 Continuing with the applications of various theoretical results that we have proved, today let me first begin with recalling the definition of Euler characteristic which we have defined for a simplicial complex. Namely, if you denote the number of vertices by Vx, number of edges by Ax and so on, in a simplicial complex we can denote f0, f1, f2, f3 and so on, the number of k simplicis by fk, then the alternating sum f0 minus f1 plus f2 minus f3, etc. was called as Euler characteristic. The exactly same way we can define it for pseudograph also, number of vertices denoted by Vx, number of edges denoted by Ex, there is no more two cells and so on in a pseudograph. So, Euler characteristic of a pseudograph is defined as Vx minus Ex. It will coincide with the definition of the ordinary pseudograph, if this pseudograph happens to be a simplicial complex, that is one dimension of the complex. Now, what it has to with whatever we have done, last time we did one important result about the fundamental group of any pseudograph. Namely, if you start the connected pseudograph, the fundamental group is a free group and its rank, the rank of free group is equal to number of edges outside any maximal tree T inside X. So, this is what we are going to apply right now and get a nice formula in terms of Euler characteristic, rate it to Euler characteristic. This is the corollary, this one line corollary, but it is very important and easy to remember. For a finite connected pseudograph, the rank R of the fundamental group is given by 1 minus chi X, the rank is given by 1 minus chi X. So, how do you do this? So, very easy. The number of edges in any maximal tree is equal to V X minus 1. What is V X? V X is number of vertices. If there is only one vertex, there are no edges at all in a maximal tree. If there are two vertices, they are connected by one edge. You cannot connect it by two edge, then it will not be a maximal tree, it will not be a tree at all. Like this, you can see if there are n vertices, you exactly need n minus 1 edges to connect them. So, if there are V X vertices, V X minus 1 is number of edges. Therefore, the rank of pi 1 of X is equal to the total number of edges minus the number of edges inside a maximal tree which are those which are not in the maximal tree. So, it is E X minus V X minus 1 which is the same thing as 1 minus V X plus E X which is nothing but 1 and this part is nothing but minus of chi X that is all. So, the formula is proved. So, now, we will give application of this corollary. So, this itself was a corollary to this corollary. So, now, we will give an application to one of the very, very important result in group theory. It is called Nissan Trier. Both of them were actually done lot of group theory also, but they were all motivated by topology. They have done lot of topology also. Every subgroup of a free group is free. That means it is a free group. Remember a free group has a basis. The subgroup may not have basis coming from this base. It is need not be a subset of the basis. Only thing what we say is every subgroup of a free group is free. It is just like every subgroup of free, abelian group, free abelian which you may have known already. So, there is a further class here. Indeed, if F is a free group of finite rank R and F prime is a subgroup of finite index, then the ranks can be related by the following formula. Namely, R prime is the rank of the subgroup is equal to 1 minus k where k is the index plus k times R where R is the rank of the free group of the larger group. So, this now we can derive by using Euler characteristic and something more that we have done everything in the course itself. Namely, we have given a simplicial structure to a covering. The simplicial structure coming from a simplicial complex of the given simplicial complex in a very particular way. So, that we are going to use here now. So, let us start with the following. A free group has a basis. Let me call that basis as i, any set. This set is the basis for free group. Then I take as many copies as a number of elements in i, the boundary of the two simplex, the standard two simplex delta R2. The boundary of the standard two simplex you remember is homeomorphic to the S1, the circle. So, it has fundamental group infinite cycling. So, what I will do is I will take as many copies as in i this thing. Then I perform this bouquet one point union, namely identify one of the vertices that E0 from each of them to a single point. Select one point, one vertex from each of these copies of the boundary of delta R2 and identify them to a single point. So, that is the meaning of this wedge of these things. A wedge of topological space is in general is defined by that method. And it is given the usual CW topology here. So, you