 Continuing with the study of CW complexes, now we take up the topic of homotropical aspects. One of the easy consequences of so much of study of the product that we did, once you have a CW complex X, X cross I will be automatically a CW complex now because I itself is a CW complex which is compact. Therefore, construction on functions, continuous functions on X cross I which is same thing as homotopy becomes easy. So, this is the first theorem that we have here take XB as CW complex. A function X cross I to Y is continuous if and only if restricted to each X k cross I, all these restrictions are continuous for each k greater than or equal to 0. What is this X k? X k is kth's character. So, all that I have to say that the coinduced topology with respect to this family X k cross I on, so in the entire X cross I, the same thing as the ordinary product topology which we know is equal to the coinduced topology on X cross I. The coinduced topology on X cross I is the general thing namely by compactly generated sets. Or you can take coinduced topology on X which is through this family 96 and then take the usual topology on I and then take the product and see what it is happening. So, whichever either two ways you have to do. So, what I do? Look at the identity map from X cross I, the coinduced topology namely X cross IW to X cross I. This is homomorphism because X's compactly generated topology and I is compact. Therefore, function f is continuous on X cross I if and only if a restricted to X cross I, the kth skeleton of this is continuous for each I because I have to verify it from this topology. The kth skeleton of X cross I is remember it is union of various we can take X i skeleton and then Jth skeleton, but I has only two skeletons 0 and 1. So, it is X k cross 0 union X k minus 1 cross I. Those are the two skeletons. In any case for each k what happens is the kth skeleton of this one is obviously contained in X k cross I which is the full k plus 1 complex and that is contained inside the k plus 1 skeleton of X cross I comma W. So, these cross subsets are trapped between these two things. Therefore, whether the coinduced topology is with respect to this family or with respect to this family they are the same. So, once you have this readymade criteria what you do is you construct homotopes on the 0 skeleton extended on the 1 skeleton extended on the 2 skeleton and so on as homotopes then you put them together automatically it will be a continuous function from on X cross I. So, this is the idea of this theorem. This is how it will be this theorem will be used now. Now, this is the central result for all homotopy aspects of a CW complex. So, which we have prepared long back right in the beginning in proposition 2.1 that U be an open subset or a closed subset where X A is a relative CW complex. Starting with say U minus 1 this is just an indexing of U intersection A put U n equal to U intersection the nth skeleton of X A n greater than or suppose the n minus 1 subset U n minus 1 is a strong deformation retract of U n each of them anyway U n is contained in U n plus 1 because this is the nth skeleton keep increasing. So, successfully U minus 1 is a SDR of U 0 U 0 is SDR of U 1 and so on. So, for each n U n minus 1 is a strong deformation retract of U n then U minus 1 will be a strong deformation retract of the whole U this is a statement. So, this U n minus 1 to U n this is one stage but this is true for all n then you can hit the U U itself will be strong deformation retract will deform should retract to U minus 1 U minus 1 is a SDR of U. The easy part here is each U n minus 1 is a retract of U n then we know composites you can take composites or retractions retractions two retractions you take composite to composite to retraction. So, you can take finitely many composites each U n will retract to U minus 1 that is easy after that U will be retract because I can define the retraction on U to be these composites depending upon where my n is where my x is every x is inside some R n sorry inside some U n right because every x is in some way x a n skeleton. It may be starting U minus 1 itself there is nothing to do there you define it as the n composition. So, that is the easier part the thing is how to get homotopy you cannot compose homotopes so easily because they are not dominant code of means are different. So, let us go through it carefully that f n from U n cross i to U n be a homotopy of the identity map and which is relative to U n minus 1 because all the points of U n are fixed okay f n of x 0 is x is identity map is to begin with and then what you have is the nth function f n of x which is f n of x comma 1 this is by definition okay that is a deformation retraction that is a retraction whatever U n to U n minus 1 automatically once you have such a thing okay. So, whole thing is taking place inside n but f n of x comma t we do not sticks for all x in U n minus 1 t is in f n of x is