 Last time, we considered a somewhat easy type of homotopic properties, easier, easier results but they are all important of course. So today we will do a little deeper into these aspects. So now we have to appeal to the coherent topology more clearly in defining the homotopies also inductively. So I would like to present a result which will be applicable to any space which has this coherent topology and then we will use it for simplicial complexes, namely the topology on simplicial complexes is coherent with the skeletons and skeletons or it is coherent with each complexes, close complexes and so on, one can use that. So here is a general result of getting certain homotopies on coherent topologies. Start with the topological space, with the topology coherent with the closed family is increasing family x0 contains x1 to the x1, they are all close subsets of x, union of x1 is x, a set is closed inside x if and only if its intersection with each xn is closed inside xn, that is the meaning of the topology is coherent. Now take u to be any open subset of x, put un as intersection of u with xn. Each un is a strong deformation retract of the un plus 1, the next one which is a subset of that, un is a subset of un plus 1. Then u0 which is u intersection x0 is a strong deformation retract of the entire u. So what we are seeing is similar to the previous part if you just do not strong deformation just take retract. So we had this kind of thing coming in the previous thing retract. But now we want a strong deformation retract means we are referring to the homotopy here. So start with a sequence of strong deformation retractions fn from un cross i to un, it is identity of un when t is 0 or it is x is inside un minus 1. And the last thing namely fn of x1 is inside un plus 1. So this is a strong deformation retract into un plus 1. So this is the hypothesis. Let us put fn of x, this is just a short notation to the fn of x comma 1 which we know is homotopic to identity keeping un minus 1 fixed. And your fn itself is a retraction from un to un minus 1. That is why it is a strong deformation retract. Let us put gn equal to f1 composite f2 composite fn. fn brings un to un minus 1, then un minus 1 to un minus 2 and so on. Finally, f1 will bring the entire thing into un. So gn is a retraction of un to un. Being a composite of retraction, it is a retraction. With each of them is strong differential attraction. I can take the composites of these retractions, composites each rank. So it should be a strong differential retract to un to un. There is no problem. Moreover, gn plus 1 which is gn composite one more fn plus 1. fn plus 1 is identity on un. So gn plus 1 restricted to un is gn. Only the first n of them will operate. fn plus 1 composite fn plus 1 part will be identity on un. So gn plus 1 restricted to un is gn for all n. Therefore, I can define gx as gn of x whenever x is inside un. So this g will be defined on the entire of un. So g is from un and takes values inside un because each gn of x lands inside un. So continuity also follows because restricted to each xn means intersection with xn. It is just un on that un it is a continuous gn. So g is continuous retraction from u to u0. Only problem here is that where is the homotopy? What is the meaning of concatenating infinitely many homotopies? See when you take one homotopy here, another homotopy here, the composition of homotopies is not composition of functions. It is composition of it is concatenating the homotopies. That is why this requires better attention, more attention. It has to be done properly. How to show that it is a strong deformation retraction? I tell you more about it. So here is what we have to do. So start with gn from un cross i to un. Let us define this one inductively as follows. The first one g1 is in the first half of the interval it is x. The second half of interval it is f1. It is a composition. Composition means a concatenation like a composition of two paths for hx, a path composition. It is that. So first half 0 to t less than or equal to half it is x identity map. It is f1 which we know what it is f1. So this is just definition given. g2 onwards there will be slight difference. That is why I have to define it deliberately namely by this formula. The n plus 1 of xt will be defined three different intervals here. The entire the interval 0 1 is divided into three portions. The first portion is 0 to 1 divided by n plus 2. Second portion is n plus 2 to 1 divided by n plus 1 and all big chunk 1 by n plus 1 to 1 by up to 1. So in the last part what it is let us look at this one. It is gn of fn plus 1 xt the biggest rectangle gn of fn of x plus t when x equal to when sorry when t equal to 1 by n plus 1 it should coincide with this one because there are two definitions. To the n plus 1 what happens when t equal to n plus 1 t equal to n plus 1 here n plus 2 divided by n plus 1 minus 1 that is just 1 by n plus 1 that n plus 1 1 by n plus 1 cancels away