 Having established the Exition and Merivitory Sequence, let us now give some immediate applications of Merivitory Sequence. The very first one itself is an important result, it's called Homology Suspension Theorem. There are similar results Homotopy Suspension Theorem, it's so on in mathematics. The Homology Suspension Theorem is much powerful than the corresponding Homotopy Suspension Theorem which will need more hypothesis and less and gives less concluvia. Whereas Homotopy Suspension Theorem is very easily obtained and there is no hypothesis at all you will see. X be a topological space, then there is a canonical isomorphism S, that's the standard notation so I have to use that from h n to the del of X to h n plus 1 to the del of Sx n greater than or equal to 0, where this Sx denotes the suspension of X. The suspension of X can be written as the join of S0 and X or you can think of it as a double cone, cone over X taken twice and glued them, glued together along X on the basis. So that is the suspension. So on the first module I have put a twiddle here and the second one whether you put a twiddle or not it is immaterial because n is greater than equal to 0, n plus 1 will be greater than equal to 1 and then we know that the reduced Homology coincides with the un-reduced Homology. So you can put a twiddle here or not, doesn't matter but here twiddle is necessary for n equal to 0, if you don't put that then this is not correct. So this is the first remark here, the suspension S is always path connected and hence h star of Sx is h star twiddle of Sx in positive dimensions, ok, even including zero also. Whereas if X is not path connected then H0 of X and you know there will be problem about H0 of X and so on. So you have to put H0 twiddle of X which will be different from H0 of X, only H0 twiddle of X will be isomorph to H1 of Sx, ok. So putting twiddle is a must because n I have to, I have include non-negative, the 0 also I have included, ok. So Sx is the double core, you can treat it as S0 join X, ok. You can instead of minus 1 plus 1 which is dangerous notation all the time, whether it is a real number and so on, minus of something or some, better to write as north pole and south pole, ok. Just like you are taking a sphere and then sphere is treated as a double cone over or the suspension over a sphere of one dimension lower over the equator. So you can take n of star X, S of star X identifying around on the common boundary. The first thing that you have to notice is that for any cone, this is the notation V star X which join of singleton point and X, we have V star X, you throw away the point V is homeomorphic to 0 and cross X, where 0 corresponds to this vertex, 0 cross X corresponds to singleton vertex. You could have the other notation, you would have pushed the 1 cross X to singleton point, then that would be different. So that is the question of notation here, which can be deformed to open 1 cross X, open 1, 0 cross 1 cross X can be deformed to 1 cross X because 0 1 can be deformed to singleton 1. So it follows that the suspension minus the north minus the south pole is, let us write, quality as a minus 1 and suspension minus the north pole, let us call it as a plus 1. These are both contractible, ok. First you can deform all the 0 to 1 to X and then whatever is left is a cone. So the whole thing is contractible. Inclusion map X, namely X cross 0 minus 1 to A minus 1 into section A1 is a deformation rate tract. You can collapse both the sides to the central equator. So the picture is exactly similar to take a two sphere and take X to be its equator and the two sphere can be thought of as the double cone over this one. So you can keep that in mind, ok. So A minus 1 intersection A plus 1 completely deforms to X, ok. So these are the few things that you have to observe with the topological aspects here. Both A1 and A minus 1 are open subsets because you have thrown away a single point, namely the the two vertices, respective vertices you have thrown away. So what we have is a Mayer-Euterie sequence now because these are both open. So they form an excessive couple. So I can take H i twiddle of I plus 1 A1, H i twiddle of I plus 1 minus A minus 1, the direct sum of these two. Then this is the union. Then the connecting homomorphism to H i, this is H i plus 1, this will be H i 1 lower, again the intersection here, then again direct sum and so on. Now these two end things here, H i of A1, H i twiddle of A1, H i twiddle of A minus 1, they are identically 0. Why? Because they are contractible. So putting twiddle even the 0 thing is also 0, that is what you know, ok. So this is 0, this is 0, this sequence is a part of a long exact sequence so it is exact, that means this delta is an isomorphism. So from suspension to the intersection of these two things which is liquid racks, you have what isomorphism already, ok. So you have to put it properly what you want to see, this here is an isomorphism like this, the delta is an isomorphism for H i, ok. Now being a retract eta for