 Last time we started the topic of partition affinity on CW complexes, we proved the major theorem there namely existence of partition affinity subordinate to any open covering assuming one of the propositions. Let me recall that proposition first here. So, it is an elaborate proposition which says that given a partition affinity, suppose X is obtained by attaching k cells to Y, then given a partition affinity on Y, you can extend it to a partition affinity on X. Moreover, the local finite ensured in the covering which ensures local finiteness for Y will get extended to the covering which ensures local finiteness for X in such a way that the sets themselves, you know extended themselves intersecting with Y will be the corresponding open subset. The open subsets themselves are also getting extended here. Before proving this one, we will need another step namely one at a time. Notice here what we have got here is X is obtained by attaching Y obtained by attaching k cells to Y. There may be infinitely many cells. So, what we do want to do is first do it for one cell at a time and that is same thing as saying that start with a open cover for DN or DAK and a partition affinity subordinate to the restricted cover on SK minus 1 on the boundary. So, you must be able to do the same thing here whatever is done here for that particular case. So, that is our first lemma we want to prove. So, let us go to that lemma now. So, this is the lemma we want to prove first now. So, you be a open covering for DN theta, theta alpha, alpha ground lambda be a partition affinity on the boundary SN minus 1 subordinate to the restricted covering U restricted to SN minus 1. Then there is an extension of theta to a partition affinity theta hat on DN which is subordinate to the given covering U. The proposition is same thing with infinitely many cells at a time the same time. And then just it is not SN minus 1 it is happening inside Y and so on. So, let us first prove this one. So, this is the major thing that we have to do. So, let beta from lambda to I be a refinement function for theta that means what support of theta alpha on SN minus 1 it is intersection SN minus 1 just to emphasize that support of theta alpha is a subset of SN minus 1 is contained inside U beta alpha intersection SN minus 1 for every alpha inside lambda. So, this is the open covering of SN minus 1 ok. Well compactness of SN minus 1 ok and because the entire family is locally finite SN minus 1 finite is compact the whole thing will be what actually finite that will be only finitely many nonzero functions here. So, it follows that we can find a uniform epsilon 0, epsilon, epsilon, SN 1 ok and opens up sets a alpha in SN minus 1 ok such that support of eta alpha intersection SN minus 1 is contained inside the these are viscous remember what this definition n epsilon of a alpha ok and then disclosure and that is contains a U beta alpha. U beta alpha is a given open subsets from the open cover. You take U beta alpha intersection SN minus 1 this you can take it as your a alpha indeed that will open subset which will contain support of eta alpha intersection SN minus 1. If needed you can take even smaller there is no need actually that itself will work. So, but instead of saying that I have said it exists as such a thing that is all. So, you take the closure of that that will contain this one that closure is contained in such one for ensuring that this closure is here you may have to take a smaller thing that is coming. Here we are using the notation of lemma 1.2 which remember this is these are collars of length epsilon of width epsilon all point coming from rays coming from the center to a point inside SN minus 1 ok. So, take a point on SN minus 1 and then take a portion of the layer of length epsilon. So, these are union of all those things as a varies over as the point varies over a alpha that will be as n epsilon. So, take the closure of that way ok. Now, let R from n epsilon SN minus 1 bar to SN minus 1 with a retraction. What is the retraction? These are points of d n remember x divided by x divided by norm x. So, that will push back push the vector back to the boundary ok. So, that is a retraction. Put eta prime of alpha as eta alpha composite with retraction. So, eta alpha is defined on SN minus 1 ok. So, now, you composite with this retraction it will defined on n epsilon SN minus 1 closure of this one ok. So, it is just eta alpha composite R. R of that will be inside SN minus 1. So, eta alpha makes sense. So, that is the definition of eta alpha prime. So, I have extended all eta alpha in a small neighborhood of SN minus 1 ok. These neighborhood I have very specifically said what it is. It follows that support of this eta alpha prime ok. This will be all these points wherein they go inside the support of a support of eta alpha here ok. So, support of eta alpha prime is also contained in the u beta. So, whole thing is contains u beta here all right. So, eta alpha primes are extended very carefully. Now, by T's extension theorem there exist continuous functions eta alpha second eta alpha people will double prime also eta alpha second prime is 1 double to 2 dashes prime second eta alpha second dn 2 I will be I mean these are extensions of eta alpha. See eta alpha prime is defined only on this closed set is a closed subset of dn. So, I extended to the whole of dn that is all. How I get extension I do not know this T section theorem that is an extension ok. And this is not good I want to control this one. So, I will do that later. So, by Eurion's lemma there exist a map G from dn to I