which is fn plus 1 of xt fn plus 1 x1 actually. But fn plus 1 of x1 is by definition fn plus 1 at that point gn of that t equal to 1 by n plus 1 gn of that is just identity. Therefore, gn of fn plus 1 is fn plus 1 itself. So that is why these two coincide. Similarly, when you can check that when t equal to 1 by n plus 2 t equal to n plus 2 is n plus 2 n 1 by n plus 2 is 1 minus 1 is 0 fn plus 1 of x0 is by definition identity. So it is identity. So these things make sense and they patch up to define a continuous function. It is in this portion, small portion it is identity. Remember as n keeps increasing this will never hit you know there will be some interval this will never hit the line 0 0 cross i. 0 cross i is always here and it will be x it is identity there. So this line this one this interval keeps coming closer and closer to 0. But for all n it is still some interval will be there. So that is the nature of this and each gn plus 1 makes sense. So here is the graphical representation of g1 is half the time it is x half the time it is f1. g2 this is half the line from here it is g1 of f2 here it is f2 here it is x this is 1 by 2 and this is 1 by 3. Next time again 3 divisions but this line will be forgotten. The entire thing here from here onwards is g2 of f3 I am taking one more division here namely this one third is here one third this will be one fourth between one fourth to one third is f3 and keep this one as x to keep going going keep doing that next time you delete this one put a line on the left of this one fifth and so on so keep going define this way so this is g1 g2 g3 and so on okay each gn plus 1 is very defined and it is a homotopy of the identity map to the last gn plus 1 gn plus 1 is look at this definition here gn little gn of this already composed composed composed g gn minus g last one fn plus 1 so this will become gn of into composition fn plus 1 so gn plus 1 so starting with x here the last thing is when you T equal to 1 this is gn plus 1 okay that is a clear all right so all gn is a homotopy of the identity map gn plus 1 each gn plus 1 is a strong different retract of un plus 1 or 2 moreover restricted to un cross i it is gn therefore inductively you can define therefore you can define a map g from u cross i to u by g1 by gf x gn of xt whenever x is inside un you may be you by chance you would have shown un plus 1 does not matter the definition will be the same because gn plus 1 of xt if x is inside un is also gn of xt okay now if v is an open subset of u then g inverse of v equal to intersection with xn is equal to g inverse of v intersection with un because the domain of this inside xn it is just un this g inverse of once it is un g is nothing but gn gn inverse of v intersection xn and then for each un this intersection is okay it is open in un this means g inverse of v is open in u okay so I have I am just showing that g is continuous as as a map from u into u cross u cross i into u not u cross i into u but u cross i into finally you know therefore g is continuous okay therefore g is a strong differential tract of u into u not okay so here you have to work much harder to get a inductive homotopy okay now we come to a standard lemma in again inside dn okay so we have to work you know familiar with euclidean space you can do whatever you like and then you want to extend them to simple shape umbrella this is the game we are playing so growing viscous this lemma you can call it as growing viscous you take a subset of sn minus 1 boundary and choose any epsilon between 0 and 1 set up n epsilon equal to all those x inside dn which are at a distance from the boundary less than 1 minus epsilon norm of x is bigger than epsilon but the modulus divided the modulus the unit vector must be inside a okay so it just means that take a point inside a and then draw a small line segment of length less than 1 minus epsilon inside the disc so that is the viscous so it is growing viscous right so n epsilon is such a thing clearly this is an open subset of dn provided a is open in sn minus okay because I have taken closed intervals on one side and a half closed interval open in the interior inside inside part is open norm makes bigger okay in any case if you intersect it sn minus 1 it is precisely that the second condition gives you that this is an open subset of dn if and only if a is an open subset of n minus 1 because intersection is this way the third thing is you can push the whole thing back to a namely xt going to 1 minus 3 times a tx divided by norm x this is a point of a okay this is a strong different retract of n epsilon on to a okay if epsilon can be chosen as small as large as you please very close to 1 depending upon your requirement this is going to give you arbitrary small open neighbourhoods of this a inside dn which is a strong different retract of epsilon so this is the germ of the idea starting point which is happening in the euclidean space from this we want to capture it to and take