in 0 okay therefore it is a function for U n to U n minus 1 and on on U n as soon as the points are in U n minus 1 it is identity. So, this is a this is the hypothesis that is given that each U n minus 1 is a strong deformation retraction of U n okay. So, the last map which is a retraction I am taking f n okay this is a strong deformation retraction for each n that is that is the meaning given by f n itself there is nothing more than that therefore if I take composites g n is f 0 f 1 f n we get a strong deformation retraction g n from U n to U minus 1 okay. I do not know that but I can definitely say that this is a retraction this is a retraction okay. So, we do not know whether it is strong deformation retraction but we will see that soon okay. Observe that the n plus 1th function here restricted to U n is just g n why because g n plus 1 is all these composites f n plus 1 but f n plus 1 is identity on U n it is a function from U n plus 1 to U n which is a retraction. So, it will be just the rest of this f n plus 1 you compose it first f n plus 1 of some x if it is an element of U n it will be just x itself and then rest of them is g n. So, this is the important thing here therefore if I define g as g n x whenever x is in U n this will be well defined function on the whole of U. This map will be actually from U to U minus 1 okay and this is retraction because points of U minus 1 all of them keep fixed okay. So, I get a retraction however to show that it is a strong deformation retraction we have to work a little harder okay all that I get is retract I do not care whether this g is a deformation retraction or strong deformation I will have to prove that one yet there is no way of directly getting it okay yeah. So, that step is a lesson here in getting how to get compositions of deformation retraction or homotopes in general what composition infinite compositions okay so you have to watch it carefully okay that itself maybe the step itself will be useful elsewhere also. So, look at here what I do I start with f naught okay and this first half of the interval I will keep it idle identity then I take f naught. So, this is my g naught times trying to define a successive way of composing okay so in the next step I compose I define g 1. So, this time up to one third interval it is idle idle means identity map no change then whatever f 1 is there one third to one by two I will put that f 1 here okay remember all these f y's are a homotopes of the identity so on this line they will be identity so they will not match up okay now this at the end I have got some f 1 or f naught little f 1 you take the homotopy of that on this side okay you put g naught here okay. So, next step I will cut down this one again identity up to one fourth itself then whatever f 2 you have got you plug in that one here and then the entire earlier stage g 1 this entire the re-parameterize from one third to one so put that one here. So, this is the way I so at the end the stage 0 to 1 by n it will be identity 1 by n to the next 1 by n plus 1 it will be f n minus 1 or f n whatever and then the earlier homotopy which you have constructed this is the way I am going to do okay this is just a schematic way but now I will have to write down exactly so define g n inductively okay g n from u n cross i to u n inductively as follows g naught of x t see this step was given due to g n so inductively defined how g naught of x t is x in the half first half of the interval then it is f naught of x 2 t minus 1 half less than t less than or that you have to say that when t equal to half there are two definitions they must coincide okay when t equal to half what is this this f naught of x 0 so it is identity it is identity here. So, for n greater than or equal to 1 having defined g n minus 1 you define g n as x in the interval 0 to 1 by n plus 2 if n is 1 this is one-third okay like this here second stage is one-third okay from 1 by 1 divided by n plus 2 to 1 divided by n plus 1 you put f n of x appropriately you know scaled you have to appropriately scale this the thing is the thing that you have to write correctly okay for example when t equal to 1 divided by n plus 2 what it this will be n plus 2 n plus cancels out so this 1 minus 1 it is 0 okay this f n of x 0 okay when t equal to 1 by n plus 1 this will be n plus 2 divided by n plus 1 which is 1 divided by n plus 1 minus 1 n plus 1 cancels away so this will be what this will be equal to 1 divided by n plus 2 I am sorry not 1 divided by f n of x 1 it will be so like that you have to see the that they match up okay so here you take g n minus 1 of not g n minus 1 itself but put f n of x t so this is the difference I which was not explained here what g 1 I have but this is just a diagram okay so that at n equal to at t equal to 1 divided by 1 plus n this will coincide with f n of f n of x t little f n of x 1 it will coincide okay so definition of g n is clear how to and one does it now