x to A1 intersection A minus 1, you can replace this one by x and by eta star, eta star will be an isomorphism, right. So therefore you take S s, the composite start with eta star from x to here and then go back by delta inverse. So that is the map S, homomorphism S, ok. Eta star is an isomorphism, delta is an isomorphism, I am taking an inverse image of that, the inverse of that one. So that S is from H i twiddle of x to H i twiddle of the intersection of these two to delta i inverse H i plus 1 of S x. So you see that is why the main root of the sequence is called the radius of use things. So all basic things you have done already, so it gives you the proof immediately, ok. There is a word canonical, why is it canonical? The inclusion map here is a canonical thing, ok. If you have another y, A, A correspond to A, B and B prime and so on, the function x to y would have, this is A cross, this is something x cross 0 1 and so on, cone construction. The cone construction and the suspension construction, they are canonical. So the whole thing will be canonical, this eta is canonical. We know that delta is canonical in the snake lemma, this is coming from snake lemma, the connecting homomorphism. Whenever a pair of closed subspaces x1 and x2 is given, introducing intermediate pairs of open cells u1 and u2 such that x i are deformation retracts of ui, so typical way excision theorem will be used in practice. Convert them into slightly larger open subsets which will deform to the closed open side. So that is what we have done. If you take the two cones, they would intersect along x and neither of the cone is an open set in the in the in the sphere in the suspension. So what you do, you take a larger one. So instead of this whole cone, just take one point, the two poles, the vertices one by one you remove, ok. So that is a much larger open set. Now excision theorem can be applied but as far as homology is concerned, it is the same thing as homology of this cone with air hold contracts. I mean it is contractable. So the whole thing is zero, homology is zero, ok. So this is the way you can apply, so this is standard way we can apply the excision theorem whenever we have closed sets. But this may not be always possible. For that we have ready made things like co-vibrations and so on which you have seen earlier. However, the the simplicity of this proof brings some kind of mystery. So where how to you know it is isomorphism but how to get this one. Whereas the topologically from x to sx is such a nice way of doing it, can one do that in principle at least directly in a sense that if you have some cycle here, can you suspend it and get a cycle? Yes, we can do it. Let us do that one not now. That means I would not like to give an explicit way, direct way of at least seeing what this s is at the homology level. We know that it is a nice homology. We have no doubt about that. But how does it look like? At least in nice cases at least we can we realize how it is. So this is here I am going to do it in much more generality but then it will be explicit. So for this you have to really know how the cones etc are formed. The first thing is recall that the cone construction is factorial namely when you form a topological space and then this cone along with that given a function f from x to y there is a map which you can call it as cone of f ok, cf from cx to cy. How it is defined? c of x comma t remember cx is nothing but the quotient of x cross i. So I am writing this notation x comma t for the element which is represented by x comma t in x cross i. So take x comma t to fx comma t the class of x comma t. This gives you the cone construction which has which itself is canonical namely if you have another map from y to z which is g then g composite f c of g composite f will be cf composite cg composite cf ok. So that is the meaning of it is canonical or factorial cone, cone construction is factorial. Now one more fact we want to use namely look at the n plus 1 standard simple x delta n plus 1. You can think of delta n plus 1 as the cone over it is substrate delta n ok, delta not a single point delta 1 can be thought of as a cone over this one this is a line segment and then delta 2 can be thought of as this is a triangle can be again thought of as a cone over delta 1 and so on. So that is what we will use it. In this one the apex of the cone corresponds to the next vertex, the extra vertex en plus 1. So here delta n up to you know even en is there. So delta n plus 1 vertex corresponds to vertex. So this is one way you can use the first vertex itself as a cone and so on the other way around also you can do. So different with things will give you different cone construction here. Given any singular and simplest tau from delta n to x. So I want to attack this suspension map directly at the chain level ok. So chains are after all generated by singular and simplex here. Take a singular and simplex that means a continuous function from delta n to x. This is tau n and tau s I am going to define two such