 Last time we defined a function from pi1 of S1 to the set of integers taking a path and its class that is an element of pi1 of S1. We lifted this path to a map into R namely exponential composite that map is equal to the given loop. In doing this we fixed the starting point to be 0. The end point of this path was the degree of the function the class F. This function we want to show it now is a homomorphism right. So, this theorem is the function degree from pi1 of S1 to set is an isomorphism 1 1 on 2 and a homomorphism. One of them one by one let us show them. To show that degree of homomorphism what we have to do? Let us take two loops F1 and F2 based at 1 ok. Let G i be the loop G i be the path is inside or map into R such that exponential composite G i equal to F i those are lifts such that their starting point is 0 ok. Then G1 is degree of F1 and G2 will be degree of F2 right. So, what we have to show is the sum of this is the degree of F1 star F2 the composition of path is inside S1 ok. So, that is what we have to show. So, after taking the lifts what we want to do is take Ht to be degree G2 of t plus degree of F1 add that number namely you shifted the G2 will be shifted by a number namely degree of F1 ok by a constant. If you shift a lift by a constant namely an integer it will still be a lift of the same function namely exponential of Ht is the same thing as exponential of G2 which is F2. The degree of F1 mean integer e raised to 2 pi i times integer is always 1. So, it will get multiplied by 1. So, it does not change ok. So, degree of F1 is nothing but the end point of G1. So, this is what my definition of Ht is. Then what I have observed is exponential of H2 is also equal to F2. Therefore, what we get is exponential of G1 star H ok will be equal to you have to take how star is defined. If you apply a function F it will be exponential of G1 star exponential of H that will be F1 star F2. So, I have got a lift of F1 star F2 instead of taking a lift of F1 star F2 arbitrarily I have here constructed the lift of F1 star F2 using the lifts of F1 and F2 ok. All that I have to verify is it is starting point is 0 ok G1 star H at 0 it is same thing as G1 of 0 but G1 of 0 we know is 0. So, this is also satisfied. Therefore, if I take the end point of G1 star H that must give me the degree of F1 star F2 ok. Degree of F1 star F2 is nothing but the end point of G1 star H alright which is H1 now because G1 star H is I mean G1 of G1 star whatever of a last point is H of that right. The second part of the loop will come H of 1 but H of 1 is G2 of 1 plus G1 of 1 which is degree of F2 such degree of F1. So, proof that degree is a morphism is now we shall prove that it is map it is an isomorphism. What happens to the identity loop such going to Z namely 0 1 if you take the as an interval t goes to e raised to 2 pi I t what is its lift at t is equal to 0 it will be just t 1 to t. Therefore, when t equal to 1 degree will be the end point will be 1 that means the degree of the identity loop is 1 ok any homomorphism into Z which assumes the value 1 must be the whole of space it must be surjective because 1 is a generator alright. So, that proves that the degree map is surjective. Now finally suppose degree of F is 0 what of that mean the lift of F namely let us call it G such that starting point of G is 0 must have end point also 0 that is the meaning of degree of F is 0 degree of F is G of 1. So, I have a loop I have a loop here in I in R I to R you have a function both the end points are 0 right G1 and G0 are both equal to the same point 0 need not be 0 any same point if they have same point we have seen earlier that such a map is null homotopic as a loop as a path homotopy 2 as constant path yes you take t times this plus 1 minus t times that that is the that is the way it works we have seen that one several times now we are using that for now. Therefore, there is a homotopy of G to the constant path namely G of T S is S times G S times G T because either point is 0 I do not have to add up ok. So, this is a homotopy of the 0 map with G relative to the end point it is the path homotopy ok. If you take exponential of this homotopy that will give you the homotopy of the constant function 1 with the original function F the loop F ok because the original function G is relative homotopy namely it fixes 0 and 1 exponential of that part will also fix 0 and 1. So, this is a path homotopy therefore our F what we have taken must be the 0 element of the group I mean additive group whatever it is the identity element of the group ok. We showed degree of F is 0 the class F must be also the identity element of the group. So, this proves that degree being already homomorphism it is injective now. The kernel of a injective kernel of a homomorphism is 0 means it is injective. So, this proves that the degree map is an isomorphism and completes the computation of pi 1 of S 1 pi 1 of S 1 is an infinite cyclic