which is 0 on the complement of u beta it is a closed set ok. u beta is an open subset of dn. So, take the complement inside dn. So, there it is 1, there it is 0 and 1 on the support of eta alpha prime. It may be eta alpha prime is contained here u beta. So, these two are disjoint closed subsets. So, I can do this one ok. Now, put eta alpha bar eta alpha bar is a modification of this extension namely G times eta alpha it is a cut off. So, this is the modification I am on. So, what is the property of this one? This will be 1 on the complement ok wherever it was defined, but on on the 0 on the complement. So, it has again it is the support will be inside u beta and 1 on the support of eta alpha prime on eta alpha prime it will not change it ok. So, all that I want is eta alpha prime with all these properties to be extended to the whole of it, but its support must be inside u beta only that is all ok. So, these are eta alpha bars ok. Next let us look at all those points where in sum of all these eta alphas is 0. Remember sum of these eta alphas is a finite sum here now ok all indexing I am writing that with that way, but the number of things which we have started with non-zero things they are only finite remaining all other things are 0 we extend them by 0 for the whole of dn no problem ok. So, this summation it is definitely equal to 1 on the boundary Sn minus 1 that is a partition of infinity ok. So, suppose this is 0 if it is not 0 this f will be empty I have no problem if it is 0 let us take all those points that is the close subset f of dn. Now, you choose a partition of infinity f a a belong to a on dn which is subordinate to this open covering some other partition of infinity ok. So, we put them these two together in a nice way so that the old thing eta alphas these these alphas they are not covering the whole thing you see they are partly 0 on f if they are if this is f is empty then I could have taken this itself divided by the sum then I have got a partition of infinity. Since, I cannot do that I have to take another partition of infinity f a ok on dn which is subordinate to you and look at all those a prime a belonging to a such that support of p a intersection f is non-empty or those are the ones which are important for me. Amongst all these so this a is actually finite set ok I do not mean I do not need that. So, look at all those points all those functions which are non-zero on f support of p alpha intersection f is at least closer to that support should not be should not should intersect f does not mean that f is actually non-zero but support of p alpha intersection f is non-empty. So, keep those functions ok let gamma from a prime to i this is subset of p be such that support of p alpha is contained inside some u gamma a what is gamma a gamma a is a point of i is a indexing point. So, it contains some u alpha because I have started with a partition which is subordinate to you right here partition of infinity subordinate to you for all a inside a prime I have a refinement function there put lambda prime equal to lambda union this extra a prime ok and beta prime equal to lambda on beta on lambda prime and this beta prime on a prime gamma on a prime ok extend beta prime extend beta to beta prime in this way ok. So, I have taken then this beta prime will be a refinement function for the new family along with it alphas it alphas. The only thing now left out is to make them into partition of infinity they are some smooth function their supports cover the whole of d n that much I have ensured. So, this is what it so this was some my f so there is some u 1 here ok for each of them there is something so I have put the whole of f inside u 1 does not matter. So, f is covered by some u are this u u gamma a a belong to a prime then there are other things which cover less rest of s n minus 1 u alpha u beta ok. You can study this picture as a model this n epsilon are this is the color of the boundary here ok. Finally, notice that on the boundary not only on the boundary on this color itself because we have defined this as attraction theta alpha bar by using this attraction some total of eta alpha bar is 1 because that is the case on this on the sphere. So, it is true on the neighborhood also ok in this this part it is identically 1 eta alpha bar sensors are 1 and therefore n epsilon s n minus 1 closure intersection f is empty because f is where some total is 0. So, these two are disjoint closed sets. So, now I take one more horizontal lemma here that phi from d n to I b a continuous function such that phi of f is 1 and phi of this n epsilon s n minus 1 bar is 0 ok. So, that the extra functions are multiplied by this function phi. So, they are not disturbed, but they do not enter n epsilon. So, there is a killing this this function is killing. So, that is what I have made phi a is all of them are multiplied by phi. So, they do not enter this color at all ok. So, that is the that is the point and in the in the inside this whatever cover for f there they are identically 1. So, their values on f that has not changed. So, that is the whole idea. Therefore, what happens is away from f this eta bar some total of eta bar is non-zero on f this summation is not 0. Therefore, this capital phi which is sum of all these will never be 0 ok and this being a finite sum of continuous functions it will be continuous on whole of the n. So, I can divide by this function eta alpha hat these are the final modification of my eta alpha we started with. First I take eta alpha bar then eta alpha prime and eta alpha double prime and eta alpha bar and eta