 Today, we shall start the study of singular homology, which is one of the most important homology theory. There are different approaches, our approach is to directly construct the singular chain complex of etiological space, discuss its basic properties such as functoriality, the dimension axiom, additivity, accession and homotopynvaryons, etc. Some of them without proof to begin with, the proofs will themselves be given in a separate section. We shall also study as some special properties, which are special to singular homology. It may not be shared by other homologies. Then, we shall compute the single homology of the spheres and the relative homology of the disk modulo, the boundary of the disk. So, this is the plan. So, let us begin with the construction directly. Take it to a political space for n greater than or equal to 0. So, fix n greater than or equal to 0. We are going to define the chain groups. So, for the first time, the chain complex is going to be a non-negative one. For negative things, we are going to take them all 0. So, for n greater than or equal to 0, a singular n simplex in x is nothing but a map from mod delta n to x, where delta n denotes the standard n simplex in Rn plus 1, which is a convex hull of the standard basis elements E1, E2, En, En plus 1. A singular n chain in x, we mean a formal finite sum of these things, which are namely integer combinations n i sigma i. Sigma i is a simplex, singular simplex and n i's are integers, some total is taken, but it is a finite sum. We can add any two chains in an obvious way, namely adding the corresponding coefficients. So, the role is, if you will be see n sigma in one place and m sigma in another place, you just put n plus m times sigma. This you do for each simplex, singular simplex sigma. Obviously, there is a empty sum and that is 0, the 0 element, wherein all the coefficients of each singular and simplex is taken as 0, and that will be the 0 element of this. Indeed, by the very definition, the set of all chains, n chains, n is fixed in x, form a free abelian group with the basis as the set of singular n simplex is in x. So, it is a very huge group and we should denote it by s n of x, s for singular. And by definition, s n of x will be taken as 0 for n negative. And finally, the total, the total group, grade group is s dot of x is direct sum of s n of x over n. So, we have defined a singular simplex and then a chain and then made a group out of these n chains and then we took the direct sum. Now, we have a graded abelian group. So, now we want to make it into a category. So, given a map from x to y, if you have a n simplex in sigma by the very definition, the continuous map from delta n to x, you can follow it by f, follow it with f, you will get a singular simplex in y. So, sigma goes to f of sigma defines a map from the set of singular n simplexes to set of singular n simplexes. Any set theoretic map function can be extended to the group theoretic map because they are free abelian groups in a natural way, namely summation n i sigma i will go to f of summation n i sigma i. So, summation you can write f inside or outside, no, it is the same thing. Okay, that will give you a graded homomorphism, abelian group homomorphism f dot from s dot x to s dot y. And the degree of this homomorphism is zero, n chains will go to n chains. Once again, it is very easy to verify that if you have map from y to z and a singular simplex in x, first you take f of that and you take g of that, it is the same thing as taking g composite f of that. Therefore, g composite f of sigma will be equal to g composite f composite sigma even at the level of dots when you put the dots here, namely from s x to s y to s z. If f is the identity map from x to x, clearly f dot is identity. So, this is all that means that the association x going to s x, f going to f dot, that defines a functor now. Okay, the morphisms are defined like this. So, not only the least we make a category. So, from the category of topological spaces to category of graded abelian groups, you get a functor. So, graded abelian groups form a category that we already know. So, this is the meaning of making this construction category. So, x to s x is a covariant functor from the category of topological spaces to category of graded groups. Now, we are not satisfied with the adjust that much because we want to do homology. So, first thing we want to do is we convert s x into a chain complex. That means we have defined boundary operator here of degree minus 1 such that d square is 0. That is what we want to do. That is our next task. But before that let us see what happens some more functoriality property here. If a is some space of x, that is a natural way s dot a will be included in s dot x by composition, namely sigma is a single or n simplex in a, but the inclusion map you can just take it as a simplex in x also. So, at the saturated level n simplex is in a will form a subset of n simplex is in x. Therefore, the free abelian group over the smaller set will be a subgroup of the free abelian group over the larger set. So, this way s n of a becomes a subgroup of s n of x. So, this allows us to take the quotient group s n of x by s n of a and we shall denote it by s n of x a. Note that s n of x a even though it is a quotient here there is a nice thing namely this is also a free abelian group. What is the basis? All those simplex is in x which are not contained in a. If a singular simplex sigma from delta n to x its image is not contained in a, then this will be generator this will be an element here this will be a basic element here this will be generator for this one. This will not be taken by s n of a at all right it will not belong to s n of a. So, this is also a free abelian group for each n. Of course, by convention because s n of x is defined as 0 for n negative s n of x a will be also defined as 0 for n negative. Once again if a is empty then s n of a is just 0 group. So, s n of x by s n of a namely s n of x a is nothing but s n of x okay. Finally, if you have a map from one pair x a to y b that means a function from x to y such that a goes inside b then s n of a would have gone inside s n of b therefore you would get a map from s n of x modulo s n of a to s n of y modulo s n of b okay. So, this consideration