 Welcome to module 14. Today we shall study what is called as cellular maps. In mathematical studies the subject of of mathematics is such that the object of study is not completed unless you study morphisms between them, the relations between them. Like if you are studying vector spaces you have to study linear maps between them. If you are studying groups then you have to study group homomorphisms between them. Like in the overall setup of topological spaces you study continuous functions but when a topological space has extra structures like simplicial complexes you studied simplicial maps between them. So here is the case wherein we are studying CW complexes. So here we would like to introduce a motion of what is called as cellular maps between CW complexes which is the appropriate functions between CW complexes. The most important property of cellular maps stems from the fact that every continuous map between two CW complexes can be approximated by a cellular map. So this result goes under the name cellular approximation theorem and this theorem is part and parcel of the algebraic topology toolkit. We shall indicate a couple of applications also in this course. So here is the definition. Take two relative CW complexes XA and YB and a function from XA to YB. We assume that the function is already continuous okay and then we say F is cellular if the Q skeleton of XA goes inside the Q skeleton of YB for every Q under the map F. Now you may recall what is the Q skeleton of a relative CW complex XA okay. It is the union of A along with all the open cells in X minus A of dimension less than or equal to Q. By definition a function of topological pairs XA to YB we mean a function which is X to Y such that A is taken inside B okay. So that we have clear definition of what is the meaning of a cellular map here. Now here is an example if K and L are simplicial complexes and Fee from K to L a simplicial map then if you look at mod K and mod L as CW complexes coming out of their simplicial complex structure associated map mod Fee from mod K to mod L will become cellular. K skeleton of mod K or K will go inside the K skeleton of L for every K that is the meaning of this. It is very easy to see okay. The cellular approximation theorem says the following take any continuous function of two relative CW complexes between one between one to another from XA to YB okay. Suppose X prime A is a sub complex of XA on which F is already cellular then there exists a cellular map G XA to YB such that this G is identically F on X prime A and G is homotopic to F relative to X prime. So starting with an arbitrary continuous function between XA to YB okay where X and Y B are CW complexes. You can replace it by a cellular map up to a relative homotopic that is the statement okay and moreover if this F is already cellular on a sub complex there you do not have to change. So it is a strong thing here okay. So wherever it is already cellular that part in the entire homotopic keeps success relative to X prime. So this is the statement of the theorem okay. So this is actually controlled homotopic already on X prime it is cellular so there you do not want to change it okay. The first step is the following lemma and that itself will take some time to prove after that the proof of the original theorem will be simple. So take alpha from DN SN minus 1 to YB a continuous map where Y itself is got by attaching a single cell EM to B okay. In other words Y minus B is an open cell EM if further M is bigger than N then there exists a homotopy H on DN cross I to Y such that H of X0 is alpha X X is in DN H of XT is alpha X for all X in SN minus 1 and between 0 and 1 as T varies between 0 and 1 okay for all X in SN minus 1 H of XT is alpha X and finally H of X1 the end map is completely inside B for all X inside here. So if the attaching cell here is of higher dimension than the domain here namely DN this N is smaller than EM any continuous map like this can be completely pushed inside B by this homotopy that is a way to remember it okay. So step by step we can do this one is what finally we will have to use this lemma namely see N is smaller than this M but this map may be going inside EM part not just B part we do not want that one we want to push it so that is possible is the gist of this lemma. So let us start doing this one suppose the image of alpha misses a point in the interior of EM that is it is not covering the whole of the interior of EM okay the image of alpha alpha is not covering the whole of EM suppose you have that one we do not know that one but suppose we have recall that there is a strong deformation retraction from Y minus Z to B. So the entire thing which is outside B the part of this EM can be pushed back into B there is a strong reference retraction this retraction is given by the standard retraction