alpha hat now ok. So, eta alpha hat is eta alpha bar by capital phi lambda a is a roll phi a into phi divided by phi. So, that now if you take the sum of these it will be equal to 1. So, you put them eta alpha hats on a in lambda and lambda a is from a is from a prime this is my extension. This is extended family ok. It is obviously partial affinity because I have divided by this capital phi which is sum total. Subordinate also we have taken because we have taken this beta prime in this exactly same way. Wherever eta alpha bar is not 0 the same thing as this one because you are dividing by non-zero function and so on. So, subordination function etcetera will not change ok. So, condition 2 and 3 has to be 1 and 4 we have verified condition 2 and 3 we have to get ensure ok. So, that theta hat actually becomes an extension there are 4 conditions remember that. So, start with eta alpha which is equal to eta alpha bar on s k minus 1. Therefore, this phi total sum total is identically 1 on s n minus 1. These things are 0. So, this will sum up to 1 here and hence eta alpha hat which is eta alpha bar ok is equal to eta alpha on s k minus 1. So, sum total is 1 I am dividing by that one. So, the sum total on s n minus 1 is this whatever you are dividing is 1 eta alpha hat will be eta alpha bar, but eta alpha bar is eta alpha that we have seen already. Therefore, these are actually extensions ok restricted to s n minus 1 they are old functions. So, this takes care of 3 ok. Finally, let w equal to w x x p not s n minus 1 been open cover for s n minus 1 which ensures local finiteness for theta. We have to get an open cover w x hat for d n which ensures local finiteness of theta hat such that w f of w hat is same thing as f of w. This f is the set of all points all indexing indexes ok wherein theta alphas the eta alphas are non-zero that is what this f is ok. So, notice that because of the compactness of d n local finiteness is obvious that is not the point, but what are the opens of sets which ensure that have you extended them that is the important point here. However, the emphasis is on latter half of the condition which has to be handled correctly. So, that has to be verified ok. So, for that I have to do this work namely define w hat of x to be n epsilon of w x remember what is n epsilon. So, in this picture if this is w x something here like w x n epsilon will be growing it up to this collar like this like that ok. Do not go out of n epsilon put up all that this much of this much of collar for that that is the definition of n epsilon ok. So, w hat of x if x is in s n minus 1 take n epsilon w x intersection with s n minus 1 it is with w x. So, that part is fine in the interior you take whatever you like actually you take the whole of d n one single set that is an open subset of d n anyway all right. So, x is the interior of d n. So, clearly the family is an open cover for d n and as a remark as you have earlier w x hat ensures local finiteness of theta local finiteness because there are no finite elements there. But the indexing sets are also correct is what we have to say we have to verify that f of this n epsilon x these are the whole things right they must be f of w x ok. There are no extra indexes coming here this is what you have to show. For this we observe first of all that if you take lambda a these are new things lambda a intersection n epsilon x n minus 1 is empty and these n epsilon w x are contained inside this one. Therefore, this lambda a support of lambda a intersection this f n epsilon is empty ok. So, w x is contained is n epsilon of w x and r home this one is retraction. Therefore, eta hat of alpha z is eta prime of z it is eta alpha r z where z belongs to n epsilon. For every point in that neighborhood we push it to the boundary and then take the value this r z is the z divided by norm z right. So, that is the condition see there from a and b it follows that both sides of this 5 are contained inside lambda the what is lambda prime it contains of lambda and a prime the members of a prime do not come here this precisely what I am saying. Support of lambda a intersection this is empty that is part a ok. So, these are both the sets are inside lambda they are ok but why they are equal because of this p from b it follows that w f of w x f of w x is what point alpha such that eta alpha is not 0 right on w x it is contained inside f of n w it is a larger thing ok because because it is retraction this is just it is contained in this and that is all it is finally take alpha belong to n epsilon f of n epsilon w f larger one take a point here ok f of that one ok what is the meaning of these there exists a set a inside this is set eta hat of alpha z is not 0 but eta alpha first z it is eta bar of alpha z that is eta bar of r z if this is not 0 but r z is a point of w x. So, I have got a point in w x eta alpha is not 0 that means this alpha is inside w x f of w x. So, every point of f of w x is there which is not actually we need to have that one one way it is enough but they are actually equal is whatever so this comes to the point of the lemma having stated that ok. So, this was the hardest part of the entire story of Paragama of partition affinity for this one ok now we can compare proposing 1.4 very easily ok and this was needed this is very important otherwise when you go to infinitely many things namely at a single vertex for example, you might have attached infinitely many one cells ok then for each of these absorbency to choose they may go keep going smaller and smaller you never know what is happening there. So, you have