will tell you that the association s n of x a a x a to s n of x a that also forms a covariant functor s n of x can be thought of as a special case of that when a is empty. So, this is another way of looking at it. So, simultaneously we have defined two functors here one is taking values on the pair of topological spaces another one taking values on just topological spaces okay. Similar to the this s n we defined s dot of x a direct sum of x a okay. And now let us make them both into a chain groups. The construction for this one is the same for whether x n of x a or x n of x okay it will be automatically done for both of them. So, recall that we denote this r omega or r infinity whatever depending upon the mode this is just the direct sum of countably infinite copies of r okay which is a standard inner product space okay. So, for each r greater than equal to 0 there is a face map or face operator fr from r omega to r omega which is some kind of a shift operator if you have heard of shift operator okay. So, it uses the basis elements there standard basis elements or coordination of r omega for example the rth operator fr does not shift the first es elements s less than r the first first r minus one element it does not shift the rth element itself s greater than equal to r is shifted by 1. So, es is gone to es plus 1 for s greater than equal to r. So, in effect it is an injective mapping which will miss one of the basic elements namely f er itself in the image you do not see er okay. So, once you define it on the basis elements you extend it linearly because this is just vector space these are called the face operators okay why they are called face operators look at this picture e naught even is my delta one okay this this edge is a delta one e naught even e2 is delta r2 okay this is a two simplex the standard two simplex. So, what does f naught do f naught start shifting e naught to e1 e1 to e2 and linearly extends. So, this edge goes into e1 e2 that is a face that is the face one face of a two simplex it is also you can see that this is opposite to e naught. So, f naught maps the the one face here to the opposite face to e naught like this f1 the even e e naught even will go to e naught e2 because e naught is fixed here by f1 but even is shifted to e2 similarly f2 will shift e naught even no shifting at all because both of them are less than 2. So, so e naught even goes to even f2 makes this map. So, this is opposite to e2 okay. So, this simple example is illustrative enough for you to understand what is the shift operators doing and why they are called face operators. These simplex is put as a face in the next one you compose it again that will be even smaller face of a larger simplex and so on. So, what we do is note that f r carries delta n minus one on to n minus one face of delta n opposite to our text e r okay at each n face goes to n plus one simplex as a face of an n minus one time dimension phase that is what we have to understand okay delta n minus one will be a always delta n always injective okay and the weight maps it is order preserving because it is just indexing are shifted by one at the most okay in order preserving or they are kept as it is. So, we shall denote each f r restricted to delta n its image will be in delta n plus one but we will not use separate notations for this that will be too much of cumbersome notations so much of decorations the exact meaning being understood where we are taking it okay so that is a picture there now the boundary operator can be defined easily the face operators we have already illustrated if we take the all the face operator they cover the boundary of this delta to here see so more generally if you take delta n the face operator coming from delta n minus one one by one it covers the boundary the boundary should be taken as in many calculus courses and so on integral calculus in an oriented fashion so for example here I would have taken e0 e1 followed by e1 e2 followed by e2 e0 e2 e0 is not the image of f1 by this one e0 e2 is an image therefore if I take this one I should put a negative sign of this so this motivates this definition boundary of a sigma the sigma is doing nothing if you think of this as identity map delta n delta n as identity map what you would have done if precisely being done after that you put sigma on the left side so minus one power r alternate sign f r okay then apply sigma so you apply sigma for the whole sum or sigma taken inside the same that is what you have seen so this is the definition of the boundary of any singular n simplex okay as a chain this is a singular n simplex this is a one single thing but on the left hand side right hand side we have a chain and extend this one linearly to obtain a morphism from sn of x to sn minus one of x because these are all now n minus one okay so that will give you sn of xa to sn minus one of xa because if sigma were in a then all its faces are just restriction maps you can see that okay so all their faces would be inside a so this entire chain will be also inside a okay so if a is m is a is not taken into consideration the sn of xa is n minus one of x so simultaneously we have this map the property of this map is independent of whether I treat it as sn of xa or or sn of x okay so remember we have defined sn of xa to be zero for n negative therefore the boundary of one also you defined for n action last time okay as soon as you hit this index be negative then there is no question this has to be zero so we define this also as zero okay so that completes definition as a homomorphism of degree minus one but we have yet to verify why daba and composite daba n minus one is zero daba square is zero is what we have to verify that is a straightforward computation nevertheless we will do it you know in a systematic way so the proposition is that sn of xa s direct sum of all this is s dot of xa together with this operator daba okay so we can write a direct sum of this all this is a chain complex and the association xa to s xa daba is a functor the functor property we have already verified for rest of them we can verify this functor property for daba also now okay so computation that daba composite daba equals zero it is a straightforward computation for which you need one of the interesting properties