 Having defined the singular chain complex and the singular homology of heterological space, let us examine today a few very very important properties of this singular homology. These are going to be the guidelines for the entire homology theory later on. The properties of singular homology, some of them, not all of them, will be picked up as the fundamental properties and declared them as axioms. So the first one, very first one is the functoriality. So first we have seen that from topological space to the chain complex level, it is a functor. S of XA-dava is a functor, it is a covariant functor from the category of topological space to the category of chain complexes. Then we have also seen that taking a homology is a functor from the chain complex to the graded ability groups or graded modules. Thus composing these two, what we get is the association XA going to H star of XA as well as X going to H star of X are functors, covariant functors. Given a map F from XA to YB, we shall denote F dot to be the map induced at the chain complex level and S F star for the map induced by F dot on the homology level. Quite often F star, F star is used for both of them and that makes it somewhat more confusing. So let us try to follow this convention. The star would be for the homology and dot would be for the chain complex. The functoriality is summed up in the following facts. I am repeating it, we have seen it earlier, there is nothing new but just to emphasize this one. F from X to Y, G from Y to Z, G composite F star, the same thing as G star, composite F star. If identity denotes the identity map of a space X to X, then identity star from H star of X to H star of X is also the identity map on the homology groups. These are the two factors you have to understand. So this is one of the fundamental properties of the homology. It is not just some group out of nothing. It is functorial association. That is what one has to understand. The next important thing is the homotopy invariance of H star of XA. For any reasonable space SN of X is a very huge group. Why we are taking such a huge group? We do not know because set of all maps from delta n to any reasonable space is going to be very huge. Of course, if the space is a discrete space or a single point and so on, then only we can know what it is and so on. Otherwise, it is quite heavy group. What is important is this each SN of X is a topological invariant, but it is too huge. It is a topological invariant, but H star of X is defined as some subquotient. You take some subgroup and then the quotient of that group. This is strange that this subquotient group is going to be a homotopy invariant which is stronger than being a homeomorphism invariant. So if it is homotopy invariant, then it will be homeomorphism invariant also because if X to Y is a homeomorphism, then it has an actual inverse instead of homotopy invariant. So it is homotopy equivalence also. So the homotopy invariance, we are going to state very clearly, namely if f and g from Xa to Yb are homotopic to each other, then f star equal to g star on h star of Xa. So this is a statement, homotopy invariant. The maps induced by two homotopic maps induced by them on the homology, they are actually the same. So we are not defining, we are not stating it in terms of the spaces, but in terms of the maps, which is a stronger result. That is what you have to pay attention to. From this, you will see that Xa to Yb, if f is from here to here, which is a homotopy inverse, there g is inverse is there, then f composite is g composite, it will be homotopic to identity. Therefore f star composite g star will be homotopy, will be equal to identity, which means f star is inverse of g star. So that kind of things you would have got. So the homotopy invariance of the spaces will follow if you prove the homotopy invariance of the maps. So this is the statement. So we will go a little ahead in proving this one, but we will not complete the proof of this one. We will leave it at a stage wherein things become difficult. So let us see what is the easy part of the proof of this one. Start with a g, which is a homotopy from Xa cross i to Yb. We just means that it is a map from X cross i to Y such that A cross i goes inside b. That is all. The relative means that Xa cross i, same thing as X cross i comma A cross i. So suppose this is a homotopy between f and g. Consider the inclusion map eta t from Xa to Xa cross i, which I have written, here I have written fully. It is Xa X cross i A cross i. What is eta t? Eta t of X is X comma t. So I am putting X in X cross i at the tth level. So of these inclusion maps, eta naught and eta 1 are some important ones, specific ones. So g composite eta naught will be what? g of X0 is fx. g composite eta 1 is g of X1 is gx, where g is a homotopy. When you pass to the homology, this will imply g star composite eta naught star is g star composite eta 1 star. Suppose you have a map like this, g composite eta 1 is f. So take star of that. This is g star composite eta 1 star. And g star composite eta 1 is g. So you wanted them to be the same. So this will be g. g composite eta 1 star is g. Passing on to, sorry, passing on to homology, you get this. Therefore, it is enough to prove that eta naught star is equal to eta 1 star. This is what we want to prove. If eta naught star equal to eta 1 star, then f star will be equal to g star. Okay, sorry, this is f star. This is g star. We want to prove that these two are equal. So for that, you do not have to worry about anything, but you have to just prove eta naught star equal to eta 1 star. So what are eta 1 and eta naught star? They are from h star of x a to h star of x a cross i. There are two maps. These two inclusion maps are there. They must be equal. So they are homotopic, of course. So the general homotopy has been cut down from arbitrary f and g for inclusion maps eta naught and eta 1, the coordinate inclusion maps. This is precisely what Poincare had done. So, in the case of differential forms. So, for this