things namely if you look at the upper cone namely x and star with n the north pole that will give you a cone construction of this tau itself c tau. So because there are two of them you may confuse which one. So it is c tau plus the upper part. So that will give you a map from delta n plus 1 to n star of x n star x that is upper part. Similarly tau s this is notation what is this? It is c tau again but the extra point the en plus 1 is going to s. So c tau minus I will write it as delta n plus x star wherein this c tau plus minus sends the last word takes en plus 1 to either n or s respectively. So there are two such cone constructions out of tau. And both of them are inside the suspension. Now what we do is we will take this sigma s instead of s sorry sigma tau s minus 1 raise to n plus 1 times this tau n minus tau s. You can ignore minus 1 raise to n plus 1 for the time being then see why it is there later on. Just do not worry about that one. So tau n minus tau s I have data. Automatically this tau n minus tau n will be a cycle okay. You can start with your x as just two points may be s naught minus 1 plus 1 and then take the double cone over that that will look like rectangle but you have to think of that as s1 now s of s1. So if you take the zero simplex there okay and take any zero simplex there and perform this thing what happens you can keep a track of this later on I will give you a picture right now. This is what it is definition tau n minus tau s is a n plus 1 chain. Tau was an n simplex okay singular n simplex. So when you make tau n it is from delta n plus 1. So this is an n plus 1 simplex. So I am taking a difference I am putting a sign here so it is a chain okay. It is a straightforward verification that this sigma gets linearly extended. I have defined it only on a singular simplex if you put summation n i delta y you just put that summation on this side also that is all okay gets extended to give a factorial chain map sigma from s dot x to s dot of sx okay but only the problem is that it is of degree 1 you have to be careful about that so it is degree 1 because tau which is delta n from delta n has gone c tau is delta n plus 1. So an n simplex has gone to n plus 1 simplex okay. Once you have a factorial chain map its chain map factorial reality is not necessary once it chain map it will give you a homomorphism at the homology level okay. You put a twiddle here we claim that this is an isomorphism this is nothing but this map is exactly same as the s there and if you put a twiddle here it will be an isomorphism. That isomorphism we have proved already so I want to say that this sigma star is same thing as our s okay so that should be very satisfactory for you because now you know exactly how sigma or our s looks like at the chain level itself okay that is a very satisfactory answer okay. Let us do that so remember this s was defined as delta inverse composite eta star. So if I want to prove this one it is same thing as proving put the delta on this side delta composite sigma star is equal to eta star remember what is eta eta is the inclusion induced map under this deformation so it is like identity map as far as the homology is concerned therefore what we are trying to say is delta star the delta composite sigma star is is like an identity map so it is one of the one is the inverse of the this eta star is just there because we have thickened x remember here this one eta star x has been thickened but in homology this is an isomorphism which is actually you can this is the inclusion induced map so it is actually you can think of this as subgroup which is the full group so it is an identity isomorphism itself okay so you can you can you right now you can don't worry about this eta star even if you don't mention over it whatever so that is what you have to think that this delta star is inverse of delta is inverse of sigma star okay so in order to show that this delta composite sigma star is eta star we need to go back okay to the snake lemma go through the whole step how delta was defined and then you get it there is no short way here to see how this is equal to this one okay other than going through the step of what was the definition of delta no theorems will give you this is a theorem only after that so you have to go back to the snake lemma okay so let me do that so I have copied that thing in a slightly different way now what are the things here now there were this this central two things were snakes part here which was given by exact sequences okay so this is s s dot of of a plus a minus intersection direct sum then s of sx okay this remember this s of sx was nothing but s of the image of a plus plus image of a minus by the by the excision theorem we are able to replace it by sx so don't worry about how sx looks like but look like what it is it is the sum of these two inside that that is what we are interested in okay so therefore the entire thing becomes very easy here now so how was the connecting over isn't defined from this kernel you come here this