 So now we shall start an entirely new topic which is again the starting point of a big topic namely relation between homotopy groups and homology groups. So we shall only deal here with relation between the fundamental group and H1. To begin with we will we can assume that x is path connected because if it is not path connected it will be the study of each path connected component separately. So you can pick up a base point because we are discussing pi1, the fundamental group. So fundamental group always should be discussed with a base point x0. Now take an element in pi1 of x namely represented by a loop, a loop in x at the point x0. A loop is first of all is a continuous function from an interval closed interval to x. The closed interval can be thought of as the underlying space of delta1, the one simplex, a standard one simplex. Therefore a closed loop can be thought of as a one cycle, it is a one chain first of all, it is just one simplex but both the end points are going to the same point so boundary of that one will be 0 as a one cycle, as a one chain therefore it is a one cycle. So if we have a loop omega, I will denote the round bracket omega to be the element represented by this omega in H1 of x, the square bracket will denote the same element in pi1 of x. So starting with an element in pi1 of x namely round bracket omega to the sorry the square bracket omega to the round bracket omega I get a map function namely from h1 of pi1 of x to h1 of x. I want to claim first of all that this is well defined. In other words if omega1 and omega2 are homotopic namely loop homotopic or path homotopic or whatever you want to say, which represent the same element in pi1, then they should represent the same element in h1 that is what we have to prove otherwise the function will not be well defined. Then we want to prove that it is actually a homomorphism and then of course we will prove that this homomorphism is subjective and so on. So this homomorphism is called Hurevi homomorphism. So let us do all these things now elaborately. The first lemma is that suppose omega i, i equal to 1 and 2 are pathis in x such that the composition is defined that means end point of omega1 is the starting point of omega2 then omega1 star omega2 is defined. As a singular one chain you can take omega1 star omega2 as a chain as a simplex minus omega1 minus omega2 this will be a chain, this chain is null homologous. So this is what we want to prove. So this is the boundary of something, boundary of a two chain that is the meaning of null homologous. So this is the starting lemma, the proof is completely obvious if you look at this picture. In this picture or in this triangle I have cut it into two parts in the megae, e0 even e2 from e2 you drop a perpendicular cut it. This interval is divided into two parts, the first part to take omega1, second part to take omega2. So this e0 even will represent omega1 star omega2 as a one simplex. I have put omega1 from e0 to e2 here in this way and omega2 here in this way. So if you trace the boundary of these two simplex e0 to e1 to e2 to back to e0 the boundary will be e0 e1 minus even e2 this way even e2 is this way so minus of that one will come. So sorry even e2 will come then here it is minus omega2 because omega2 is this way and e2 e0 is again minus omega1. So point is that I can fill up this whole triangle there will be a continuous function defined on these two simplex into x such that boundaries like this. So how do I do that? Take any point here join it to omega join it to e0 and this line you define omega1. So as point keep moving from this bottom to that point to the point e2 as the point keeps moving this will fill up this half triangle. On this side you put omega2 this way so that continue omega2 is from this point to this point here. So keep omega2 like this so over. So what this proves is that omega1 star omega2 minus omega2 minus omega1 is null homologous. So if you are not convinced with this kind of argument actually you should argue like this to get the idea how to fill up. But finally what you have to do all the time I have been training you that you must be able to write down this formula sigma of any point in this triangle is defined ok. Every point in this triangle is uniquely expressed as t0 e0 plus t1 e1 plus t2 e2 where summation ti is equal to 1 and t0 t1 t2 are all between 0 and 1 this is by definition the geometric realization of a standard to simplex. So use that expression sigma t0 t0 t1 e1 t2 e2 t0 bigger than t1 and t0 less than equal to t1 is the dividing line here. All the points on e1 e2 sorry e0 e1 are given by t2 equal to 0 ok. So this part t1 bigger than t2 and t2 bigger than t1 you see these two are two different things so that is where first part I am taking omega1 second part I am taking omega2. Then you have to write down these expressions for these lines and then put them there. So t0 bigger than t1 omega1 and omega2 is like this ok. So check this whether this is correct there may be typos and so on. So you have to check that this is a correct definition this is not a unique way you may have different things ok. For example why should you divide this way you can divide here also but then what you are doing with this division that is what it is. So this is a nice way of doing it whenever. So what we have proved is that subdivision take a path subdivide ok then think of this as a composition it is same thing as this plus this and so on integration theory integration over this one minus integration of the first half minus integration of the second half is the same thing as this one that is the kind of thing it is happening here ok. So in homology this is what subdivision