purpose we construct a chain homotopy H from S of X A to S of X cross I A cross between eta naught star and eta 1 star ok at the chain level. So, I should put here eta naught dot eta naught 1 I have myself not bought eta naught dot and eta naught eta 1 dot at the chain level. So, if they are chain homotopy then we know that chain homotopic maps induce same map at the homology level. So, this chain homotopy is called the prism operator ok. It is from X A to X cross I A cross I ok. So, it is a prism operator for future reference which will state this as a separate lemma. So, that we have to prove this lemma not the whole theorem ok. What is the lemma? There exists a funtorial chain homotopies there is a whole lot of chain homotopies H from. So, you can write it as H of X A. So, that is the way to write a funtor ok. This is itself a funtor H of X A it depends upon X A H S Q of X A to H Q plus 1 of X A cross I Q greater than or equal to 0 between the two inclusion maps eta naught and eta 1 ok from X A to S dot of X A cross I. We shall post from the proof of this lemma to the last section and take it for granted that the homotopy invariance is proved for the time being ok. So, what we will now start using it and first corollary is homotopy equi-l-topoliacal spaces isomorphic homology groups in particular contractible spaces ok have homology groups of a single point ok. So, both of these we will use keep using again and again contractible spaces as far as homology is concerned they are just like single point ok well do you know the homology of a single point. So, that is what we have to do now. So, luckily we can do that ok though we cannot do much of computation here. So, take n to be any positive integer what are all maps from delta n to a single point space the star bracket single where I denote the single point space ok. What are all continuous functions here what are all functions is only one function ok and it is continuous alright. Therefore, a sin of this space is nothing but a group generated by one element a free group generated by one element. So, that is in finite cyclic this is true for all n greater than equal to 0 delta naught is also a single point ok for negative things of course, we have defined them to be 0. Observe that now comes the crucial thing from s to n plus 1 star to s to n look at the boundary operator this map from 2 n plus 1 to n I want to say it is 0 map why because look at s 2 n plus 1 let us say delta 1 n equal to 0 so s 1. So, then you have a one simplex here which is a function from delta 1 to star what is boundary the boundary operator will take sigma of 1 minus sigma of 0 both sigma 1 and sigma 0 are just this point single singleton function so the singleton function minus singleton function is 0 ok in the free group. So, if you look at now delta 2 ok if you take sorry so delta 1 I have done. So, then next next thing I have to take delta 3 delta 3 is tetrahedron how many phases it has 4 phases. So, how does the function the boundary operator will look like to alternate you some of constant function all the time how many of them will be there 4 of them will be there. So, they cancel out each other and give you 0 more generally from delta 2 n plus 1 there will be 2 n plus 2 of them terms each occurring with plus or minus 1 alternatively. So, they cancel out each other that is why from from odd degree term to even degree term this map is 0 the same reason from an even degree term to the odd degree term there will be odd number of phases alternatively occurring only one of them will be left out which sign it does not matter it will be a generator plus or minus right. So, what we have is this sequence here this is this is S naught this is S 1. So, this map is sorry this is yeah what what we have what here this map is 0 this S 1 to S naught this map is an isomorphism because it goes to generator again this map is 0 this map is an isomorphism alternatively 0 and isomorphism are there. Now, when you take the kernel of this here you have to take the kernel of this which is the whole thing image is 0. So, H naught will be infinite cycling here the kernel is the whole thing the image is also whole thing the quotient will be 0 here the kernel is 0 image is also 0 quotient is 0 here the kernel is whole thing image is also whole thing again quotient is 0. So, everywhere the quotient will be 0 except the first one namely the 0th one the 0th one the kernel is the whole thing the image is 0. So, this is infinite cyclic group. So, what we get is H naught of star is infinite cyclic H i of star is 0 for all i greater than 0 of course, for negative things it is automatically 0 we do not have to mention that at all. So, only the 0th homology of a single point survives and rest of the homologies are all trivial all 0 alright. So, let us go ahead homology long exact sequence of the pair just for this one we made a lot of preparation namely the snake lemma ok. So, we will use that one and see how we get this here already we have done that. So, by definition S of X A is the quotient of S of X by S A. So, therefore, we have a chain complex here sorry a short exact sequence of chain complexes here S dot A to S dot X this is the inclusion induced homomorphism which is again inclusion homomorphism itself. So, S X to S S dot of X A this is the quotient map quotient homomorphism ok. So, this is a short exact sequence of chain complexes the snake lemma and then the theorem following that we have we have deduced a long exact sequence of the homologies what are they this is the homology of this one is just h n of A h n of X h n of X A and then delta going to h n minus 1 h n minus 1 h n minus 1. So, I have written h i h i h i h i h i h i minus 1 the last four terms you can write down h 1 of X A then h naught of X h naught of A h naught of X A after that there is everything will be 0 ok out of this ok for singleton point we know h naught of X we do not know even these groups completely. So, better to start computing this there is a long exact sequence which is