 Following the computation of the homology groups of the spheres and some specific generators in Hn of Sn for each n. Let us make a little more computation here with one definition first, take any continuous function from Sn to Sn, look at the induced map f star homomorphism f star from h star hn of Sn to hn of Sn, this we know that they are both this infinite cyclic group anyway. Fixing a generator for this infinite cyclic group, let us denote by gn, let f star of gn will have to be some integer times gn because these are infinite cyclics generated by the same gn. This integer k is called the degree of f, so this is defined whenever you have a map from Sn to Sn, the degree would be any integer in general, let us try our hand in computing this degree in some very special cases, obviously if you take f to be identity map, then f star will be also identity map, so gn will go to gn, so k will be 1, so identity map as degree 1. Another thing important thing you have noticed is that the homomorphism f star depends only on the homotopy type of f, if you change f by a homotopy, the degree does not change because f star does not change, so let us now consider the functions which are reflections, reflection in one of the planes, let us take the first simplest one namely in the first coordinate itself, the zeroth coordinate itself, x0 goes to minus x0 and all other coordinates are going to the same x1, x2, xn, only one coordinate changes the sign, so that is a reflection. On Sn itself which is a subset of r, Sn remember is just the unit vectors in r which means minus 1 plus 1. So we have carefully avoided writing minus 1 plus 1 because that can confuse with the real number or the integer which we are going to use. So we have denoted them by u and v, what makes, what does r make, the effect of r on that one, u will go to v and v will go to u, so therefore r star of g0 will be minus of g0, it just interchanges the u minus v is the generator g0, remember that, so it will be v minus v, so it is minus of g0, therefore this degree is equal to minus 1 on Sn. Now on Sn what happens, the map is the same thing, minus x0, x1 going to minus x0, x1. If you think of this as the complex number, it is somewhat similar to the conjugation but the conjugation is taken not when the first one is the minus of the conjugation. So on S1 it is again changes u and v but the north pole and south pole are kept as it is. So go back, let us go back here on S1 which is sigma star of g0, it is u s minus v s minus v u v u n minus v n, so u and v are interchanging, so the sign of this one changes again. So therefore on S1 also the degree is minus 1, inductively you can see that each time u and v are the only two things change, all the later the vertices which come they are fixed, it follows that degree of r is always minus 1 in all dimensions. So next we can take other reflections instead of the first coordinate, the 0th coordinate, like if you take the second coordinate in S1 then what is it, it is actually the conjugation, does not matter what it will do, it will keep u and v fixed and n and minus, n and s it will interchange, therefore again the degree will be minus 1. So only one coordinate can change, what happens is the generator changes its sign, therefore for all r i, r i of x naught x n which is by definition x naught x i minus 1 kept fixed z, i th coordinate become minus x i, x n plus 1 x n kept fixed z, so it is the reflection in the yet axis. So they all have the same property, namely r i star of g n will be equal to minus g n. Now let us consider the antipodal map, antipodal map changes the sign of all the vectors, all the coordinates, v goes to minus v means all x naught x 1 x n going to minus x naught minus x n minus, therefore you can write it as the composite of r naught r 1 r n, n plus 1 reflections, there are n plus 1 reflections. By punctual reality alpha star will be the composite of r naught star r 1 star r n star, therefore it follows that degree of alpha is minus 1 raised to n plus 1. So we have completed the degree of alpha, it is very effortlessly here. This could have been done and it is done in all differential topology courses also. The degree concept comes from different topology, it actually comes from all the way from complex analysis. Here is a quick and a very interesting theorem which you must have learnt elsewhere also, alright. After fundamental group you have proved this one. So I do not waste so much of time but indicate once you know the degree of the antipodal map you have this theorem which is called Heribol theorem in differential topology which is roots in differential topology. What it says there is no continuous function f from s into s n such that f of v is perpendicular to v at every point at every for every v, v and f we are perpendicular. So such a function is actually called a non-vanishing vector field. Actually it is a unit vector field because I want it to have f of v is also a unit vector. So there is no continuous function, no continuous vector field which is a unit vector field on an even dimension sphere. So it is the statement, okay. The proof