having inductively we have now we know verify that each g n is a strong deformation retard of u n into u minus 1 so this was the claim there okay but there we knew only retraction now it is proved that each u n retracts to u minus 1 in a strong deformation way moreover g n restricted u n minus 1 cross i is g n minus 1 whenever it is inside u n minus 1 it is it is g n minus 1 therefore there is a well defined map g from u cross i to u given by g of x t could g n of x t whenever x is inside it if v is an open subset of u I have to show that why this is this is continuous but this is more or less obvious from the co-induced topology peak topology what is the inverse image g inverse of v g inverse of v intersection with nth skeleton the same thing as g inverse of v intersect with u n but that is g inverse of v intersection u n is what x a n skeleton okay and at g n level it is it is a continuous function so this is an open subset of x a n for each n so this means g inverse of v is open inside here okay therefore g is a continuous function so that g is a strong deformation retract of u into u minus 1 is directly verified because once from something is in u u minus 1 all the time no matter what the second coordinate are there these functions will fix the point no matter what they are all of them are identity g n of x this f n of x is also x g n of that one is also x and so okay so that's why it is a strong deformation attraction of of u onto u minus 1 now we can have nice theorems like this which we want to remember forever you may not be able to remember immediately how this is constructed you may just whereas you should remember this way the picture here then these formulas nobody remembers them you have to work it out by yourself so that things match up correctly what you want two functions are two continuous functions when they define continuous function on the intersection of these two on the neighborhood on the set of points where in that and they must agree that's all that kind of things is what you have to do okay so what is the theorem theorem is that every c w complex is locally contractable right in the beginning we told you that these c w complexes must be having lots of euclidean property every euclidean space every open subset is locally contractable okay inside r n so that properties here also later on when we define manifolds you will see that manifolds are locally contractable because they are locally actually euclidean spaces so c w complexes are that property it's a very important property so how do we improve this one what is the meaning of locally contractable first of all even a point x and an open subset we contained in containing a point we have to have a subset of that which is open and a containing x and that must be contractable there is slightly weaker versions also but we are not satisfied with weaker version we are going to prove this stronger version so every open set contains an open set okay contains a neighborhood of that point which is contractable that that point that neighborhood is contractable this is what we have to prove all right so using the previous lemma this this this homotopy lemma whatever okay what we want to do is we want to construct it inductively okay skeleton wise so this construction is done inductively to begin away there is a unique open cell ek in x to which x belongs to so this is again and again used what is that namely a c w complex is disjoint union of its open cells so take the open cell to which the given point x belongs to okay first choose a contractable neighborhood uk of x in ek because ek is a it is some defumorphic to some homeomorphic to some is a dk right there you can find intersect it to be that will be an open set inside dk x belongs to that open set inside that you can actually choose a neighborhood which is like a bar homeomorphic to a say an epsilon bar that will be locally contracted to first choose a contractable neighborhood uk let us call it uk of x inside ek so that even the closure uk bar is contained inside p intersection ek bar you can take as small as you want closure should be there the closed wall it should be there that it and now for n greater than equal to k having constructed a neighborhood un so that the closure of un bar is contained inside xn intersection v okay so you can make this one the yi to closure so that this closure is compact and such that un minus one the other thing is a strong different tract of un okay so this is an inductive step the inductive step is carried out as follows so you have to how to how to take the un right once i do that i am done because then i put union of u and the whole thing will be over so how to construct this un so let lambda be the indexing set of all n plus one cells from xn you see i started with xn itself here or because sorry ek ek itself k great now n up to n i have un minus one i have done now i want to un i have done sorry un i have done this one un