to control that on the other hand this part was not at all necessary if we had assumed that the CW complex is locally finite we would want to prove it for all CW complexes not just locally finite CW complexes ok. So, the last part is not very difficult at all. So, start with a open cover for you open cover for X ok theta beta alpha lambda of partition affinity on Y and beta be the infinite function and W be the open covering which ensures the local finiteness of theta ok. For each J in J let us have FJ from SK minus 1 to Y with attaching map for the Jth cell DK that is the attaching map ok and FJ from DK to X be the characteristic map ok this is defined from whole of for each each DG or each DK. Now consider theta J ok J is fixed at J is a theta J to be eta alpha composite FJ ok. So, I am pulling back the partition affinity defined on Y I do not want the whole of Y I want it only on Sn minus 1 Sn minus 1 to Y I have map F continuous function you will pull it back on on Sn minus 1. So, FJ eta alpha complete FJ. So, these will be partition affinity on SK minus 1 and support it to each one support network F inverse of FJ inverse of UIs FJ inverse of UIs which is a restriction of an open cover FJ inverse of UIs to take FJ inverse of UIs that will be open cover for DK ok. So, this is a restriction. So, now I apply the previous lemma ok for each of these now I have my entire attention is one single DK one single cell ok. So, by applying above lemma we obtain extension theta J hat let us denote them by psi J alpha where alpha belongs to lambda union lambda J. There I had lambda union lambda prime but that was only one now I have to index it it will depend upon which FJ I have taken. So, I index it by lambda J that is all ok of theta J on DK which is subordinate to FJ inverse of UIs what is the cover? This is the cover it is open cover for each eye ok with the refinement function again beta J from lambda union lambda J to eye which is an extension of beta all these the lemma guarantees me for each J I have an epsilon there but that epsilon I have to index by epsilon J 0 less than epsilon J less than 1 as in the lemma for each W inside W ok and each J inside J look W epsilon J to be take FJ inverse of Ws ok for each W FJ inverse of W take then take N epsilon J of that ok all these if you drop out this J here and F here and so on this is the this is the lemma. So, instead of arbitrary W now F inverse of J FJ inverse of Ws there that is all ok let us take this one. So, this is shorter notation epsilon WJ instead of big this one. We then have an extension of the members of of the open cover FJ inverse of W of Sn minus 1 to opens of set this W epsilon J W belongs to W these are open subsets of dk such that their intersection with Sn minus 1 is precisely FJ inverse of W because it is N epsilon J of that is an extension of this that is the property that for each J the index the indexing sets F of set of all indexes here F of W epsilon J is simply as F of the intersection with Sn minus 1 FJ inverse of W ok and that is precisely equal to FW because I have taken FJ inverse of FJ the function to be precisely what is eta J of FJ ok that is why it is FW ok whatever happens inside Y all those only I have kept there here this is the restriction this is the pullback of the cover of the functions on Y ok. Put lambda prime equal to lambda union all these lambda Js as J varies over the indexing set now I indexing set as J ok that beta prime with extension of these all these on lambda it is beta and lambda J it is beta Js. So, take that one all the beta Js we have extended for each alpha belong to lambda prime define rho alpha prime from the disjoint union of Y and all the dk's on the disjoint union you define first namely on as follows on Y it must be always eta alphas no change there ok on Y let rho prime of alpha eta alpha where alpha is inside lambda for all the new things it must be 0 ok that is what we want anyway we do not want the new things to enter Y right. So, on Y it is this these are the functions ok on dk alphas what are they on dk alphas for each dk J they must be rho prime of X must be the new psi alpha J of X if alpha is in this one for other lambda Js of course they must be 0 because they are disjoint anyway. So, from all other lambda Js must be 0 ok. So, this is now the collection of this all rho alphas ok they are defined on dk the additional dk union Y but when you attach along the boundary what happens to this one they will agree with the corresponding function in Y here ok because this is just eta alpha composite the functions where nothing but what eta alpha composite f alpha f f Js ok. So, they all these functions will now factor down to define a continuous function from X to I for each of them there will be a rho alpha prime I will drop that prime rho alpha from X to I if you look at a quotient map. So, rho alpha of Q X X is some point there is equal to rho alpha rho alpha prime of X that is the way we have defined. So, it is well defined because of on the boundary s n minus 1 f they are f Js given by f Js ok. So, we claim that this family is required partition of infinity which extends the order in a partition of infinity on Y ok that is very clear because that these on Y they agree with this one by the very definition that part is fine ok. So, let us verify one by one if alpha is in lambda then rho alpha restricted twice eta alpha ok on each phi J of the decay of J. So, rho alpha composite phi J is rho alpha prime which is psi J of alpha then support of this one is contained with u beta, but on lambda it is equal to beta prime