take this one, boundary of i delta R2, i for each of them is a copy of boundary of delta R2 which is homeomorphic to S1. Then we know that this is being a bouquet of circles we have computed. The fundamental group of this k mod k is nothing but that mod k I am putting delta x is nothing but the free group over i and that is f. So, f is isomorphic to pi 1 of x. So, starting with a abstract group free group, we have realized it as a fundamental group of a space. The space is constructed in a very particular manner. So, it is a pseudograph force. Now, given any subgroup of the fundamental group, we can construct a covering corresponding to this. This is our general covering space theory. Let P from x prime to x be the connected covering corresponding to the subgroup f prime. What is the meaning of this? That x prime, let us say little x prime as a base point sitting over this x naught here. The x naught is a common base point that I am taking here. Then pi 1 of x prime comma x prime is isomorphic to the subgroup via this the covering map P check. Covering map P induces homomorphism. P check is injective map. The image is precisely f prime. So, this is what the general theory of covering space is that I am going to use it here now. But now, x prime itself has a simple shell complex structure, one-dimensional simple shell complex structure because this is a one-dimensional simple shell complex. So, what in the structure that also we know exactly, where k prime is a one-dimensional complex, in particular it follows that it is fundamental group because this means it is free. So, the first part of the theorem is over. Now, for the more elaborate part namely, suppose f is the field group of finite rank R, then what does it mean? The set I must be chosen such that it has exactly R elements. Then this bouquet will also have exactly R copies of boundary of delta 2. R empty triangle, triangles will be there. How many R of them? All of them having one single point as common. One vertex is common. So, therefore you can count the number of edges and number of vertices here. Then for each vertex in k, in k prime, there will be how many vertices exactly depending upon the number of sheet. So, how does the number of sheet is determined? The index of this group, index of this group is what we have taken as k. So, there will be k many vertices for each vertex below. So, that is what I have to write down now and in the counting. So, let me repeat for the more elaborate second part, let I be a finite set with coordinate f i equal to R and let f prime be a subgroup of index k in f. Then, p is a k sheet recovery. The p is a covering projection. It follows that the number of vertices in k prime is k times the number of vertices in k. But number of vertices in k is precisely equal to 1 plus 2 R. In each triangle, there are 3, but one of them is a common. So, that should not be counted too many times. So, that is only once. I mean for other two, for each of them, R of them are there. So, 2 R will come. So, k times 2 R. Similarly, number of edges inside k prime is precisely k times E R. So, k times E k. What is E k? E k is 3 times. Each triangle has as many 3 of them as edges. So, there will be 3 times R. So, it is k times 3 times R. So, if you take vertices, it is k times edges also k times. So, Euler characteristics just k prime is just nothing but k times k of k. You do not need all these elaborate things, but I have just written down. But then R prime by the earlier formula is equal to 1 minus chi of k prime, which is equal to 1 minus k times chi of k, which is equal to 1 minus k times chi of k, it says it is 1 minus R. So, I get 1 minus k plus k R part of the formula is written here. For example, if your group is say free group of rank 2 and you have an index again 2, then what will be the rank of the subgroup? It will be 1 minus 2 plus 4. It is quite a good formula. So, it will be 1 minus 2 plus 4 is 3. It is readily, both of them are 2, 2, but the subgroup has order, subgroup has rank 3. So, this is used in several differential geometry, different algebraic geometry contests to prove many other results. It is extended to what are called Riemann-Hurwitz formula and so on. So, now I will give you another important application of this G covering that we have done. So, here let X and Y be connected locally contractible spaces or you may assume less things like locally, semi locally, one connected, semi locally simply connected, some such thing you can assume. So, locally contractible it is easy. Let F from X to Y be any map such that let us you know C F, you know the mapping cone of F. Then the inclusion homomorphism eta from pi 1 of Y to pi 1 of C F is surjective and the kernel is a normal subgroup N generated by the image of