is an inclusion map then you pick up something then you push down push it down then pick up something and push it down right so take an element here you come here okay what is the element I am taking I am starting with delta I want to show that of sigma star so I start with sigma star of something okay and then apply delta and see what is it whether I get back the same element that's what we have to do okay starting with an element tau in h n of x instead of a simple instead of a singular simplex now I have a chain which is actually a cycle and then the representative class of that in tau in h n so this tau is represented by an n cycle tau cycle means dab of that is 0 right it's in the kernel of something okay and then we have sigma star of tau okay and then I want to show that delta of that is again tau or eta eta star of tau whatever if that's what you are so sigma star of tau represented by what we have actually constructed what is sigma star of tau it is minus 1 raised to n plus 1 tau of n minus tau of s right so this is an element of s n plus 1 of the sum which is a sub of s x okay already it is represented by in this way as a sum of one in that upper part and one in the lower part therefore j of tau n comma minus tau n this j is what I am j star etcetera I am just making it a simple notation j of this now I have come here this element it is j of something here already you I don't have to pick up something myself it is already there namely it is j of tau n minus minus tau of s remember j j was a1 a2 was a1 plus a2 remember that therefore this is this element is nothing but j of tau n comma minus tau n okay I am ignoring minus tau n minus 1 raised to n plus 1 for the timing just looking at this one okay you can change the sign up to plus minus later on so now look at the boundary so I have come here I have to take the boundary of that the boundary of the direct sum is nothing but direct sum of the boundary so I have to apply boundary here to get an element here so that is what the boundary of tau n comma minus tau n nothing but boundary of this plus minus boundary of that now now you must know how to compute boundary of this one this this tau is some summation for each tau what happens boundary of tau n the last vertex is has gone to always to en plus 1 okay so you have to how do you take boundary you have to apply where is minus 1 raised to i the f i the phase f i is composite with tau sorry tau composite f i okay right that is what you have to do so when you apply these things to first part part of this part you will always get boundary of tau and the last one will be en plus 1 okay but when finally when you apply the the last phase which will omit en plus 1 namely f of n plus 1 f of n plus 1 yeah that will omit en plus 1 so then whatever is remaining is just tau so this is minus 1 raised to n plus 1 times tau because the the sign has to change so this is why that there is a common minus 1 raised to n plus 1 here minus 1 raised to n plus 1 tau because this is the coefficient of tau composite fr in the in the definition of daba similarly on other side there is a minus of tau tau of this one it is boundary of tau extended by s minus minus 1 raised to n plus 1 again tau okay now i started with the tau which was a cycle therefore boundary of tau itself is zero therefore these two first terms here cancel out i mean they drop out so minus 1 raised to n plus 1 is common this is tau comma minus tau tau comma minus tau is in the precisely in the image of i so here the i is a comma a going to a comma minus a j was a a 1 a 2 going to a 1 plus a 2 those were the maps which we have defined in the mario sequence right so this is already i know exactly what element coming here it is nothing but tau the proof is over actually minus 1 raised to n plus 1 cancels minus 1 raised to n plus 1 what your sigma star of tau when delta when you apply is that tau is 1 so this is the procedure tau you see in the procedure of defining delta starting from an element here push it here pick up an element here push it here pick up an element here push it here pushing means here whatever element you got it must be read as modulo the image of the above so when you pass down it is tau represented by tau itself okay and these kernel and these co-kernel are interpreted what are they they have to carefully say that in the in the definition in the application they have interpreted as as what as certain homology namely this will be nth homo n plus 1th homology of of this complex and this will be this nth homology is n minus 1th homology in my notation i should have started with n plus 1 here okay because tau all these things are n here but actually n plus 1 and n minus 1 in the analysis here h n i have started okay so that is the difference here so proof is over okay this will also will give you now now because you have shown that this isomorphism actually is nothing but sigma star and verification that sigma star is canonical is very obvious very easy so you get a another proof that the suspension