of DN minus Z to boundary of DN the boundary of DN it is a map which attaches to B. So from this standard deformation retraction you get a strong deformation Y minus Z to B okay. Now if alpha we see the point Z namely image of alpha is contained in Y minus Z follows that alpha is homotopic to R composite alpha see image is contained here I can composite R otherwise I would not have been able to do that and R being a homotopic to identity between the strong deformation retraction R composite alpha will be homotopic to alpha. Therefore it is enough to prove that there is a homotopic H of alpha to a map alpha 1 relative to SN minus 1 such that the image of alpha misses a point in the interior of AM. Once the image misses from there I can take this homotopic so the first we observe that if it is misses a point then the then the lemma is over okay. So now what we want to do is to establish that there is such a possibility that we can convert the original map into another map alpha 1 which has this property namely one point at least one point in the interior of AM is not in the image okay so that is the strategy all right. So now instead of using the round disc dk we would like to use the square thing namely square model namely jk jk is what minus 1 to plus 1 raised to k okay for all k. So put j equal to minus 1 plus 1 first of all by fixing a homomorphism jk to dk we shall assume that the characteristic map of the m cell em in y is phi from instead of dm it is phi from jm to y and similarly alpha is a map from jn boundary of jn to yd instead of dnsl minus 1 okay recall the lattice structure that we have introduced in the example 1.4 I am just recalling it so I do not need to go back here it is in capital Rn so this lk n denotes all those points of arc capital N with their coordinates xi equal to some integers divided by 2 power k their RIs are integers okay for all i 1, 2, 3 up to n. Now consider l2 mk equal to 2 and m equal to n equal to capital N is for m now okay if you take 2 then 2 power 2 okay k equal to so all these things will be 1 fourth right. So this will cut jm which is minus 1 to plus 1 raised to m into each side will be cut into 8 parts 8 raised to m little cubes of side length 1 fourth each by starring each phase of these little cubes and the cubes themselves okay phase and its cube phase means what 1 dimensional phase, 2 dimensional phase, 3 dimensional phase, m dimensional phase itself is also there the total thing so you star all of them you get a simplest structure on jm so that each little cube is now a sub complex okay so such sub complex structure also we have studied before okay so similarly I can do the same thing in the codomin also where this jm is going to be attached to the to to b to get y to put bi equal to i by 4 times jm where i equal to 1, 2, 3 up to 4 1 by 4 2 by 4 3 by 4 4 by 4 is jm itself let ki equal to alpha inverse of phi i of b remember phi is the attaching map phi i of b is bi or b 1 b 2 b 2 they are all subspecies of y and alpha is a map from jn to y so I can talk about alpha inverse and these things will be subspecies of jn then each ki is a compact subset because it is the inverse image of a compact set a closed subset the inverse image closed so they will be compact subsets of j power n and we have by because b 1 is the smallest contained in b 2 contained in b 3 actually interiors contained inside the next one and so on okay so k 1 will be contained k 2 contained k 3 contained in interior of k 4 each k i is contained in interior of the next one okay and this entire thing is contained inside alpha inverse of phi of entire interior of b 4 okay that is contained inside jn okay so because alpha is a map from map from jn to the whole of y so I am looking at just phi of interior of b 4 which is an open subset alpha inverse of that will contain these things by definition because b 1 b 2 b 3 are all inside itself okay interior of k 4 is contained in interior itself okay so put beta and say a subsidiary notation that is all put alpha composite phi inverse these are actually maps from ki's so largest ones interior of k 4 into interior of jm okay so now the whole idea is now I am looking at maps from one map from where from jm to j jn to jm okay okay so y b etc has gone away so now we have come to Euclidean you know Euclidean spaces this my beta is a map from jn to jm so we will do something here some analysis here and then go back to our space historical spaces via the map phi okay so we shall construct a homotopy h from interior of k 4 cross i to interior of jm such that this homotopy h of x t is beta x if t is 0 it is the starting point is this map beta or if x is outside the interior of k 4 minus interior of outside interior of k 2 inside k 4 interior of k 4 minus interior of k 2 okay so it will it will not be changing outside interior of k 2 