plus one i want to do is in this inductive step is like that okay u minus one here is actually equal to the disk this disk itself that uk bar that we have taken below that there is nothing because x itself is in the kth skeleton right it is in the open part of the of a kth skeleton of a cell there okay so from n to n plus one i would like to do this step now to look at all the n plus one cell index them by lambda for each alpha in lambda we have phi alpha inverse of un bar which you have constructed un bar is compact set phi alpha inverse of un bar is a closed subset of sn so it is a compact subset of sn okay contained in the open set phi alpha inverse of v so your compact subset contained inside a closed subset of a compact so compact inside v so therefore you can find an epsilon which depends upon alpha okay between 0 and 1 such that if you construct remember this step before the lemma before 2.1 this n epsilon phi alpha inverse of u alpha that is some a its closure is contained if we alpha inverse of v so all the epsilon viscose will be contained in that is one will be taken so this c is an open subset its closure is contained again okay with this way for each epsilon for each alpha I have chosen an epsilon so I have function here okay then you define un plus one to be n epsilon of un which is union of all these things along with un that is the definition as defined in 0.1 now we take u equal to un from lemma 2.1 follows that it is a contractable neighborhood of this one okay so basically it is the lemma 2.1 which is employed here okay and prior to this proposition 2.1 it actually proposes 2.1 the next theorem is about the cofibrations and so on so let me just begin this one with a lemma and then we will stop and take a break okay so x be obtained by y by attaching k cells okay see it seems that all our lemma start with this one the inductive step for CW complexes so that we can carry on with inductive steps inductive constructions so x is obtained from y by attaching k cells then look at x cross 0 union y cross i is a strong deformation retract of x cross i so this is a proof so let me go through this one and then stop again let us have this family of alpha alpha and lambda with attaching maps for the k cells one of the standard that we have proved in part one is that there is a strong deformation retract dk cross i where dk is a cell I mean dk is just the unit disc cross i dk cross 0 the bottom and only the boundary cross i so it is like a tub it is like a cylinder one top top open and bottom close cylinder that cylinder is is a strong deformation retraction of stronger to retract of the full dk cross i the the full cylinder okay so this is what we have this was one of the central results in part one let h from union the disjoint union of dk alpha cross i to disjoint union of these things is in the union of all such h that means I am taking the copies of h on each alpha it's just the disjoint union of copies of this h okay let f from disjoint union of sk minus 1 alpha cross i to y cross i be defined by using the attaching maps on the first coordinate is f alpha of x the second coordinate undisturbed t okay so these are the attaching maps for all all k cells once you have this what is the what is the space x x is the quotient of y x cross i is a quotient of y cross i disjoint union dk of alpha cross i right because x is the quotient of y disjoint union dk of alpha so x cross i is a quotient of this one this is this needs a a proof by the way in general this is not true that if x to y is a quotient map then x cross z to y cross z is a quotient map okay so you must have some hypothesis on z namely locally compactness and so on so this i is a compact sense this this is okay here this is general this is a fact by the one single identification namely this f is just composed of all f alpha right so f alpha is just x comma t t factories undisturbed x is identified as fx so this will give you the identification on the product x cross i as this as a quotient of this one that is just q cross i in any case okay so therefore what i have is y cross i disjoint union dk cross i all the cells to q cross identity x cross i i want to define a map here okay i have a map here already on the on the top on the product and disjoint union of these things okay on each on each part here it is a it is a strong distribution retract on the part why it is just identity map okay this on the part why it is just the identity map all right so what you do you have to just verify that this factors down to through identity quotient map that's all why because on the boundary wherein f alphas are being identified it is identity map the entire homotopy is identity map there that is why it works so you have got a map here to verify that this is a strong literature you have to verify it here so let us stop here and continue next time and we will come back to this important result how we are going to use this one thank you