F check in pi 1 of X to pi 1 of F. F is any map. So, if you want to be very careful here, you have to select a base point here, say X naught and then you have to take F of X naught here as the base point for Y and then you have to write down this. Since those things are obvious and also always have to be remembered what is more essential and simple I have stated this. Okay, instead of writing all the base points and so on. So, those things will be done in the proof carefully. Moreover, the entire statement is independent of what base point you take. If you change suppose somebody takes X 1, X 2, X 1 and X 1, X 1, X 2 it will be still true. Therefore, what you have to remember is this statement. But a more elaborate statement has to be also remembered always whenever fundamental group is involved, you if you want to be very careful, finally you have to put the base points. Okay, so that is the summary of this one. So, I have deliberately omitted mentioning the base point so that it will be easy to remember. Not that you should forget it. You should make a practice not to mention it yet. Remembering it. Okay, so let us recall because this is a long time we have done these things. Let us recall what is the mapping cone and mapping cylinder and so on so that we write down a proof carefully. So, given any topological space X, C X is defined as the quotient of X cross unit interval wherein we identify all the points X comma 0 to a single point. The mapping cone is defined similarly but little more elaborately you take X cross interval disjoint union with Y and then you identify X cross 0 to a single point. Also you identify X comma 1 to F X for every X in X prime. So, this way F is involved in this way. Okay, then the mapping cylinder is defined in a different way wherein this part of the identification is not taking place at all. Namely, it is X cross unit interval disjoint union Y and X comma 1 is identified as F X for every X in X. Okay, I am just recalling this I cannot run the whole thing again this theory. There is a standard notation that all the points X comma 0 are identified single point that point is denoted by star. Okay, and that star is called the apex of the cone. The whole C X is star shaped at this star. Remember all this. Therefore, the C X is always contractible. Moreover, if you throw away that point, then the remaining portion is nothing but X cross open interval 0 to 1. Therefore, it is strong differentiated tracks to X. You can bring it back to X comma 1. So that is X. X can be thought of as a subspace of X comma T where T is fixed for any T other than that 0. At 0, there is a not the whole of X, but only single point is there. Okay, there is also an obvious inclusion map. From C X, you throw away the base. That is an open subset anyway. X comma 1 is a base. The whole thing is contained, C X itself is contained inside C F. Okay, you can see that C X X cross I disjoint even why I have taken whatever identification is there that those those things are also there here. Right? Right? Except that X cross I is fully not identified here. X cross 1 is not identified here at all, but there is some identification here. Therefore, if I throw away X cross 1, the open part, then that is contained inside this one. Okay, so C X cross C X minus X cross 1, there are identification here. If you throw away that part, that is included in C F. So let us put, let us just denote it by V. This is temporary notation. That is an open subset. Also look at the open subset obtained by throwing away the star from C F. If you throw away that, that is same thing as mapping cylinder minus X cross 0. If you identify the whole X cross 0 single point that is C F. So you are throwing away here, you are throwing away this part also that is M F. Okay? So U and V are both opens of sets of C F. And C F is U union V. Okay? So, so far what I have done is setting up notations for application of cypher and companse theorem. Like you had a sphere, then you removed North Pole, South Pole and then you took the intersection and so on. That is the kind of thing we are doing here. Okay? So intersection, let me denote it by A. This is all just convenient temporary notation. That is nothing but X cross 0 1. Okay? So let us have these notations also, eta 1 from A to U, eta 2 from A to B. These are the inclusion maps. All right? So let me show you the picture first. So picture is here. This is the entire mapping cone. So this is the star. So this is X cross, let us say X cross half, half way. This whole thing is 0 to 1 here. So at 0 there is one identification. At X 1 also there is identification. Here X will be identified with F of X. So I have taken X naught as a base point for X. But this is actually, I have put it X cross half, half way. I