only inside interior of k 2 we are going to change it and such that the last map beta 1 which is I have called as h of x 1 this has the property that the interior of jm minus beta 1 of the interior of entire k 4 is non-empty that means beta 1 misses at least one point in the interior of jm once you have that beta 1 you compose it with phi then you get the hypothesis here that the u alpha 1 whatever we wanted has this property so that will complete the proof of the lemma okay so we have to prove that such a beta 1 x is okay so let us carry on we can then extend c composite h to a homotopy h pi putting h of x t equal to alpha x for all x outside k 3 because outside k 2 itself its identity so put alpha x all the time compose composite phi and it is alpha is what phi composite beta why beta is defined as phi inverse composite alpha okay so so that will complete the proof all right now choose k large enough so that 1 by 2 power k is less than 1 by square root of n times the distance between k 2 and jn minus k 3 see k 1 is contained inside k 3 so k 3 okay jn minus k 3 is completely disjoint from k 2 so this distance is a positive distance all right so they are all bounded subsets within everything is within inside jn okay take 1 by square root of n times that take k such that 1 by 2 power k smaller than that note that the diameter of any little cube in ln k ln k cubes have length 1 by 2 power k therefore diameter is square root of n by 2 power k that is why I have divided by 1 by square root of n here so that if l is the union of all little cubes inside ln k which intersect k 2 then that k 2 will be covered by l but l itself will be inside k 3 it will not go it will not intersect k 3 okay that is the whole idea so this is a inverse image of so here I have drawn this one in in the in the domain b 1 k 1 will be inverse image of this k 2 will be inverse image of that and so on right they are arbitrary subsets they are not nice subsets but but we have divided it in the domain we have divided it is so fine that if I take all the little cubes which intersect k 2 okay then the union will be contained inside k 3 okay it will not go out of k 3 again as before we studied in the lattice structure we can give l a finite simplicial complex structure okay because this is the union of finitely many n cubes of size 1 by 2 power k okay so that can be also be given a simplicial complex structure okay so in proving something some theorem in c w complexes we are now using the simplicial complexes here okay so given the map is already given on l namely this beta okay it is a continuous function we want to approximate it whenever you have a function from one simplicial complex to another simplicial complex you may have to subdivide the original simplicial complex to get a simplicial approximation so after taking a subdivision l prime of l if necessary we get a simplicial approximation gamma from l prime to j m okay so this gamma is a simplicial approximation to beta so beta is defined in a larger state but I am taking beta restricted to l and then I am replacing it by a simplicial approximation gamma now what we know about gamma mod gamma is a continuous function from model means underlying topological space of l prime to j m okay and that will you homotopic to beta we know much more about it so we are going to use all these properties of simplicial approximation here okay so here is the technicalities of the analysis that we have to use namely that eta from j n to i be a continuous function to set eta on this compact subset is identically 1 and outside the interior of k 2 is identically 0 so this is sure you can you can make sure such that eta is there by by what Yuri Zones lemma because these two are k 1 and j n minus integrals k 2 are disjoint closed subsets okay so fix some continuous function like this then define this h of x t I would prefer to do simply this gamma and beta and take the linear combination but that will not work I need to do a bit of circus here what I do I take I take this gamma x mod gamma x but then I multiply it by eta x okay so now I take linear combination of gamma eta x and beta x even that will not do so I have to do one more twist here so this is the final function namely t times eta x mod gamma x plus 1 minus t times eta x into beta x for x comma t go down to l cosine okay now suppose x is l minus interior of k 2 everything is in l for this h but suppose it is not in the interior of k 2 it is not in the interior of k 2 eta is identically 0 this t times eta x into this one this first term will be 0 this term will be also 0 it is 1 minus 0 is 1 times beta x so it will be exactly beta x h of x t is beta x that is what we wanted one of the condition was that remember that yeah outside in the such that outside this one outer interior k 2 it is identically beta x okay so we have