is one line proof namely if there is such an f and I can consider a homotopy h from s to n 0 pi to s to n given by h of v, theta equal to cos theta v plus sin theta f v. Remember v and f we are perpendicular to each other. Therefore if you compute the norm of this one this will be also equal to 1 because cos theta v plus sin theta f v is there, okay. The norm of h theta v will be again 1. So this will be all the time taking value in s to n because f is continuous this will be continuous also, okay. So at theta equal to 0 it is v which is identity, identity map and theta equal to pi it is minus v, it is anti-portal map. That means that identity map and anti-portal maps are homotopic to each other and that is a contradiction because identity map has always degree plus 1 and when n equal to when this is 2n, so the degree here of anti-portal map is minus 1 raise to 2n plus 1 which is minus 1 and that is a contradiction, okay. So let us carry on with one more comment here. In part 1 we had introduced a degree of a map from s1 to s1 using just the fundamental group. Actually in computing the fundamental group itself even before computing it we had the notion of degree and we used that one to compute the pi1 of s1 that show that it is isomorphic to that. That degree and this degree are actually not two of two different ones, they will be the same though the definition is different. So we will see this one later on, okay. So now let us compute go back to Mary Wetter sequence and do some more computation, computation with the this time with the torus. By a torus I mean the product of s1 with itself, s1 cross s1, okay. So to apply Mary Wetter sequence just like in the case of s1, so this time we will break it up, break up the second factor, okay. The first factor s1 I will keep it as it is. The second factor s1 I write it as u plus minus where u plus minus is s1 minus minus plus. So you subtract from the minus 1 then you call it as u plus and subtract plus 1 you get u minus. So u plus and u minus are biggest neighbourhoods of of plus 1 and minus 1 respectively, okay. Then s1 cross u plus, s1 cross u minus, they are open subsets they cover the whole of s1 cross s1. So we can apply Mary Wetter sequence here because they are excessive couples, okay. So Mary Wetter sequence 17 can be applied to get an exact sequence, okay. But before writing down this we will make some simplifications. Namely s1 cross s0, you see s1 when you subtract the two points minus and plus 1, okay. You have the other 0 1 and 0, 0 1 and 0 minus 1, okay. So they do not s0. So s1 cross s0 is sitting inside s1 cross u plus minus s1 cross u minus is a strong deformation retract, right. This we have seen in the computation of homology of s1 itself. So therefore we can replace the h star of the intersection by h star of s1 cross s0. So as if it is two copies of s1 cross s0, s1 cross s1 itself, s1 cross 0, s1 cross minus 1, s1 cross plus 1, okay. There are two copies of s1. So this we can do in the exact sequence. This is one of the terms. So that term I will not write down. I will write h star of s1. Okay. What is the exact map is? Exact map is inclusion map here from here and from here whatever i star we are taking. Okay. So we get such a exact sequence. So h2 of the total space then you have the connecting homomorph h1 which comes delta h1 of the intersection which I have replaced by s1 cross s0. Then you have the i star here which is defined as a comma minus here, right. So h1 of s1 cross u plus 1 copy. Other copies s1 cross u minus. Okay. Then you have g star which is a1 a2 going to a1 plus a2. Remember that again into the entire space namely h1 of s1 cross s1. Once again there is a connecting homomorph delta to h0 again to h0 of the direct sum to h0 of the total space. So this is where the exact sequence stops. What are these terms? The term before this what is it? This is h2 of the direct sum. What are h2 of s1 cross u plus? We have computed h2 of s1 cross u plus the same thing h2 of s1 because u plus is contractible. h2 of s1 we have computed 0. So like that all the terms before that they are all 0. We do not get any other information and that is the information. That is the good information. All the terms beyond this they are all 0. In particular all the homology groups hi of s1 cross s1 which we want to compute for i greater than equal to 3 they are all 0. Because of the same fact about h1 of s1 they are 0. Hi of s1 they are 0 for i greater than equal to 2. So there are some more information we have to gather here to compute what exactly this group is and what exactly this group is. These two things we want to compute. Other things we know but we have to unravel what are these i star, j star and so on. Each time depending upon what information we have conclusions may be different. The first term in the above sequence actually corresponds to h2 of s1 cross u plus h2 of s1 cross u minus that is why it is 0 which I have told you I have repeated. Not only that every term beyond that for the same reason they are all 0. Now in particular the first entry is 0 as we have indicated above. For the same