gives you correct the corresponding things all these things we have been doing and we will do more sophisticated things soon ok. So next theorem is that the assignment omega bracket square bracket to ground bracket defines a factorial surjection ok from phi from pi1 of xx0 to pi1 of h1 of x and it is denoted by phi of omega of omega whose kernel is precisely the commutator subgroup of pi1 of xx0. So this is a function of surjective homomorphism is what it says thus phi defines if it is surjective homomorphism the kernel is precisely the commutator subgroup means what pi1 mod commutator subgroup is nothing but abelianization of pi1 is isomorphic to h1 by the first isomorphism theorem. So h1 is abelianization of pi1 that is very easy to remember ok h1 by definition h1 is a is an abelian group ok. So if you abelianization pi1 you precisely get h1 instead this is a relation. So let us go ahead and prove this one the first thing to do is phi is well defined we have not yet defined that one right. So let h from i cross i to x be a path homotopy from omega 1 to omega 2. Consider the following triangulation of i cross i the vertex set of the simplicial complex A consists of these four vertices i cross i have to triangulate and then take a homotopy and then think of it as a sum of two simplicies so that is what I am trying to do. So a function defined on this one restricted to this triangle will give you one simple x which I am going to simplify this plus that will be a chain in x. So I am defining v0, v1, v2, v3 four vertices but I have to make it triangulation so I introduce v0 to v2 an extra edge here. Now it is cut into two triangles ok. Now what I am going to do take a homotopy take a homotopy here that means what i cross i to x I have a continuous function restricted to this triangle it is one simple x restricted to the other triangle it is another simple x the sum total this plus this what happens that is what you have to see right. So the boundary of that is going to be omega 1 minus omega 2 that means omega 1 minus omega 2 is null homologous in x they represent the same element in h1 is the boundaries omega 1 minus omega 2. So how do you get omega 1 minus omega 2? By the very definition a homotopy of two loops has a property that on this vertex on this edge here it is identically a single point namely the starting point of omega 1 ok starting and end point of both of them are omega 1. Here it is again starting omega this is single point this is a single point is single point ok this is omega 1 single point this one may be any wild arc I do not know this is omega 2 but when you take the boundary of this what is the boundary of this one boundary by definition you have to first omit v0 so v1 v2 then we will omit v1 v0 v2 will be with a negative sign then we omit v2 v0 v1 will come ok. So this omega 1 will come this is a single zero so we can throw away that one this will be in this direction when you do for this one it will come in the opposite direction so they will cancel this will cancel out this is zero and that is minus omega 2 so that is what is happening. So this I am writing details here the edges v0 v1 v1 v2 v0 v2 v0 v3 and v2 v3 these are the 1 2 3 4 5 edges the two simplex is in k are these two so alpha beta from delta to k defined by alpha equal to v0 v1 v2 and beta equal to v0 v2 v3 so these are the two one simplex two simplex I am taking a two chain in x is given by tau equal to h composite alpha and h composite beta alpha and these v0 v1 v2 are inside this k this is a simplex complex h of that will be inside x ok. So when I take some of this one this triangle plus this triangle is the what I am taking h of alpha plus h of beta ok we claim that boundary of tau this is tau boundary of tau which omega 1 minus omega 2 and that will show that the class is represented by omega 1 and omega 2 are the same ok. So restricted to 0 cross i and 1 cross i they are constant maps at x0 therefore h of this v0 v3 and v1 v2 they are equal and they will cancel out. Now boundary of tau is h of so I am writing the boundaries here both the boundaries v1 v2 v0 minus v0 v2 plus v0 v1 and then h of v2 v3 see you have to take this order ok. So when you omit this one the sign is positive when you omit this the second one here sign is minor when you omit the third one again positive that is what you have to do v2 v3 v0 v3 v0 v2 v0 v2 cancel out ok v1 v2 will give you omega 1 and this v2 v3 ok v2 v3 will give you omega 2 ok therefore h of h of this will be h of v0 v1 plus h of v2 v3 will give you omega 1 minus omega 2 ok. See v2 v3 is minus ok v1 v0 v1 is in the correct direction omega 1 so this is this opposite direction so this is minus omega 2 so that is why you get minus omega 2 ok. So this verifies that it is a homomorphic well defined it is well defined. Now in the first lemma when you cut down omega 1 star omega 2 namely concatenation is omega 1 my omega 1 plus omega 2 right when you take the brackets that will tell you that phi is a group homomorphism 5.1 lemma can be used now that is for more generally for pathes but for loops it is this loop is the composition inside pi 1 and that is adding is some position inside h1 ok. So therefore the bracket omega 1 omega 2 phi of that will go to omega 1 phi of omega 1 plus phi of omega 2 round bracket omega 1 plus omega 2 so that is homomorphism. The functoriality is something which is totally obvious here but I will leave it to you to think about that ok. So you think about it and write down what is the meaning of functoriality and you to verify that ok. So now just for writing the proof I mean it is a lazy notation just for temporarily pi 1 of xx naught I will write it as pi 1 h1 of