 Homotopical aspects of simplicial complex, by which we mean how to construct homotopies and what are certain basic properties, not topological in nature, but homotopical in nature, like local contractibility, co-fibration and so on. That is the topic for discussion for this module as well as the next module. Last time we did some combinatorics of links and stars and so on. All of them were what are called as closed subspaces, closed links, closed stars and so on. So, today we have a slightly different version here. Start with any simple x in a simplicial complex. Then the open star, let us say I am denoting, we try small stf. The capital stf remember was the closed star. So, this is open star of stf. This is an open subset of mod k. It is not defined as a combinatorial object, not a simplicial complex or something. It is an open subset of the simplicial complex consisting of the union of all open simplexes g. Remember, this notation is used for open simplex of g. Namely all those alphas such that support of alpha is actually equal to g. So, take all those points. Take for every g, take such a thing where g is a simplex containing f. In particular, g can be equal to f also here. So, this containment sign I am using for subset equal to also. So, the claim is that this is an open set just because you called the open star it may not be open. So, claim is that it is an open subset of mod k and it is a star shaped at every point of the open simplex f. Notice that when you take g equal to f, this open simplex f comes here. So, open simplex f is a subset of this one. Now, you take any point there, the entire set this one will be star shaped at a point is the claim. These are all easy not difficult, but you have to just verify and be sure of them. Let us all to show that any subset is open in mod k what you have to show is that its intersection with each closed simplex mod h is open in mod h. So, this should happen for every simplex. Of course, if h simplex is a single term, then this is automatic beyond that you have to verify this. So, clearly if f is not in h, then the star of f intersection h will be empty. Intersection h is empty because only things which are remember the whole space is union of such open simplexes. This was the starting point of point setcological thing that we have seen. So, here we are taking only those f which those g which contain f. Therefore, either this entire open simplexes are there or they are not there. Therefore, if f is not contained in g, open g will not intersect this at all. So, star of f intersection h is empty. So, empty set is open that is fine. So, let us take the case where f is inside h. Then I want to show that this intersection is open in mod h. Now, if f is subset of h, you can write h as f disjoint union f prime, where f prime is h minus f. Both of them will be simplexes, h is a simplex, f is a simplex. So, f prime is a simplex. But what happens is now star of f intersection mod h will be precisely equal to mod h minus this subset namely the boundary of f star f prime and its modulus. In open simplex f remember boundary of f, the modulus of boundary of f is not there. So, you take a subset of f that will represent boundary part, right? Subset of f then union with f prime, star with f prime means what this is your the join f prime is a simplex complex now being a simplex. So, any subset here any subset here union all of them union you take that is the simplex complex here modulus of that I am taking. So, all that you have to do is f prime does not intersect h. So, that part is not there none of them are there their union will be also not there because this is boundary of f part. So, none of them will contain f completely. Therefore, all these things they are subsets of subsets of h by the way h is a whole thing is inside h right everything is inside h now. So, you have to throw away this part then this precisely equal to star of f intersection modulus all other things are there. Ok, namely all those subsets is f contains inside say some h prime contains at h there open simplex will be there ok. But instead of writing all that I can just write it as mod h minus this one and this is a closed subset therefore, this is open. Now, take any x belonging to open simplex f I want to show that star f is star shape at this point what is the meaning of that take any point y in star f the line joining x and y should be completely contained inside star f ok. The line joining x and y makes sense inside both star f as well as mod g where g is the unique simplex for which y belongs to open simplex g remember every y will be inside some unique open simplex ok. Then x is also inside g because g must contain f ok. So, y is inside g this line segment makes this will be inside g because g itself is a convex open side. Similarly, star f will be there because of the definition of this ok. So, this x y makes sense inside star f intersection g ok where g is the unique simplex this just means that star f is star shape that is all every line segment for each point inside star f ok every line segment x y must be inside of that. Now, we make a definition which which we might not might have made it earlier, but now it is more relevant anyway. We say a topological space is semi