reason the third, fourth, sixth and seventh terms are isomorphic to z direction z because of our computation which I have named computation 19. What are these? Third, fourth, sixth, third term is h1. This is h1 of s1 h1 of s1 so z direction z. The fourth, this one we want to compute. We do not know what it is. This star h1 h0 of s1 cross s0 there are two copies of s1 therefore it is h0 is the h0 of a connected thing is infinite cycle. There are two copies so this is again z direction z which is h0 of s1 and h0 of s1 again z direction z. Finally this is h0 of s1 cross s1 which is connected which is z. Not only that this map take just this one this one as 0 this part for example 0, some a where does it go? This is the inclusion map from h1 of s1 cross s1 cross u2 s1 cross s1. Therefore on the connected component this is an isomorphism this part itself. It just means that it is this part is surjective. Together both if you take this this direct some 0 or 0 direct that they will be isomorphic j star but together they will be just surjective you cannot be isomorphic because this is just z and this is z direction z. Both the generators here 1 comma 1 they both go to the generator here that is the meaning of this one. So having seen these things what we have is little more information here. This we do not know this is h1 of s1 cross s0 this is the third term. There are two copies right so h1 of s1 cross 1 h1 cross minus 1. So both of them contribute so z direction z. This also z direction z because this is direct some of two things. This we do not know this is again two copies z direction z. These two copies z direction z is only one connected thing it is z all right. Now this is surjective any surjective map into infinite cycle it will split. Therefore this will look like the kernel of this directs on this itself. But the whole thing is z direction z therefore the kernel must be infinite cycle. So kernel of j star is infinite cycle by the exactness that must be equal to image of i star. Now this is z direction z and its image is z image of i star z exactly similar thing here instead of this whole thing you can write i star going inside an infinite cycle group. The same reason what happens is the kernel of this when you do infinite cycle which will be equal to image of delta star. So I have come some information all the way. Image of delta star is infinite cycle group once again the same kind of thing happens so that infinite cycle group becomes a direct summoned of this one. This I do not know now okay its kernel will be equal to image of j star. So this h1 of s1 cross s1 is what is isomorphic to the image of delta star directs from the kernel of delta star which is equal to image of j star. So I have to see what is the image of j star okay. So let me repeat it whatever I have given j star is surjective on h0 level and hence its kernel is isomorphic to z therefore the kernel of i star is also infinite cycle. Hence delta okay. So I have come up to infinite delta here from h1 of s1 cross s1 to h0 of s1 cross s1 infinite cycle image okay. Now let us look at what this this map j is self is j of 1 cross 0 and j of 0 cross 1 you can think of that through generators here okay generator here 1 let me write it as 1 and this part this part directs from 1 cross 0 0 cross 1. So we can think of this way okay the property that I am going to use is j what is j star j star of a1 a2 is a1 plus a2. So it follows a j star of 1 cross 0 has to be non-zero and j star of 1 comma minus 1 which is 1 cross 0 plus 0 cross 1 which is which will be equal to 0 okay 1 cross 0 goes to 1 0 both of them go to 1 and minus 1 that will be 0. So the two things have the property that here there are two generators z and z okay each of them if you restrict z cross 0 or 0 cross z it is injective together okay 1 comma minus 1 would have gone to 0 right. So kernel is precisely infinite cyclic therefore the image is infinite cyclic because only one component survives okay. So image of j star is infinite cyclic h1 of h1 cross h1 we observed is equal to image of j star direct sum the kernel of delta star which is infinite cyclic. So this is isomorphic to the direct sum z it is important that the delta the image itself is infinite cyclic so it is subjective map on to infinite cycle so it splits that is important otherwise you may not say that this is a direct sum there are other kinds of groups also okay just subjective map does not mean that it is a direct sum this is infinite cyclic group it will split up. So h1 of h1 cross h1 in infinite cycle we have computed this one after all this now the effort is easier because we have already come up here this we have shown that this is z direct sum z its image is just this side here. So kernel of j star is again infinite cycle which means image of i star is infinite cycle okay now here is a again it is ranked to z direct sum z image is infinite cycle therefore its kernel is infinite cycle the kernel is image of delta image of delta is infinite cyclic so you can write 0 to whatever this group and then to image to 0 and then this will be isomorphism so delta gives you