factorial this sequence comes handy in many computations as we will see ok when you have information here and information here this group will be trapped between them. So, you can get a lot of information on this one now if not completely ok. Suppose these two are both 0 then this will be of course 0. So, such things we will have to keep using. So, these four properties are so fundamental ok they have been raised to the status of axioms there are some more of them which we will take care of a little later, but now we will come to some other properties of the of the singular chain complex which are special to the singleton chain complex itself singular homology itself ok this may not be true for other homology groups ok some of them may be true but it may not be true this is not part of the axiomatic kind of thing. So, this is an extra property of homology. So, homology of path components. So, I have put it as an example rather than properties 1 2 3 4. So, you will have some more property later on. Suppose you have a space x which has path components xj index by j clearly x is disjoint you know of xj. Since delta n is path connected if you take any simple x singular simple x namely a continuous function from mod delta n to x it will be inside one of the xj's therefore the set of all singular simplexes of x is just the union of set of all singular simplices of xj the union taken over j to set itself is partitioned like this. Therefore, the free group over that free abelian group over that will be the direct sum of the free abelian group over each of the xj's. So, Sx is the direct sum of xj's S of xj's the very first thing for homology of a chain complex that we have proved namely if we have a direct sum of chain complexes then the homology is also the direct sum of the corresponding chain complexes. Therefore, the homology of x is a direct sum of homology of each of its path components xj's. So, here we have used the construction of the singular homology not just some functional properties and so on. It comes out of continuous functions from delta n and therefore this is true. Alright, now we will do some more things these things are concerned in the general construction also but this is motivated more motivated by combinatorial aspects the reduced homology and H0 of a path connected space. Now why suddenly I have come to call path connected space because to compute the homology of any space x by this discussion you have to do the homology computations of each path component. Once you know how to how to do for a path connected space then you would know it for general things also. Therefore, you can concentrate on computations of path connected spaces homology of component path. So, this we have gone to do at least for H0. So, let us compute H0 of x for any path connected spaces. The first thing is H0 is what S0 then you have to take the function next function is 0 there and the previous function is S1 to x which is dhaba. Therefore, H0 is nothing but S0 of x divided by the boundary namely dhaba of S1. What is S0 by the very definition you have to take functions from delta0 to x delta0 is just a single point functions from a single point to x is just just singletons namely constant functions. Therefore, how many of them are there precisely as many as number of points in x. So, S0 of x is a free abelian group over the underlying set x itself for any one singular simple x sigma in x the boundary of sigma is nothing but sigma of E1 minus sigma of E0. E1 and E0 are the respective edges the vertices of sigma1 remember that of delta1. So, this is a phase operator. So, we define an augmentation map this is called epsilon is called augmentation map this itself will be that is what I was telling this itself will generalize, but here we are defining this motivated by computation for this one. So, what we are going to do epsilon of epsilon is defined for S0 of x to z by this by this formula epsilon of ni xi is equal to summation of ni. See this is a free abelian group over x z is just the infinite cycle group. So, just add all the coefficients and that is the function epsilon. Okay, it is the same thing as saying that all the points here are sent to one here all the points are sent to the generator one here automatically summation ni xi will go to summation n. So, then epsilon is a rejective homomorphism because x is non-empty I have to start with a non-empty space. Okay, then epsilon complete daba 1 of sigma is 0 because there is one one vertex here another vertex here two two two points are there but with negative sign. So, here is plus 1 this minus 1 so some total is 0. So, epsilon of daba 1 of sigma is 0 for all one simplex is therefore epsilon composite daba itself is 0 because these sigmas are the generators so they are the basic elements. Okay, so what we have what is a rejective map with epsilon composite daba 1 equal to 0. So, this will be taken as a definition later on for a for augmentations right now what we have done is this epsilon not how does this help to compute the homology we will see okay. Finally, what we want to show is that kernel of epsilon see here kernel of epsilon contains the image of daba 1 but I want to show that kernel of epsilon is contained in the image of daba 1 okay. So, for this considered an element summation ni xi which is in the kernel of epsilon, summation ni xi summation ni is 0 like this element here sigma 1 sigma e 1 minus sigma e 0 the sum total of the coefficients must be 0 take such as take such an element take any point z naught in x okay x is path connected therefore you can join z naught to all these x i's okay by a path what you do declare a path from z naught to x i and call it as sigma i okay and that sigma i be a continuous function from the closed interval can be thought of as a one singular simplex okay now you look at ni sigma i boundary 1 daba 1 of this one is nothing but ni sigma i of 1 which is starting sigma i of 0 or sigma it does not matter which way so starting with where we have taken sigma ni of 1 will be xi sigma