group. So, that is the conclusion ok. So, the first non-trivial group fundamental group that we have computed is infinite cyclic group. All other things where we have contractual spaces convex subsets and so on they were all trivial ok. Let us give just a few applications here. This will be keep coming again and again it will have lot of applications. For that we have developed other things. Right now we can give one very important application namely the Prover 6 point theorem for disks two dimensional disks not in general. In general we will take more time ok. So, the boundary of the disk D2 is not a retract of D2 that is what we will first prove. The boundary of D2 is not a retract of D2. What is the meaning of retract that there is no function R from D2 to S 1 which is identity on S 1. S 1 is a boundary right. So, this is what we want to prove. There is no map no continuous function ok. No continuous function from the entire disk to the boundary such that is identity on the boundary. Suppose there is such a thing namely R from D2 to S 1 a retraction. Let us denote the inclusion map of the boundary S 1 into D2 by i ok. Then what we say it is retraction means R composite i is the identity of S 1. So, you can take the base point as 1 comma 0 the unit complex number in S 1 ok on both sides. Then what we get identity of the integers is R composite i check. See identity homomorphism identity map induces identity of homomorphism right on the fundamental groups. But identity map is R composite i. So, when you take the check of that it is R check composite i check. i check goes from pi 1 of S 1 to pi 1 of D2. Then R check goes from pi 1 of D2 to pi 1 of S 1 back right. And this is identity because R composite i is identity alright. Now you see that there is a contradiction to the algebraic fact that we have proved that the two end groups here this one and this one they are infinite cycle groups. The middle one is what? This is a disk so it is convex set. So, pi 1 of this one is trivial group. If you write A it is 0, if you write manipulative it is 1 you can write a trivial group. So, from a infinite cyclic group any homomorphism trivial group that is trivial then whatever you compose back here it is a trivial map. So, composite will be trivial because it is going through trivial group. But we have identity of Z identity this is remember this is infinite cyclic group pi 1 of it. So, identity map is trivial that is a contradiction. So, look at how a topological fact here topological hypothesis here converted into algebraic facts and then use that algebraic fact right to conclude that whatever we started here is wrong because this does not fit here. So, you see these are illustrations of how algebraic topology works. So, we will have many such things. So, pi 1 of a disk is trivial group is used, but if we did not know how what this one group is we would not have been able to tell anything ok. If we should know to have this conclusion we should know that pi 1 of S1 is non-zero group non-trivial group that much we should have known we did not know that. So, you have to compute it now we actually know it is infinite cyclic ok. All that we used is that it is non-trivial ok. Let us see how to get a very good theorem out of this one M. Lee Brower's fixed point theorem ok. So, this corollary was a step towards that one. So, Brower's fixed point theorem says for any disk ok a continuous self-continuous map has a fixed point which you have studied for closed interval 01 or any closed interval AB also in your real analysis course by just using intermediate value theorem. But, intermediate value theorem will not work in the case of D2 because there is no order here. D2, D3 etcetera it is not work. So, what is involved here? So, that will come out if you look at this kind of flows ok. So, how to show that any continuous function from D2, D2 has a fixed point. This again we go by suppose there is a map such that fx is never equal to x for any of the axis. X and fx are different points of the disk right. Any two points inside R2 will determine a unique line right. So, look at that unique line. Inside that line you can take only the portion between fx and x. So, that is a segment ok. Line segment fx and x. Now, you can give it a direction namely trace it from fx to x. You can trace it from x to fx fx to x. So, look at trace it to fx trace ok. Keep extending this the same line till you hit the circle. The unit the boundary of the disk ok. The boundary of the disk will be hit by whether you go forward or backward in two different points. I want it the forward point. So, from fx to x that is nearer to x than to fx ok. There are two points on this line segment the two points on the circle intersection of the circle with the two the entire line. So, I want the point which is towards the point x rather than towards point fx that is all. Now, let us call that point gx. So, I have constructed a function x going