an isomorphism of h2 of h1 cross s1 with its image because it is after its injective mapping okay and what is that that is infinite cycle. So this will complete the computation of h2 of h1 cross s1 also the second homology of h1 cross s1 is infinite cycle the first homology is infinite cycle cross infinite rank 2 okay the 0th homology because it is a path connected space it is infinite cycle so that completes the computation of of the homologies of s1 cross s1 all other groups are 0 because beyond h2 h1 h i of s1 are all 0 so you go through it again and again because this is a typical way not all the time you will get all this you are lucky enough to have many things here these things but this is the kind of argument you have to use whenever you are you know stuck up with mayor vitre sequence and you want to extract some information okay. So let us sum it up h i of s1 cross s1 is z i equal to 0 and 2 in the middle h1 it is z direction z and 0 otherwise so the example is a typical way mayor vitre sequence can be employed in specific situations of course there are more elegant proofs of this this result they are called connex formula and so on which will not be able to do in this course okay. So I want to make a general remark here the functoriality or the first thing that means the homotomy invariance or the second property of all properties of the singular homology the homology exact sequence of the pair or the third one then the excision property or the fifth one okay these things are so so important that they have been put into you know they have been raised to the status of axioms for homology. Of course this is genuinely by one single author I mean set of authors namely island vergand steenrod this is done at around 50 when they established the entire singular even singular homology itself is because of them the most of one in deriving a certain result concerning singular homology we need not appeal to the actual construction we can just use these properties in a genuine way in a in a clever way and get many many results therefore all such results will be true for any other homology whatever quote unquote which satisfies these axioms remember we have assigned homology for any chain complex that is not what they are at you could have got any homology which arises maybe without chain complexes you need a category of topological spaces that may be differ that may differ it may be smaller or different kind of topology you know instead of all topological spaces you may have to take some special class of topological spaces then you can create your own homology in a very different way what do we mean by homology it should satisfy some of the very first thing is functoriality nobody cares about anything which is not functorial that is genuine homotopy invariance is another thing the homology exact sequence because a chain complex the our homology arose out of chain complex because same plan now it was a free gift but if you are not coming through a chain complex it may not be there there are such homology theories which don't satisfy the property three accession property this is again a very very crucial one in you know what in computing homologies of from open subsets small open subsets to larger sets so that also is is quite there I have followed up so far talking about property four which was the computation of the path connected component computation of path connected components right the homology of a topological space was a direct sum of of its path connect components there is a more general result namely if you have just direct it is just a disjoint union of topological spaces then the homology is direct sum of each topological space there okay that is very obvious that we have we have also seen that one so one can make that kind of additivity remember additivity as an axiom because you have to make an axiom because for homology singular homology that we have done it comes because of definition okay so that axiom is called dimension axiom okay there are are many homology theories which don't satisfy the dimension axiom and some of them don't satisfy property three here that are very rare but many of them interesting ones don't satisfy dimension at all those things are called extraordinary homology homology theories the examples of such things are called like k theories and Buddhism theory and so on which we cannot discuss here okay so what Eilenberg Schindler did it yes that you you take all these axioms suppose a homology theory satisfies all these axioms then on the category of compact simple shear complex polyadromes they will all coincide so this was a fantastic result we will not be able to prove that one because you have developed quite a different kind of tools here and what I do is I will show several of them are equivalent whatever was interesting to us by hand and when I when we do that there are techniques there if you master that one then you can go back and read Eilenberg Schindler's results then it will be easy for you so that is the idea so let us stop here next time we will try to do more and more different kinds of homologies thank you