xx naught x will be written as h1 ok he is just writing pi 1 h1 and so on in the diagram if you write all big things it will be difficult for. So psi pi 1 to pi 1 of a b denote the canonical quotient homomorphism by taking the quotient by the commutator subgroup. So pi 1 hat ab denotes the fundamental group abelianized namely the quotient by pi 1 pi 1 the commutator subgroup h1 is an abelian group it follows that any homomorphism from any group to an abelian group always factors to the pi app. The commutator subgroup goes to triviality here therefore pi 1 by pi 1 by pi 1 pi 1 to h1 there will be a map ok. So pi 1 to psi this is a quotient map here this phi whatever we have got the Hurwitz map gives you a map here gives you a homomorphism here phi prime and we claim our claim is that this is an isomorphism ok. So you have to prove one one and onto instead of that what I am going to do I will produce an inverse here and then verification that it is actually inverse is easy. So h1 to pi 1 app itself I will define a map ok a homomorphism. So how does it wonder? So not pi 1 app I am going to define I cannot define a map from h1 to pi 1 there will be problem but abelianized part I will define. Ok we shall construct an explicit inverse of u prime. So for this we use the typical calculus experience fix a point x in x and use the path connectivity of x. Actually for each x0 it is already fixed that is the base point is there there is no need to do that. For each x take a path from x0 to x and call it omega x. There are many paths is you take one and fix it let us solve ok. Only thing is do not make life complicated. So when when you joining x0 to x0 just take the constant loop there you could have taken any other loop but do not do that for simplicity even if you do that there is nothing wrong but do not do that by simplicity you take x0 to the loop at single point that namely the constant look at it. So you have chosen all the every point is connected by a path ok. Now each singular one simplex sigma in x what is a singular one simplex this is just a path inside x. It has a starting point and an end point ok. So both of end points and starting point there are path is from x0 right. So I am going to define theta sigma to be the element in pi1 of ab ok represented by a loop omega a a is the starting point compose sigma sigma is the path goes all the way to b. So come back by omega b to x0 ok omega b in a month. So this will be a loop at x0 it will represent an element in pi1 of x. Don't do that go to the abelianization and take that element that is theta sigma ok where a is sigma0 and b is sigma1 that is the end end points. Then theta extends to a homomorphism I am writing it as theta from the whole of s1 of x s1 of x is what the free abelian group overall one simplex. Once you have defined them on the generators sigma is all the sigma is a generator it will give you an extension to any abelian group. So these are advantage if I did if I take pi1 here pi1 is need not be abelian group if it is abelian group it is fine but then pi1 ab it is same thing as pi1. If it is non abelian then there is no need to have any map any homomorphism here. A free abelian group will not give you a map homomorphism into an arbitrary group ok. It is bigger this because pi1 ab is abelian it is a unique extension of these maps which are defined on the generators ok. So I am denoting it by theta itself ok. So the definition of the map is over now theta. So this phi prime and this theta from here to here they are going to be inverses of each other over. So let us verify that ok. It would follow ok what you have to do theta from s1 of x is defined we have to show that theta component boundary of s2 whatever suppose sigma chain if it is a chain is fine but when you homology when you go to homology it must be 0 ok. Then only it will define a map from h1 right. Right now it is s1 to here we have what go modular h1 ok go modular to h1 means what boundary, go modular is boundary it must be 0 more. Then it will give you a well defined map from h1. So that I am going to as a theta to it ok and then you can verify that this is the map theta to it is going to be inverse ok. So let us verify this and then verify that it is actually the inverse. Take any singular to simplest gamma from delta to x. If I show that theta of the boundary of this is 0 then the whole of s2 is generated by all these sigma 2's that will be also 0. Because any element of sigma 2 is a finite linear combination of 2 simplexes. For every 2 simplexes I am going to show theta of boundary of gamma is 0 ok. So this is as easy as the earlier thing namely let gamma of e0 equal to a even equal to b and e2 equal to c ok. So these are 3 points in x. Let f i be the standard phase operators that we have been using all the time ok. Then theta of boundary of gamma ok look at boundary of gamma is some chain. So how is it got? It is gamma of f2, gamma of f0, gamma of f1 and so on. So you have to join them right. But they are starting points are all the same. So omega a wherever they start means a this a ok. Gamma f2 star gamma f0 star gamma f1 you got back to the same point namely omega a inverse. So come back by omega a inverse because boundary of any 2 simplex is is a cycle is a loop. So already so starting at a so ending at a so you take this one. So psi of this is by definition theta of boundary of gamma ok. But this is what this is filled by triangle right. Therefore this is null homotopic in x. So it is conjugate is all null homotopic. Ok it follows that the term on the RHS is actually 0. You see I have take this element in pi1 in pi1 itself which is 0. So when you go to pi1 of a, b, a, b ok. Ability