i of 0 will be z naught for all of them so it is summation ni z naught here summation ni xi okay so but this summation is 0 so therefore this is summation ni xi so for each such thing we have got a one simple one chain such that its boundary is this one which means the kernel of epsilon contained in the dabao image of dabao therefore boundary operator s 1 of x is precisely equal to kernel of epsilon okay so hence we get what first time of in theorem is what h naught of x which is s naught of x divided by the boundary of s 1 of x is s naught of x divided by the kernel okay that is isomorphic to z because we have a map from s naught of x to z okay which is surjective so s naught of x modulo the kernel is isomorphic to z and this kernel is equal to the boundaries what we have verified so this is this is by definition this is h naught thus we have computed that h naught of x is isomorphic to z okay whenever x is path connected just connectivity is not enough by the way you can give easy counterexamples okay the homomorphism s naught of x to z is called augmentation okay using this we extend the chain complex by altering it at the negative minus one level that is solved okay s n of x is s n hat of x is twiddle of x is s n of x if n is greater than or equal to zero no change it is z with n equal to minus one zero again if n is less than minus one no change accordingly the daba operator is same thing as daba and n greater than or equal to one from s naught to s naught twiddle to s minus one you take this epsilon so this is the s naught to this z is becomes s minus one now right that's why this epsilon everywhere else is zero so only this path this was zero and then further zeros instead of zero we have put a z and this map was zero instead of that we have put epsilon so that is called the extended chain complex or augmented chain complex from s dot okay so the idea is if you look at the chain if you look at the homology of this one okay that will differ only in h naught level okay everywhere else it is the same thing so that is also of importance h naught twiddle of x it's called reduced homology groups clearly h i twiddle is zero for i less than zero h i twiddle is same thing as h i for i bigger than one because homologies and maps haven't changed but in dimension zero h naught twiddle of x direct sum with z will be h naught of x so the homology of this reduced thing is reduced by one factor z so it's called reduced homology in particular for a path connected space h naught twiddle of x is will be also zero without a twiddle namely un-reduced it is z that's what we have proved but that z factor goes away here so h naught twiddle of x will be zero okay so one factor down one z factor down there for h naught twiddle that's all you have to remember okay now i will tell you something about why this why this kind of thing is there for a contractible space we have we have just seen that the homology is the same thing as homology of a point for a point space we have seen that h naught of z h naught of star is infinite cyclic group z and everywhere else it's zero if you take the reduced homology what happens now the entire homology will be zero even at h naught it will be zero and this is what a you know psychologically what mathematics wanted out of a contractible space all the homologies are zero they wanted to say but that was not the case with the usual homology that we have defined that h naught will be always surviving there as infinite cyclic so you make a slight modification like this somewhat unfunctorial okay later on we can make it a funtorial also in some cases okay to make it so that the homology of a contractible space completely vanishes okay so this was the perhaps the motivation for making this augmentation and reduced homology okay so note that for any non-empty subspace a of x if a is non-empty then s twiddle of x a remember s twiddle of x a is by definition what it is given by an exact sequence s twiddle of a s twiddle of x to s twiddle of x a right so both of them have increased in minus one only in minus one dimension there is a z z factor they cancel out so what you are left with minus one factor of s twiddle of x a is nothing the same thing as s twiddle of x there is no change at all okay therefore the reduced homology for a relative pair when a is non-empty is the same thing as homology for the ordinary thing without reduced there is no change at all okay now what you can do is you can go back to the long homology exact sequence here you can put a twiddle everywhere here no problem but when you come here you have to be careful h naught of a okay this is different from h naught twiddle of a to h naught twiddle of x this is different but this is the same thing okay h naught of x so does that fit here or not you have to verify if we are h naught twiddle of a to h naught twiddle of x both the z would have canceled out here itself and so this remains as it is so this should still be an exact sequence the exactness for h naught twiddle twiddle twiddle is not a consequence of this you know snake lemma and so on so you have to separately see that one for only for this part rest of the part it was obvious because there is no problem okay that is easy to see so we can now we know what is h naught of a to h naught of a both of them are what if a suppose a is a and x are path connected both of them are z z and this is an isomorphism so like that you can compute this one even when a and x are not path connected any arbitrary path components so this can be done so it is not difficult so you can verify that this part is exact the rest of them are as as usual yeah so that brings us to the another very important property which is much more topological than whatever you have discussed out of these we have discussed the one which is very topological was homotopy invariant and the other one was this one was also somewhat topological path connectivity okay so next one is much more topological and that is called excision we will study it separately next time thank you