it to all the simile complexes okay so this is a picture dn is the disk a some subset I have shown it as an arc here this may be open arc or a closed arc you don't know if it's closed arc these end points will be there which is an open arc even these end points will not be there and this n epsilon a will be an open subset okay so this remark is very important starting with an open subset of dn such that a bar is inside you now a bar is inside you allows you a bar becomes compact we can choose 0 less than epsilon less than 1 such that this n epsilon a is contained inside you so that is because this becomes compact okay so I think this much topology you know now we have the big theorem let l be any sub complex of a simile complex let u be an open subset of mod k which contains mod l then there exists an open subset v of mod k contained inside u and containing mod l such that this l is a strong deformation retract of v okay in particular we know that this will imply that the inclusion map mod l to v is a co-fibration in particular it gives you many other results first let us let us work out how you got this one okay we shall construct this v the open set as well as the strong deformation retract inductively we start with v not equal to l in the lemma in the general lemma or a theorem whatever it is this one where this theorem this u was an open subset of x we did not have to construct this one okay here we are constructing that the subset we are constructing okay put v not equal to l assume that we have constructed an open set v n of k n such that v n minus 1 is contained inside v n contained inside u this is u is given there we applied the whole thing to u intersection u n and so on now we have to construct some v n inside that and then apply it to v n itself that is our idea okay suppose we have constructed an open set v n of k n such that the whole thing is contained inside the new one the the open subsets are getting extended here okay this is not actually open subset this v not equal to l is not an open set so I am not leaving that these this is just like a getting extended to an epsilon we can just think of that way okay v n minus 1 contained here v n but v n will be contained inside u v n is retracting v n minus 1 is a strong differential attraction of v n that is a strong differential attraction v n to v n minus 1 so this is the inductive step we want to do okay we have not yet done it so now index all the n plus 1 simplex is f of k such that f is not in l because l part we do not want to trouble but v f intersects l they are nearer they are not far away v f does not intersect l you ignore them you do not they do not have to if v f the boundary intersects l put them there okay provided they are already not there that means they are not inside l for each such f n for each of f which is a n plus 1 simplex okay in the previous lemma we get an open set v f inside f such that v f intersection l is contained inside v f contained inside u and a strong differential attraction r f from v f to the boundary v f intersection l okay so how did you get this one this is by growing viscous on each of these okay this is just growing viscous lemma okay so put now v n plus 1 equal to v n union all these n plus 1 viscous v f okay f is inside n plus 1 the new n plus 1 simplex which are not already inside l okay now r n plus 1 will be defined by patching up these r fs so each r f r n plus 1 is equal to r f on v f so this will give you a retraction of v n plus 1 to v n strong different retraction okay now put v equal to union of v n this v is open in mod k none of the v n's are open okay they are open inside k n that is the point here okay so now I have to just appeal to this big theorem that we proved to say that once you have retractions like this you can patch them up so v is open inside k model is contained inside v contained inside u because each v n is contained inside u you can now appeal to this theorem to obtain a strong different retraction from r r from v to model okay the conclusion is every polyhedron is locally contractible so this is the stronger conclusion we have come up now namely which you couldn't do earlier we could only only semi locally contractibility now it's every polyhedron is locally contractible okay how do you do that so take any point x make it into a vertex by choosing an appropriate subdivision once it's a vertex singleton v will become a sub complex okay so one now you can apply when sub complex we have done this one okay so there is a strong deformation so there is a neighbor v v strongly deformally retracts to x that means v is contractible okay this time we are not using the strong difference inside u at all but inside v cross i to v itself okay so this way this is just a corollary to this strong namely every simplex every sub complex of a simple complex has this property namely it is locally n dr n dr pair there is a neighborhood which retracts strongly to model okay so let us stop here