. Okay. Whole this is like an ice. So the ice melts out but only the only the container is left out. There will be number of implications now one by one. At a, b any close of space of s. Okay. Then the pair x a has homotopy extension property with respect to every space. If and only if the subspace a cross i union x cross 0 is a retract of x cross i. What is the meaning of it as homotopy extension property with respect to every space? That is the inclusion map is a co-fibration. Okay. So this is what inclusion map is a co-fibration if and only if a cross i union x cross 0 is a retract of x cross i. So this is the direct application to homotopy theory of that simple construction of str that we have done. Okay. This is a picture. The bottom thing is x cross i is this picture. A is the triangle is a subspace there. A cross i is standing here. You have to take A cross i union the bottom x cross 0. This is a strong deformation retract of the entire x cross i. The hypothesis is that a to x the inclusion must be a co-fibration. The stronger conclusion is that this is if and only if. So that is a fantastic conclusion. Why? Because to verify something is a co-fibration you have to verify for every space every map blah, blah, blah, blah, right? As homotopy extension property you have to establish. It is an impossible task. Here you have to only verify that this particular subspace whether it is strong deformation retract of x cross i or not over. So this is going to be extremely useful theorem. Okay. Small, small results. Okay. But they are going to build up the whole theory what is called a homotopy theory. Right? So let us work out how this is done. To prove the if part that means suppose you have the deformation retract. Then I want to show that this is A to x is a co-fibration. Co-fibration means what? Let us assume that there is a retraction R from x cross i to z. This is just a retract. Sorry, it is not even a strong differential retract. Suppose you have just a retract. Retract means what? Restricted to z, it is identity. It is continuous function from the whole space to subspace. What is z? Z is I have introduced this notation namely A cross i or x cross i. Okay. Suppose there is a retraction. All right. Now suppose there is a homotopy extension data as in definition 1.4. What does this mean? This means that you have a map from A cross i to some y and on x cross 0 is inclusion map. The restriction map can be extended. Then I want to have an extension on the entire of x cross i of the homotopy h. So suppose you have such a homotopy data. Okay. So I have treated it as a map theta which is the union of these two things namely theta restricted to A cross i is the homotopy f theta restricted to x cross 0 is the map G which is extension of f on A cross 0. Okay. Define theta from z to y by taking this restriction namely restriction A cross i is f and on x on x cross 0 is G. Okay. So what you have is a map from z to y. Then there is a map from x cross i to z which is r. Take the composite that G be theta composite r. Okay. All that I want to say is G is the required homotopy extension. Okay. Because r is identity on this part then you take r composite theta it will be when you restrict it to x cross 0 it will be the required map x cross i and A cross i that is all. And it is an extension it is defined on the whole of x cross i. Okay. So this comes as a by as if by magic. All right. Now this retraction we want to construct if the inclusion map is a co-fibration that is the converse. Suppose x a has homotopy extension property with respect to every space every space y in this case now I take y to be z itself. Okay. And f and G be the corresponding inclusion maps. What is f A cross i to z its inclusion what is G? G is x cross 0 inclusion. All right. So that is the data. So with this data there will be capital G from x cross i to z which is a continuous function which extends this one that just means that on z it must be identity. So it is a retraction. Okay. Go back to the statement. Statement says that starting with A any close of space of x then being a co-fibration or result being a retract these two are equivalent. Inclusion map co-fibration or this A cross i union x cross 0 retract these two are equivalent. Where did we use that A is a close of space? Why do we have that hypothesis? Look at this one. I defined this theta by patching about two different functions. On this set itself that is continuous. On theta restricted to x cross 0 it is G that is continuous. These are hypotheses. But why theta is continuous? On the intersection namely A cross 0 these capital F agrees with G. G restricted to A cross 0 because G is an extension of that A cross 0 map on A cross 0 f. Okay. As a function it is well defined. But why it is continuous? You need some hypothesis namely if A is closed then the intersection A cross 0 here of these two will be a close of space. Okay. Look at A cross i and x cross 0. What is the intersection? It is A cross 0. That must be a close of space. The same thing as A must be a close of space. So the hypothesis is required there to say that theta is continuous. Okay. It is not a very costly hypothesis because finally what you get is the corners namely A to x is a cofibration. Then what has happened? A cross this one z is a close of space of the whole thing. If x is closed off any retract of a closed off space is closed that is what we know. If z is closed then A cross 0 will be close of space which is very easy because you are just intersected with x cross 0. Z intersection x cross 0 okay or x cross t or x cross 1 actually is A cross 1. A cross 1 is closed the same thing as A is closed in x. So under half darkness the closeness of A is a must. So we are not assuming too much. All right. So I have told you already that this proposition comes extremely handy in determining whether the inclusion map is cofibration or not. Okay. It reduces the practically impossible task of verifying the homotopy extension property with respect to every space to just one task of checking whether the subspace x cross 0 union A cross i is retract of A. Okay. So this itself has many bits of i bits applications. Okay. So let us take a look at some of these applications now. If the inclusion map is a cofibration then for any space z the inclusion map z cross A to z cross x is also a cofibration. So z cross A to z cross x what is the map? It is the inclusion map again because A to x is an inclusion map. How do you do that? Of course here all I have for all or A I have assumed that A is a closed subspace because that is you have to apply that theorem you need a closed subspace. Okay. So why is a cofibration? I have to just verify that z cross A cross i union z cross A cross 0 sorry z cross x cross 0 is a deformation is a retract of x cross z cross x cross i from the hypothesis that from for x cross i the corresponding things are true. So everywhere you just take z cross cross with z. Okay. So this is very straightforward. So this is what I have done with a retract from x cross i to x cross 0 union A cross i. Okay. Right. r of z comma x comma t keep z as it is you can just second factor is r of x t. Just taking identity cross r identity of z cross r that will give you the corresponding thing for the product. Next hypothesis next corollary is also as easy take any tocological space inclusion map x inside x cross 1 okay of x into cx remember in the cone x was identified with x cross 1 as subspace. So now I am thinking of x as a subspace of the cone that inclusion map is always a cofibration at least here you do not have to assume something close automatically x is a close subspace of the cone okay. So now the x itself is a cofibration inclusion map is a cofibration okay. Can you see this how to get this corollary just like in this case okay you have got a product now right now you can take the cone over that one. In the previous thing you got a product you can parameterize once you have this r you can parameterize by any other another factor okay. So I will leave it to you to think about these things this may be you know unless you write down something you may have this writing the cx the cone over x contains the bottom space x x cross 1 as a inclusion map will be a cofibration is the statement. In particular we have seen that 1 is the bottom of the cone over 1 in the interval 0 1 right the single turn 1 inclusion into the interval is a cofibration you can directly prove it also okay but actually the first picture itself proves that at the starting point but now I have something more here corollary take any point in between also this also a cofibration I told you how to make it reflect it and then the center making the center and so on so is the trick so you reflect it on the other side so one is a was a retract no you retake it on the other side so instead of of end two end points you will get a central one an edge here okay the above construction can be used to to get a picture for this one house method of construction can be used to prove that any map inclusion map is a cofibration go back to that theorem that proposition and try to prove that retract is possible that's all once you have this take any space that at any point in i the inclusion map z cross t in z cross i is a cofibration this is a direct consequence of two of the previous theorem first you can use this one and then you can use its product product here to combine these two corollaries you get its corollary for any point z cross p inclusion into z cross i is a cofibration let us stop here and take up more further applications in the next module thank you