negation it will be 0 element ok. Thus theta will well define ok by definition of psi is what is represented by this element take the loop here and then take the element of the element itself here itself is 0. So this will be 0 here ok. So you have got a very defined map from h1 to pi1 ab which is which is what going to be the inverse of 3 prime. So let us verify that one last thing ok. So it remains to verify that theta 2 it is 3 prime inverse. Given tau in pi1 of xx0 ok phi of tau is represented by a one cycle tau itself right. You see what is definition of phi? Phi is just square bracket you remove and make it round bracket that is all right. So that is what that is what it is one. Therefore I do not have to search for some omega etcetera. Theta rule of phi tau it is already a loop at that point ok theta rule of phi tau is just psi of tau. The same tau will be I can check. So it is psi of that one in pi1 of a, b. You get the point in defining in defining this map theta arbitrary I had to take all these pathes and join them and come back and so on. I do not have to do anything like that here now because it is already a loop at the point x0 right. So that is why I said do not make it complicated by taking arbitrary loops and so on. When a starting point is x0 itself you take the constant loop there therefore it is same phi which is already given here ok. So that will come to the corresponding element here starting with an element here which is represented by an element here you go the same omega here that will come to this point. That means what the image of this when you come here it is identity ok. So sorry where I am here on the other hand given an element lambda in h1 I have to work hard I will not put it starting with an element of h1 represented by a cycle now it is complicated thing it need not be at the base point x0 summation lambda i ok n i lambda i whatever you want to lambda 5.1 shows that phi of theta lambda i what I am doing lambda i all of them I am introducing starting a path from x0 to the starting point then lambda i then coming back keep doing that each time go up and then come back by another loop ok. So this is theta phi of theta lambda i this represent same element as lambda why because the extra edges keep cancelling out. So this is like representation theory you know it is complex analysis when you have keep adding extra edges so they keep actually cancelling out here ok. So all the extra edges introduce along for this cancel out in pairs. So this means the phi prime of theta lambda is lambda. So that is a entertaining and easy but this is very useful result in algebraic topology. Let me take a few more minutes and give you one small application of this result now ok a familiar result which is not new to you. So to view this that pi1 abelianized is h1 helps in computing a number of things ok. So let us compute the CW chain complex of s1 cross s1 without going to elaborate mayor vitri sequence we have done that one already right using mayor vitri sequence ok. Recall that treating s1 as a CW complex with zero cell and a one cell ok the product CW complex will then have a one single zero cell two one cells and a one two cell right that is a product structure. So it follows that the CWs I should write C0 CW here I am not writing that C0 of this the 0th part is z C1 has two simple axis so it is z direction z C2 is again z it is only one and CQ is 0 for Q bigger than equal to bigger than 2 all right it is also clear that the boundary operator C1 to C0 is zero map because the single point is there and then you are attaching a attaching a one edge or one simple edge with both the end points going to the same point. So when you take the boundary the point minus point cancels out ok so from both the generators here the boundary map will be zero so the entire boundary one from C1 to C0 is zero map what is the boundary map from C2 to C1 so this is given by the degree two maps obtained by composing the attaching map S1 from boundary of I cross I namely S1 versus S1 with the two projection map S1 cross S1 versus S1 I have elaborately extended how to compute the boundary map in a chain in a CW chain complex ok so look at the attaching map there and project it to S1 and come to the degree ok for both of them then you can you can determine what is dabbato ok we have seen that the attaching map is actually the boundary of the product of two characteristic maps that is also we know that if you remember how the characteristic map for the product is defined what is the product of this cross this so boundary you will have to trace it first along say x the one coordinate and along then the y and then x in the opposite direction and y in the opposite direction so that is how a torus is defined by identify identification from a I cross I from a piece of paper ok so whatever you wanted to whichever way you wanted to see so in H1 of wedge S1 wedge S1 ok in H1 if you just take the projection to S1 one of factor this this element ok will be give you x and then x negative x x negative so it will cancel out so degree will be 0 similarly due to y it is y and then y inverse so again degree is 0 so boundary map from C2 to C1 is also 0 therefore the CW chain complex of S1 wedge S1 here it must be Z here but the map is 0 it looks like Z2 Z2 map is 0 ok Z2 to Z again the map is 0 and after that it is 0 therefore the kernel at each level is Z and Z2 ok and Z and the images are all 0 so homology is just Z Z2 Z ok H0 is Z H1 is Z2 H2 is Z again ok so let us stop here next time we will be doing so several of the proofs that we have postponed in while doing homology theory so we will go through some of the proofs ok thank you very much sir