could have put it at any level t, not equal to 0 or not equal to 1. Anywhere I can put it, same thing. Okay? So from here to here, this entire thing, this is my V and from here to the rest of them, this is my U. The union is the whole space. See here, intersection will be, remove this point, remove this part Y and this open part, which is nothing but X cross open interval 0, 1. All right? So we will come back to this picture again. So right now this is the, this is what I have said up the notation. Therefore, so one more, namely I have chosen X naught as a base point, whatever X naught, does not matter. Y naught is F of X naught. But X naught is X where it is, it is X cross half I am taking. Therefore, I would put it inside X naught cross comma half, that is X naught hat. Okay? So that belongs to inside A, X cross 0, 1. Okay? X itself is nowhere there, X cross, some t is there after all. They have everywhere. So you can identify X as this one. Okay? Inside A. Once you have this, you have the one compass theorem. The C F is written as union of U and V. It is a common base point X naught belonging to all of them. I take pi 1. Intersection of these two is A. So pi 1 of A X naught, eta 1 check, eta 2 check. And then these are also inclusion maps. I am not writing them, separate notation is not necessary. Inclusion induced maps here. Okay? The statement of one compass theorem is that pi 1 of C S is the amalgamated free product of this path. Namely, it is free product of this group with this group, modulo, the normal subgroup generated by the image of this, minus this and so on. Right? So that is what we have written down here. It is pi 1 of U star pi 1 of V. Modulo, the normal N is the normal subgroup generated by eta 1 check G multiplied by eta 2 check G inverse. That is one single element as G belongs to pi 1 of A. Take all of them, then take the normal subgroup generated by that. The quotient of this is pi 1 of C F. I can put the X naught hat here. Okay? X naught hat, X naught hat here always. The base point can be always written down. It is written down here. Okay? That is one compass theorem. But now we know all these things. So we have to figure out what these things are. Okay? Now use the fact that V is star shaped at the star. Every star shaped set is contractable. In particular, it is fundamental group is trivial. Therefore, this statement here, whatever we have to, this gives you this pi 1 of V, there is no need to write, this is a trivial group. So it is only pi 1 of U and eta 2 of G, there is no need to write. They are all trivial elements. It is normal subgroup generated by eta 1 of G. G varies over. It is the same thing as normal subgroup generated by the image of pi 1 of A. So image of pi 1 of A under eta 1 check. So this is the neat expression that we have got. Pi 1 of CfA is the quotient of pi 1 of U by the normal subgroup generated by eta 1 check. We are not yet over. So what is this U, what is this eta 1, etc., you have to figure out. Okay? Because these were not in the beginning, our statement in the statement is only x and y. So everything you have to convert it to x and y, what is that? So this just means the inclusion means homomorphism, pi 1 of U to pi 1 of Cf. Remember this is giving this isomorphism, it is not 50 T as it is given by the inclusion of the homomorphism that is subjective and its kernel is the normal subgroup generated by pi 1 of A. So if you go back to this part, you can ignore the upper part here. This part is an exact sequence. This is subjective. Kernel of this map is this one. That is the meaning of this. Okay? So we have to do one more step here to understand in terms of x and y. That is the that is the major part of the thing is over. We have to figure out that the correcting is arrived. Okay? So this is the picture going back to the mapping cone. X cross half is sitting somewhere in between with X naught as base point. This is a deformation retract of A. Therefore the inclusion induced map is an isomorphism here. A itself is contained inside U. Okay? So this is pi 1 of U here and this is that is an identification here. Right? So that is Q check. The quotient inclusion followed by the quotient restricted to the subspace that is the inclusion map. So here it is just forget X comma half going to just X. This is the projection map. So this is a homeomorphism. So this is again an isomorphism in pi 1. So now it is pi 1 of X X naught. So this is nothing but a copy of this here. All right? From here I have f check to pi 1 of Y. So what is the map here? Remember the mapping cone whole mapping cone there is a diffusion but here I have removed just the one open part here that is all. So mapping cone there is a map any bracket X comma T going to fX. This is a very defined map