isomorphism is factorial so take a representative in the lower in x suspended is what I have to do so take tau n and tau minus n tau s and take the difference whether you take this minus that that minus this depends upon but that will only change up to sign so that won't matter much okay so now you understand why why i have minus 1 plus 1 i have put so that i will just get it out here next let us do computation of the homology of s n itself so first time we will be doing something non-trivial here now okay again i have been telling that s n is nothing but the suspension over s n minus 1 s n minus 1 suspension over s n minus 2 so finally it is iterated suspension of s naught s naught is what just two point space for which the entire homology modules we know okay all the higher homologies are trivial the zero thomology for s naught because it has two components is that direct from z but when you take the reduced homology it will be infinite cycle okay immediately you will see that h 1 of s 1 is infinite cycle h 2 of s 2 is infinite cycle h 3 of s 3 is infinite cycle and so on you would have immediately got that h n of s n is infinite cycle k equal to n whether you put twiddle or not it is okay only thing is the twiddle is necessary when you put n equal to zero when n equal to zero is s naught which has two points h naught of twiddle itself is z but if you do not put twiddle it is not z it is z direct from z okay for that reason you have to start exactly okay so what happens to other homology let us look at h 1 of s 1 okay we know h 1 of s 1 just now we said it is infinite cycle but what is h naught h naught is there is no information it is not coming from coming from this mayor vitro sickness h naught is depends upon number of path connected components is path connected so h naught is infinite cycle but h naught twiddle I have put here so it is zero what happens to h 2 h 3 and so on they are isomorphic under the under the homology to corresponding h i minus 1 of the the single point I mean s naught so they are all zero so above they are all zero therefore we have proved this one for s equal to n equal to 1 now you repeat this what you will get is all the homology up to nth stage is zero nth stage it is z and again everything else is zero beyond that so this is a picture for s n once you have this you can actually write down this is what I meant by you should have this picture so this is u comma v is represents minus 1 plus 1 inside s 1 s naught the zero sphere and this is north pole south pole this is a suspension of that okay you can think of so think of this as a circle now the sigma star of I want to see sigma star of the zero thing so for s naught what is the generator of h naught twiddle it is just u minus v or v minus u whichever one you take okay under the augmentation it must go to zero that it must be cycle okay u minus v this will generate the h naught twiddle of s naught so that is g naught so you pick up this cycle which will represent the generator in the homology okay take sigma star of g by definition now it is minus of g n minus g s write down it completely g is u minus v so it is u s minus v s I am now writing a test just like an hgc v s minus u n plus v n so these are the one simplex how do they look like look at u s u have I put an arrow here okay minus v s v s is like this minus v s will be s to v okay then minus u n u n u n is like this but minus u n will be like this plus v n v n like this so they make perfect cycle so the identity map you know trace the circle exactly once that is the generator for h one of s one okay so you have to think of this as a cycle one cycle that means it is some maps from delta one into s s one so this is the picture okay like this now you can go from here to make a cone over double cone over this one you will get s picture of s then you will get some of various triangles there oriented triangles correctly oriented triangles okay so when you study simplex homology which we will do that you will come again about this picture next thing is we can now compute the homology of the pair d and s n minus one this time we don't need marvita sequence we just need long homology sequence of the of the pair okay d n is contractible therefore if you take long homology exact sequence hi of d n this will be 0 hi of d and d n minus one hi of s n minus one the next thing again will be hi minus one of sorry what is this so this is a typo here this is hi minus one here hi minus one of d n so this must be hi minus one okay so what we have is following the long homology exact sequence hi of d n hi of d n comma s n minus one hi minus one of s n minus one hi minus one of d n this term and this term will be both 0 therefore this middle arrow will be an isomorphism since we have computed this thing now we know hi of d n s n minus one is 0 if i is not equal to n and z if i is equal to n okay so when you are treating with the narrative thing there is no need to put quiddles so next example i think we cannot do that example now it's time so let us stop here