defined interior k 2 k 4 yet but we have defined it only on l smaller part but once this is there we can extend it by beta x that is the whole idea okay so that was one of the conditions now come to therefore h can be extended to a map that is what I told you it is only defined on l cross i but now we can define it on interior of k 4 cross i to by just putting it equal to beta x ignore t h of x t equal to beta x for all points outside interior of k 4 minus interior of k 2 outside interior of k 4 k 2 okay so now look at the last map here that is our beta 1 beta 1 of x is h of x 1 we claim that this beta 1 of interior of k 4 makes sense because we have extended the whole thing to interior of k 4 cross i beta 1 of this one we seized some points in b 1 itself this will complete the proof of the map the whole thing b 1 is not covered the smallest thing there that is not covered by this b 1 is not covered by entire of entire of this beta 1 this is a claim suppose x is in k 1 let us see where it has to go beta 1 of x with given by gamma 1 of x let us see x is in k 1 then eta is identically 1 okay this is 1 here so it is 1 minus t times beta x here eta x is 1 again 1 again yeah t times gamma x plus 1 minus t times beta x okay so that is the meaning of when even x is in k 1 all right and t equal to 1 so 1 minus t is 0 therefore beta 1 of x is equal to just mod gamma x okay x is in k 1 beta 1 of x is mod gamma x since gamma is a simply shall approximation to beta okay beta x must be also in beta x is always in b 1 by definition because x is in k 1 and this k 1 is nothing but beta inverse of b 1 okay it follows that this gamma of x is also in b 1 because b 1 is a sub complex okay so both beta x and gamma x are in b 1 therefore the entire of k 1 is contained inside b 1 okay so this observation is not the end of it but this helps us to finally we want to say that the whole of b 1 is not covered by vice okay so first of all is now gamma is a simple shell map is what you have to use the gamma is initial map its image is contained in the nth skeleton of b 1 because dimension of k 1 is n so the whole nth skeleton is contained inside the nth skeleton of b 1 but b 1 is of dimension bigger which nth skeleton is a very small subset the complement is an open subset non-empty open therefore if you put u equal to b 1 minus beta 1 of k 1 the same thing as b 1 minus mod gamma of k 1 this is a non-empty open set we wanted just one point this is not the end of this so what we have shown here is that k 1 misses the whole u okay but we want to say the whole of interior k 2 will miss something so let us see what happens to other points okay just the k 1 cannot cover it there are there are a lot of open subset outside gamma k 1 okay now let x belong to l minus k 1 for points in k 1 we have seen what happens take a point in l minus k 1 by the definition of k 1 beta x is outside b 1 because k 1 is inverse image of b 1 under beta so j m minus b 1 the beta x will be inside that say it is in one of the little cubes a of l 2 m but not contained in b 1 okay so in this picture now I have come outside here it may be here it may be here it may be here it may be here does not I do not know but it is not inside this shaded part the image is somewhere here here somewhere okay in one of these little cubes so that is a part here now okay so it is in one of the little cubes a of l m of 2 not contained in b 1 note that this implies that that cube does not intersect the interior of b 1 okay now since each a is a sub complex of j m it follows that we have assumed beta x is in that a mod gamma x is also inside the sub complex a simplicial approximation as this property we have again used it twice but then the entire line segment because a is convex subject entire line segment beta x to gamma x this line segment is contained inside a therefore this linear combination it is a convex combination okay eta x plus 1 minus eta x times beta x okay this thing is inside a okay so this implies that beta 1 x which is in that line segment is belonging to a but a does not intersect interior of b 1 the entire of l minus k 1 is going out of interior of b 1 okay finally now suppose x is in interior of k 4 minus l so first we took k 1 then we took l minus k 1 now we are taking interior of k 4 minus l so what happens to that which is interior of k 4 minus k 2 because l is smaller than that one then by definition beta 1 of x is beta x and hence belongs to by definition it will be inside b 2 it is outside b 2 so again this will not intersect interior of b 1 it follows that u which is non-empty is completely contained inside j n minus beta 1 of the whole of j n okay so this completes the proof of the lemma next time we will complete the proof of the simplicial approximation theorem which is much simpler than this one actually the proof okay thank you