locally contractible at x belonging to x if for every open set u such that x is inside u that is an open set v such that x belongs to v contained inside u and a homotopy h from v cross i to u such that h of y 0 is y which means the homotopy is a deformation starting point is identity. And the end function is a constant what constant precisely the point that we are interested in the only thing here this is like a contractibility of v, but if v is this is u this is a larger open set. If you replace u by v this is nothing but saying that v is contractible to the point x it is stronger than saying that v is you know v is just contractible that contractible to the point x is important for us, but the whole thing is taking place inside u the homotopy can go out of v, but it is it has this property ok that is the why this is semi locally contractible. However, if you replace v this u by v itself in in this one namely h is taking values inside v then we actually say that this capital X is locally contractible at x the same condition the only thing that you have to do is replace is u by v then you delete the semi locally contractible ok. So, that is the meaning of locally contractible if h can be chosen to take values inside v itself ok. So, locally contractibility is stronger than semi locally contractible ok. Now, suppose this happens at every point of x just like a function is continuous it is continuous at every point if it is happens at every point then we say x itself is semi locally contractible or in the latter case x itself is locally contractible. So, that is the meaning of this definition ok. So, now next theorem is that every polyadron is semi locally contractible take a simple shell complex take mod k I have to show that at every point there will be a given any neighborhood u there is a smaller neighborhood v of that point such that that v contracts to single point h inside u this is what I have to show alright. So, let us go through the proof you see this is not so difficult at all starting with x in mod k choose f such that x is inside open simplex f remember every every point is inside a unique open simplex because mod k is partitioned into open simplex s. So, that is used again and again the previous lemma says that star f is an open subset of mod k which is star shaped at x. So, this itself is a contractible neighborhood of the point x, but we want to do it inside a given open set. So, that is the catch here ok. So, how we how we go about it this is how therefore, first of all we have a contraction h from star f cross i to star f which is just given by h of y t equal to t x plus 1 minus t y this x is the star is apex of the star shape thing. So, any y can be joined to x by t times x this is the line segment this whole thing is inside star f. So, this is a continuous function put t equal to 0 this will be y put t equal to 1 this will be x ok t equal to 0 it is identity map t equal to 1 is a constant magnet x alright. Now, start with an open set u such that x is inside then look at this h h of x cross i x is x is the given point right h of x cross i if y is equal to x what is this map what is this one it is just x it is singleton x that is inside u u is an open set. So, this whole interval closed interval is mapped onto mapped inside an open set this is compact. Therefore, you know standard methods in point set topology like tube lemma there exists an open set v such that x is inside v and h of v cross i is contained inside u one single open set cross i instead of singleton x cross i will contain inside u. Now, you look at the same h restricted to this one what we get is this condition. So, given any u we have found a v such that and h is capital H here like that it is a little h in the application this special H here. So, this proves that every polyadron is semi locally contractable okay very easy statement very easy proof. Hello sir. The u that you choose the open set u around x you cannot be subdivided to make the star of that point inside u then we can contract it. What do you solve you subdivided? Yeah I mean if we subdivided then it will be small can it be small enough to be inside any u? How do you do that? We know that there is always one. I mean can we can we can we can take any point. k is not a finite simplicial complex. Oh okay okay. Even the point x you have taken is not at all first of all a vertex even if that is the case that may be infinitely many simplices coming and meeting at x. Okay. So, it is not even locally finite even after that you have to make x to be a vertex etc. Okay. But but cannot we have a simplices structure that makes x a vertex? That is the next thing that we want to do. Even then you know the problem is by the bare centric subdivision that we have taken that will not do the job. Okay. That is why this one has to be here. That is that will do only if it is finite simplicial complex. Okay. Okay. Once you have this semi locally contractibility later you will see that such spaces admit a simply connected covering. Okay later. So, right now you assume that then what we have proved is every connected polyadron admits a simply connected covering. Okay. That is an application of this one. Okay. Now let us come to the second point here namely the co-fibration