and that is what I have written as f hat. So this is f hat check and f hat itself is a strong deformation retract onto Y. So this is an isomorphism here. So what we know is this is the CF here we are sitting here. There is an inclusion map here but here I have taken different map here. The inclusion map is factored by this way. Okay, goes into Y and then comes here. So this is a commutative diagram here. This is, okay, it is surjective and it is kernally precisely this one is what we have seen from pi 1 of A. Pi 1 of A is isomorph to this, isomorph to this one. All right? So the problem here is you see this is X naught twiddle here. Okay, this f check this is Y naught. So what is this base points? Where do they go? This is what we may wonder. If you don't have any such troubles in your mind you have completed a proof. This part is over now. Okay, the normal this is surjective, this is surjective. The normal subordinate by this one is isomorph to this one is over. However, I have taken trouble to exactly what happens to this map here because this is some other map here f twiddle of that comes to Y naught. That is fine. So you have to understand that this map is a homotopy equivalent. So it is an isomorphism. But why it should be the same thing? Why these diagrams are commutative etc? You will have to learn. Okay, so here I have elaborated this. Let us go through this one. The unlabeled arrows indicate homomorphism induced by inclusion maps here. What are they? This one, it is the only one. This one here. This is something I will explain it. Okay, and q is the restriction of the quotient map q x comma t is our notation is x comma t bracket, round bracket change to square bracket. p of x t is just x, f hat of x t is fx. All these I have told you. Since p is the restriction of the projection, it is a homeomorphism and hence p check is an isomorphism. As seen in our theorem, namely about the homotopy properties of the mapping cylinder, if you do not remember, this is what it is, namely f hat composite J's identity of Y. And f hat is a strong deformation attraction. So that J I have not included here. This J is from Y to the mapping cylinder. So Y to the mapping cylinder up above. Okay, so J and then f hat that is the identity map. Okay, and this is the strong deformation attraction. This is what we have used. I am going to use that one now. Okay, so but I will be write down what is that strong deformation attraction here. Let us take h from mf cross i to mf given by h of x t s equal to x comma 1 minus t times 1 minus s times t plus s. Okay, this defines homotopy of identity map of f with f hat relative to Y. What are the points here when t equal to 1? Okay, that is a point of Y. So on Y it is identity map. You see the meaning of this. Okay, when t equal to 1, okay, t equal to 1, it is 1 minus s plus s which is just 1, x 1. So you can extend it by identity on Y. Okay, the rest of the time it is homotopy. When t equal to 0, what is this one? x 0 s is t 0s x comma s is identity map of the mapping cylinder itself. Okay, so when s is 0, s is 0, this is x comma t plus s. Yes, s is 0, x comma t, x comma t goes to x comma t. So in particular, f hat check is an isomorphism. Okay, so I have repeated it three times at least. Commutativity of the left side rectangle is obvious at the topological level itself. Look at this, what is it? x comma half goes to x. Here x comma t, bracket x comma t goes to x. Here x comma open, x comma half open, there is no identification, goes to a bracket here. So this is obvious. So this f of that. Okay, once some diagram is commutative at the topological level here is there. Induce map diagram will be commutative at the homotopy groups, at the pi 1 level. Okay, only commutativity of the triangle on the right side involving the dash arrow has to be justified. What is this one? So this is remember h tau, what is this tau? So I will explain that one now. Okay, which changes the base point x naught to y naught, x naught hat to y naught hat. Okay, remember this is subspace of that, but this is not an inclusion map here. Base point has been changed. So this is what it is. So x naught hat is here. Okay, and y naught is there, f of that. And you have the arc here from half to one. You trace it this way, Uta, the other way around. That is tau. That is a path. That tau is a path. Whenever you have a path, you have h of that path, h tau, define how take a loop here, start with the base point here now, go via tau, trace the loop, come back by the tau inverse. So that is the definition of h tau. If you remember that, that is fine. Otherwise I have just recalled it. Okay, so what I am going to do, use this thing and homotopy here to push this x naught by a homotopy to this point. So that is precisely what