ok capital lambda it is same thing as this one. Now, suppose alpha is in some lambda J and such a thing is unique right alpha is either in lambda or in one of the lambda Js. Then support of rho alpha is phi J of see support of psi J is with psi J of alpha ok I have to take phi J phi Js are the characteristic functions right phi J of support of that is inside u of beta Js ok. So, because we have pulled back this the inverse image of these things for the original cover ok. So, this u of beta J, but beta Js are by definition beta prime is beta J when alpha is in lambda J. So, it is u of beta J. Therefore, the family theta is subordinate to the open cover u with beta prime as the refinement function ok. Now, let us take w belong to w the old local finiteness steering covering for wire put w epsilon ok I want to extend each of them how equal to q of each w epsilon alpha this is the w epsilon J epsilon w epsilon Js are inside happening inside each disk d J right. So, each disk you take this column take the union of all that and q of that the image of that ok. So, that is my w epsilon automatically this w epsilon becomes open subset in x because intersection with each d J if we if you take the inverse image it will w epsilon J ok. So, each w epsilon is an open subset of x also for each J let us take w J denote the interior of tk phi of that. Remember phi J is a homeomorphism on the interior. So, look at the copy of that that is my w J ok. Now, you take all the w of silence w is coming from capital W or curly w add this extra interior of each cells that is an open cover for the whole of x. And that will ensure local finiteness for theta add as far as all this y's are concerned it will come from w epsilon inside these things inside each d J only things which are non-zero in in this interior of this one are coming from d J that is will be only lambda J. So, it is a local finite this will ensure local finiteness ok. Finally, we have to look at this patching of the index in sets themselves f of w J for each J f of w J set of all alpha and lambda J lambda prime support of rho alpha intersection w is non-empty it is a definition ok. As soon as you write J here this will be happening in lambda union lambda J right for each J this is what it is it is it is inside this particular thing. Now, by the compactness of d K and local finiteness of theta J it follows that only finitely members of theta J's are non-zero here. So, therefore, it follows that this f of w J is finite. Similarly, for w x or w first observe that f of w epsilon is all lambda inside all affines lambda prime support of rho alpha intersection w epsilon is non-empty. So, this may be because w epsilon's are infinitely many w epsilon J's are there ok. But, the whole thing is none of them will contribute anything from acquire anything from these lambda J's this is the whole important thing. So, it is a subset of lambda itself this is this is the key here. So, this is the final thing here. So, we have extra thing if you allow one of them then you do not know how many you have allowed. So, I do not allow any of them. So, whether they are finite any finite does not care. So, no no things come here. So, it is considered lambda. But, then f of w epsilon is same thing as union of all these w J's because that is the definition of w epsilon ok. And, but each of them is f w's here ok. So, finally, this summation is similar to the final theorem also we add this one for any x in x suppose x is inside y then rho x is 0 for its lambda alpha x is not in lambda. All these lambda J's they are all those functions are 0 right. So, only the summation here in lambda prime summation only alphas which are inside lambda will contribute. But, then this is a partition of unity on y therefore it must be x. So, it is eta alpha of x. So, it is 1. Otherwise x is inside one of the phi J of interior of d k alpha d k d k of J ok that a J is unique also ok. Then x equal to phi J of x prime for some x inside that because it is in phi J of this one. In this case rho alpha x is 0 for all alpha not in lambda J prime where J prime is for all of a J inside J prime where J prime is not equal to J only when it is in lambda J ok you will get something. So, this some total will be either it is in lambda J or in lambda ok. So, all this big thing does not come. So, only lambda in lambda J will come summation because all other things are 0 rho alpha of x is 0 right because x is infinite in d k of J that is why. But, these are by definition psi J alpha ok on this one on this d k it is a partition of unity therefore it is equal to 1 ok. So, this completes the proof that theta hat is an extension of theta. So, this part is easy it is more or less the repetition of the final theorem that we have proved in some sense except that this is the key thing here which ensures that the local finiteness is censured here because of this one ok yeah. So, I have told you already that the existence of partition of unity with horse dorseness actually ensures that the space is a paracompact. If you do not know paracompactness you do not have to worry about that right now. Paracompactness is essentially in practice it is essentially to prove the existence of partition of unity. That is the key in all the analysis everywhere in algebraic topology or differential geometry wherever you want ok. But, some basic points of topology has to be done like that that is why Dio Donner thought about this partition of unity ok. So, that was the theorem that I was telling corollary every CW complex is a paracompact. I think this is where we stop today. Thank you.