property. This theorem says that every sub complex that L be a sub complex of K then the inclusion map model to model K is a co-fibration. Remember that we had a proposition which says that how to verify co-fibration namely you must produce a detraction from mod K cross i onto mod K cross 0 union this mod L cross i and that is precisely what we are going to prove. Retraction should be from mod K cross i to mod K cross 0 union model cross i. Okay. So, this one we are going to get such a retraction in a inductive fashion. Put ln equal to L union KN where KN denotes the nth skeleton of K. Then by the very nature L is a sub complex, KN is a sub complex, union is a sub complex and each ln is a sub complex of K. And we have the inclusion maps here L is contained inside L0. What is L0? L0 consists of all of L and all the vertices which are already not in L. Those are extra vertices coming from K0 and so on. So, ln will contain ln plus 1 and so on. Finally, if you take the union of all of them that will contain all the skeletons therefore it will be the whole of K. So, I have written K inductively as a countable union starting with L. If K is a finite simplicity complex, there are many other methods to do this. There is no need to go through this. This is precisely to take care of infinite simplicity complexes. Once we have, you know we have seen this one several times, namely if you look at a disc cross I that will be deformation retracting on to the base of the disc union, the boundary of the disc cross I. Just topologically, there is no need for going to simple simple complexes. I have written it in terms of simplicity here delta n plus 1 is representing modulus of delta n plus 1, the underlying topological space is dn plus 1, the disc cross I to modulus of delta n cross 0, the base union, the boundary which is a sphere of one dimension lower is Sn cross I. So, there is a standard retraction. I am going to use this one now. For each n greater than equal to 0, we have such a thing. So, now just you know index the set of all n plus 1 simple complexes which are in n plus nn plus 1 obviously, but which are not in L. Don't worry about things which are in L because they are going to be here which part. So, let us index them. So, let us call them as FS. For each S there is a row S like this. That is just same row, a copy of row, but I am written by row S for each S. Then what do I do? I start R naught, there is a slight different definition. Inductively, I am going to define this R naught, R 1 and R naught. R naught is L naught cross I to L naught cross 0, L cross I, that is fine, but on L naught cross 0 or L cross I, it is identity. Either X is in L or T is 0, it is identity. Then there is a disjoint union of vertices cross I which are which correspond to all the vertices which were already not in L. All those vertices cross I, their line segments, you push them to just the base point see X cross 0. Instead of X cross T, all of them goes to X cross 0. So, X is a vertex not in L. So, this is obviously a retraction of L naught cross I on to L naught cross 0 union L cross I. That is a starting point. So, in this inductive picture, we have got a retraction corresponding to this L naught correspond to L. It is not a retraction to L naught to L, but it is L naught cross I to L naught cross 0 union L cross I. So, that is the kind of thing we are going to do here all the time. So, once you have this one, you can assume that you have defined this one for up to N and you can go for N plus 1 stage. So, what I am going to do is for R N I am defining L N plus 1 cross I to L N plus 1 cross 0 in L N cross I. It is similar method. The first part it is identity namely all L N cross I or L N cross 0 to the identity X T. The second thing when they are inside a delta N, delta N plus 1 cross I, there I take this rho S because they are all indexed by rho S on F S, rho S of X T. Rho S of X T will push delta N plus 1 cross I into the boundary of that which is already inside this one, boundary of that cross I and the base, base is here. So, take X belong to F S 1. So, on the boundary we already its identity. So, these patch up. So, there is no conflict here. So, each R N is a retraction. But R N is a retraction from one index higher to one index lower. What we want is from entire K to L. So, how do we do that? Now, take any point X in mod K. This will belong to a unique n simplex inside the unique n simplex namely X inside open F. This open F may be a singleton also. Open simplex for a vertex is the same set as a vertex. In all other cases it is different. For a, for an edge it will be a open interval and so on. So, X belong to open F. That means it is inside the nth skeleton in any case where F is dimension of F is N. So, you define R of X T start with R N that will bring it to L N minus 1 then R N minus 1 and so on come up to all the way R naught. So, that you come back to all of it L cross 0 union K cross sorry K cross 0 union L cross I. So, that will be the final picture. So, really R agrees with R N. You see once a point is inside K N then it is R N and then all these are R R naught. The next stage R N plus 1 will agree with this one if it is already inside R N inside K N. K N plus 1 from K N plus 1 the R N plus 1 will