I am going to do. Namely, as if the base point is slowly moved along this point to this point. That is the homotopy I am going to write it down. And then you are confused. These two are the same thing. Okay, so otherwise this map is there something else changing from base point here. We know it is some isomorphism. But why the isomorphism should give you the same normal subgroup etc. That kind of doubt can be there. So that is why I am taking elaborate time to explain this part. Now let us look at what we have written down here. Okay, so check. For 0 less than s less than 1, put tau s of t equal to x naught comma 1 minus 1 minus s times t by 2. Okay, so I have taken various tau s now. Let us look at s equal to 0. Tau 0 of t is x naught 1 minus, this is completely s is 0. So 1 minus t is 1 minus t by 2. So it is 1 minus t by 2. That is precisely this trace delta from 1 to half. That is tau naught. What is tau 1 when s equal to s equal to 1? This 1 minus s is 0. It is x naught 1, which means it is just this point. So I have taken, first I take this full part and slowly I take only trace up till here, up till here, up till here. Finally, this is the constant factor. Why not? So that is tau s family of arcs. Okay, and h s of x tau, you just write h of x t that whatever you have written earlier this h, h this h, you take this one, that definition. h s of x tau is h of x to s s. Check that tau naught is a constant part at y naught equal to x naught comma 1. And tau 1 is tau in the segment x naught cross half 1. Trace in the opposite direction. Now given any loop omega in M f like this picture, this is omega here at base point x naught, okay, inside the whole of C f. That will be converted into, by pushing it at base point here in the right in the beginning. Finally, it will be pushed into a path inside omega loop inside y itself. Okay, there are two of them, two things you have to do here. So to begin with what we have, now start it, okay. Now given a loop omega in M f, okay, at the base point x naught hat, you want to prove that h tau omega is homotopic to f hat composite omega. F hat composite omega is a path, is a loop in y at the base point y naught. So if you show this one, then this diagram commutativity of this path follows, this h tau. Okay, when you push it, this loop here, this loop here, inclusion map followed by this one, that is fine. Okay, so this is what we have to prove. And here is the formula, that is all. Consider f of T s equal to, see remember the first thing is tau star, okay, tau inverse of omega itself. So h s starting the point will be just identity. So that is what you have to write, tau s star h s composite omega star tau s inverse T. Check that f we say homotopy as required for h tau of omega to f hat of omega. So it follows that the inclusion is homomorphism from y to pi 1 of c f y naught is subjective with its kernel equal to the normal suffrage generated by this one, once you have established so we shall come to the last reverent of this series now. As a special case of this general theorem, take the space x to be a circle and f be any map into y, that is a special case. When you take the cone over that, it is nothing but the entire space is obtained by attaching a two cell to y, okay, because the cone over s 1 is nothing but the disc d 2, right. So that is a two cell being attached via the map f to y. So that is a special case. So we shall slightly generalize this one, not just one and that is the next result here. So let z be a space obtained by attaching two cells d 2 alpha, you know not one but several of them indexed over lambda, capital lambda, okay, to a path connected space y via the attaching maps f alpha from s 1 to y, okay, these are f alphas. Let tau alpha be a path in y from the base point y naught to f alpha of 1, this 1 is indicating the base point in s 1, this complex number 1 there. Let lambda alpha, you know the element in pi 1 of y y naught represented by this tau alpha, tau alpha is a path, right, I have chosen from y naught to f alpha 1. So we take f alpha tau alpha, now trace f alpha, f alpha is a loop after all, it is a map from s 1 to y, right. So f alpha composite tau alpha involves, okay. Then the fundamental group pi 1 of z y naught is isomorphic to the quotient group of the quotient of pi 1 of y y naught by the normal subgroup generated by all the elements lambda alpha which you have defined here as alpha runs over lambda. So take the normal subgroup generated by this one and go by go by quotient out by that one that is pi 1 of z. So the missing thing here if you look at is that in the previous theorem we had x is connected, y is also connected and we have chosen a base point there. So pi 1 of x naught to pi 1 of x naught to pi 1 of y y naught could have