of these phase operators the composite of two phase operators if you do it in a different order namely see r bigger than s first the smaller one here than a bigger one then you can bring them back f s composite f r not f r but f r minus one so when you commute this one there is an index change here okay so r bigger than s you can bring the smaller one first but the other one will become r minus one one reduced okay so this is the formula so let us verify this one if k is less than or equal to s minus one then it will less than or minus r also right r minus one also therefore both f r and f s will not move e k at all f s of e k both both both the sides sorry both the both of them will fix e k here the here always the same thing because f r minus one will be bigger than k so f r minus one fixes it f s also fixes it so now what we have to assume is s is less than or equal to k and look at its effect on e k okay now s is less than k where the first one even if when s is equal to k f s of e k will be e k plus one for all of them f s of e k c k plus one so i have taken left hand side area f s of e k e k then what happens f r now f r will depend upon what k plus one is right f r of is f s e k if k is still less than r minus one then e k plus one will remain e k plus one but if r minus one is less than equal to k this e k plus one will be shifted by one e k plus two so this is a combined formula for f r composite f s I want to say this is equal to f s composite f r minus one of e k okay so what is f r minus one of e k in the top top case when k is less than r minus one okay f r minus one of e k is just e k but now when you apply f s it will become e k plus one okay on the other hand if r minus one is less than equal to k f r minus one will push e k to e k plus one then f s which is even smaller than e k will also push it once more it will be e k plus two okay so this proves the lemma which I usually leave it as an exercise but now I have proved it now we can prove the proposition in a very elegant way the first part of the proposition we shall show that daba square is zero for any singular n simplex then by the linearity it will it will follow that daba square of any chain is also zero okay so this makes sense for n greater than or true if n is n is one then daba sigma is already zero simplex and daba another daba will be automatically here there is nothing to prove so you can assume n is greater than or true okay so by definition daba square of sigma is daba of summation minus one to the r sigma f r r is summed over finitely many entries but I do not care which where are the other entries okay these finite sum okay this depends upon n so n plus one entries will be there okay so once again if you apply daba minus one trade so r will be there but now minus summation minus one raise to s will be also there so minus one raise to r plus says sigma of okay first f r sigma composite f r is there this is now treated as sigma f s will come on the other side sigma of f r composite f s so now it is a double summation all right so break this double summation into two parts all those are less than equal to s and are strictly bigger than s okay all those are less than equal to s f r and f r they are in the order r is smaller and c is smaller but here it is it is the other way around so I interchange them r bigger than s what is this formula f r composite f s is f s composite f r minus one that is what we have okay now what happens is s is here r minus one is here the total is one more than the r plus s whereas here the total is r plus s precisely so if you make this one total r plus s minus one then one of the minus sign will come out on the other hand what has happened to the indices indices r bigger than s r bigger than equal to s it becomes because now r minus one is replaced by r so do that systematically namely by putting s equal to i and r minus one equal to j then I can write it as a f i f j minus one is equal to i plus j i greater than equal to j okay now you see which is the i greater than equal to f i f j and r less than equal to the f r f j okay so these two sums are the same and there is a minus sign so this is zero okay given the map f from x x a to y b the fact that f hat defines a chain map from follows from the fact that in defining daba what we have done f r where is the definition of daba f r is on the far side on the right side right and how is f of sigma is defined f is f of sigma is defined uh f is in the left side so whether you take the boundary operator first or sigma first it's the same thing because they don't collide you have to take one on the left side other one on the right side first you take left side and then right side or right side first and left side side it's the same thing that is the meaning that is how the the daba commutes with the the map f okay that makes it daba functorial so f composite sigma of both f r is f f comma sigma f r and then you take the summations here that's all okay so the relative singular homology group of a pair x a are now you have made a chain complex s dot of x a tape the homology of the chain complex okay then instead of big something like this we'll just write it as h star of x a when a is empty we'll just write it as h star of x so these are called the homology groups this is relative homology group of x a the pair x a so construction of the homology groups is over okay once again if you have a chain map if you have homomorphism x a to y if you have continuous map x a to y b that gives you homomorphism chain map from s of x a to s of y b a chain map gives you a homomorphism of the homo of the homology modules homology modules therefore this association becomes functorial okay one more thing I can I make a remark if we had started with a commutative ring instead of taking summations n i sigma i you allow n i's to be varying over the ring r the scalars from r what you would have what is a free module over r so then you would have what s dot of x r s x a and h r of x r and so on instead of z you would have what r here when the ring r is z the way I have done we are not going to write that z here that's all okay so construction of this one would have been possible over any ring okay so this is where we stop next time we will start the properties of homology groups thank you