already agree with R N because R N plus 1 of that part is already identity and then only R N will up to R naught R N will change it. Therefore, this compatible family of this one that is why R of X T makes sense. Restricted to each K N it is only this part if K N plus 1 if I have taken one more composition will come R N plus 1 will have come that is all. But they are all well defined and when you want to check the continuity or check the continuity on this part. And then this is the composition of finitely many continuous functions so it is continuous okay. The retraction for retraction path it follows very easily because for each point this is why this one. So, every sub complex the individual maps are co-vibration okay. What we want to do is arbitrary open subsets containing inside model. Inside that you must get a smaller open subset such that the inclusion of that pair model to you that itself is a co-vibration. So, these are the harder results all right. That is what we want to do. We would like to strengthen all these results. The first step let us have one of the exercises that we had earlier namely that every point can be thought of as a one text inside a simulation of complex. The first step is again this we have discussed several times I will discuss it again. Let mod L to SN be a triangulation mean L is a triangulating SN that is a homeomorphism. H is a homeomorphism that is the meaning. Take any point we in the interior of D N plus 1. D N plus 1 is a disk bounded by a SN. Then this triangulation extends to a triangulation L star V which is a cone. This is the the join of these two simplex is that is the another definition cone of L over V okay cone of L with V as vertex. So take the modulus of that this H hat defines a homeomorphism of that okay. This is given because given any interior point of V take any point on the boundary SN say call that as U. Then every other point has a unique expression namely it will exactly on one of these line segments U V. Only V will be in all the line segments that is why it becomes a cone okay. I am just recalling that this we have done earlier right. So this process let us call it as starring a simplex. This L is the boundary of the simplex which has been subdivided okay L could be any subdivision not necessarily delta N plus 1 boundary of delta N plus 1 that that triangulation extends to triangulation of this starring a simplex okay with whatever boundary triangulation is there that gets extended. So this process we will call just starring a simplex just temporarily if do not worry about other people may call it differently and so on. So now we can solve this problem given any polyadron x and a point x there exists a triangulation in which x is our text okay start with a simplex complex that means what I am taking about now x is now mod k right choose a point x inside mod k again choose the unique n simplex f belong to k such that x is inside open simplex f define a subdivision k prime of k as follows okay we are going to define subdivision and that subdivision will have this x as a vertex that is all okay. So what we do this this vertex is in some simplex it is in complex if you work out all other everywhere else okay there is no need to subdivide and so on somewhere something near this part you have to subdivide. So what I do first of all for all simplex is of dimension less than n choose k prime to be just k whatever it is there so k prime m m skeleton is k m up to m less than n when it is n you have to be be careful what is that except this f which is also dimension n don't divide anything other than this this f keep them as it is all the n simplex is in k are in k prime also the simplex is are as as they are that means we are not dividing dividing that one okay all right so far so what you have done is you have got into a simplex you have concentrated one simplex and f p n little x is inside in the interior of that the boundary of that is not divided okay it is any whatever it is boundary is actually a simple a simple complex now you use the previous lemma star this simplex at that point instead of barycenter you choose any open any any point in the open simplex that is all okay for example this becomes a vertex and all the lines joining these two vertices of the boundary they will be the one simplex says and so on so that is the star right okay so this will complete the subdivision of k n the nth skeleton after this there may be many simplexes which contain this f which has been divided therefore you don't know what to do and so on so don't worry divide all of them by starring at the barycenter beyond beyond this n dimension star this we have done so up till here we have done the last thing is inductively as soon as m is bigger than n extend the subdivision of this k prime whatever up to m we have got to the next one by subdividing k m plus 1 okay each of m plus 1 by starring at each m plus 1 simplex at the barycenter keep starring them put them all together you had k prime so that will complete the subdivision okay achieving that the given x is a simplex is a vertex is done at the stage c okay this does not complete the subdivision so for that you have to do the stage d also this we will use next time so today we will stop here thank you