taken, right, connected spaces. But now here d alphas by definition they will be disjoint, they will not be connected. So except that it is similar to the previous theorem. So we have to be careful how this will this can be done, we cannot directly apply that. If d alpha consists of just one family, one single element, one element in this family say lambda is a singleton, then it is direct application of the previous theorem, okay. Then it is image of image of this pi 1 of s 1, right, pi 1 of s 1 which is infinite cycling. The normal subgroup generated by that one that is precisely what it is, okay. Now I want to question you here namely we shall give the proof of this only when f alpha is a finite family that means lambda is finite. The general case follows by what is called as a direct limit argument which we shall teach you in the second part. Not only that in the second part the attaching cells etc itself will be formalized into what is called as CW complex. Not only just attaching one cells and two cells you will do attaching k cells in general and then we study those things very elaborately, okay. So you may think that this last result here is a motivation for the second part, all right. So let us work out this one when f alpha, the family f alpha is finite family, okay. The first of all I want to repeat the post lambda is a singleton. Then we can apply the above theorem by choosing the base point y naught to be equal to f alpha of 1. Choose that one because we have, we feel to choose the base point. Here x will be s 1 and z can be viewed, z is mapping, mapping cone now. I mean you can think of this mapping cone of this map f alpha. You can call it as obtained from y by attaching a two cell yi f alpha. These two pictures are the same, right. Since phi 1 of s 1 is fine infinite cyclic group generated by just the identity inclusion map s 1 to s 1 to take the identity map that will be denoting an event in the phi 1 of, that loop phi 1 of s 1 and that is generator. It follows that its image and f alpha shape is nothing but the class f alpha as a loop. It is generated by f alpha, the loop f alpha because f alpha composite identity I have to wait. So, it is just f alpha in phi 1 of y y naught. Therefore, phi 1 of z with the base point f alpha of 1 is the coefficient of phi 1 of y f alpha 1 by the normal subgroup generated by f alpha, okay. You do not have to take all the elements because all other elements are powers of f alpha, okay. So, case lambda is just singleton is over, okay. Now, what we are going to do? We do a simple induction. The only problem is each time the base point will have to be kept changing and therefore it is important to realize our earlier theorem was base point free. It is true for even if you change any other base point, okay. That is why I had taken so much of trouble in introducing that one, okay. Let us do it. How we do it? Let us see. So, now consider the case when f alpha 1 may not be equal to y naught, okay. We then appeal to the isomorphism h tau alpha, okay. What is tau alpha? Tau alpha is any map from y naught to f alpha 1. Any path from y naught to f alpha 1. Then h tau alpha from phi 1 of y f alpha 1 a loop here get converted into loop here and this is an isomorphism which we have, okay. Once this is an isomorphism and normal subgroup will go to normal subgroup here and what is this normal subgroup? It is lambda alpha which we have defined elaborately here in the statement here. Tau alpha composite I mean this is path composition star f alpha star tau alpha inverse. So, f alpha will come to this one. The normal subgroup generated by f alpha will come to the normal subgroup generated by this one, okay. Under isomorphism normal subgroup will go to normal subgroups, okay. So, this will give you now the statement for y naught where y naught is a fixed point, base point for y independent of what I have called. But now this is applicable for all the other alpha solves when lambda is not a singleton. So, you can apply one by one. First you get identify say alpha 1, alpha 2, alpha n. First attach alpha 1 through alpha 1. Next attach alpha through alpha 2. So, what do you get? The quotient of the second one is quotient of the normal subgroup generated by f alpha 2. So, together f alpha 1 and f alpha 2 will give you the entire normal subgroup. So, this is a finite case is got by a simple induction, okay. The infinite case won't conclude like this one. Even the countable case you won't conclude like this one. So, for that you need what is called as direct limit arguments. So, that will be part of the next course, second course in this one. So, you are welcome to attend that. Thank you. I have enjoyed teaching you people. Hope you have learned something from this course. Thank you.