to gx out of the function f. The only thing is only after assuming that f has no fixed points ok. The whole thing you know geometrically it is easy to see that whatever you have done is a continuous operation namely g will be also continuous, but you do not have to rely on your geometric intuition here you can actually prove that g is a continuous function where because f is a continuous function ok. Let us see how. So, first of all let us have a look at what we have done in this picture. x is not equal to fx, but both of them are in the disk. I am not saying that they are inside the disk they are in the closed disk they are some of them may be on the boundary also does not matter the picture shows both of them are inside that does not mean that they should be on the inside ok. Extend that line towards the fx to x and hit the point g which is on the boundary. So, this function g is expressed in this picture by geometric means ok. What I want to say is that this g you can actually write down a formula for g in terms of x and fx ok. Finally, it is a function of x only because fx is already a function of x. So, let us see how. So, a parameterization of this entire line is 1 minus t times x plus t times x where t range over r given any two vectors u and v t times g plus 1 minus t times that gives you the entire line passing through the end points of these vectors u and v right. So, that is the vector algebra. So, t times 1 minus t times x this is the line segment this is the entire line. But I do not want the entire line I want what g x ok is 1 minus t time t naught times x plus t naught times fx where t naught is the root of the equation quadratic equation in terms of t what is that equation the norm of this must be equal to 1 because it is on the circle. The norm of this vector is nothing, but you say you can rewrite this whole thing t square t naught square norm v square plus twice t naught v dot index plus x norm of x square minus 1 ok. So, that is the solution of this when t equal to t naught you get this one. So, this is the equation it is a quadratic equation norm v is given ok what is norm v namely fx here ok plus 2 times t v is fx dot x x is another vector plus norm of x square minus 1 equal to 0 ok. So, I have put what is this v v is the vector fx minus x ok you reject what we have done here 1 minus t times x plus t times fx can be written as t naught times fx minus x that is why I have taken that you rewrite that t times fx minus x as v then the computation of this norm is here ok. So, solutions of a quadratic equation wherein the coefficients are functions of x automatically will be or smooth functions if the coefficients are smooth functions. If they are continuous the solutions will be also continuous because we can give them by formula v naught plus minus square root of v square minus 4 a c divided by 2 a whatever right that is what we have to do. So, since x square minus 1 is negative or non-negative because x ways inside the circle right in the disk of Bayer radius 1 it follows that the discriminant of this quadratic equation is non-negative and identically 0 if and only if this x square is equal to 1 this is 0 ok and v dot x is also 0 v square minus 4 a c so that is 0 means we both of them must be 0. But then what happens is modulus of fx square will be equal to modulus of x square plus v square v is fx minus x ok if v is perpendicular to x then a norm of fx square will be norm of x square plus norm of v square right norm of x square plus norm of v square will be greater than 1 which is absurd because this whole thing is less than 1 ok that will not happen therefore the discriminant is strictly positive and hence the two rules are continuous. See whenever equality equal to 0 there will be a problem about discontinuing functions are continuous or discontinuing and strictly positive you can take the square root it will be positive function it will be a continuous function right. So, I have given you full justice why the roots of this polynomial of this quadratic equation are continuous functions of the variable x here. So, discriminatively positive and hence the two roots are continuous in particular the t0 which is the solution of this equation is a continuous function of the variable x it depends upon x we are writing it as a constant but as x changes the point also g will also change ok this g is nothing but you know g is nothing but 1 minus t0 times x plus t times fx ok finally. So, g is continuous function ok now you can verify either by using the quadratic equation or by directly using the picture if x is already on the circle fx may be anywhere what happens to the line segment when you extend it x itself is a point on the circle beyond that it will go away. So, this type of intersection is gx will be equal to x gx is equal to x if and only if x is on the circle ok. I have given a point g I have given function g which is identity on the circle and it is a continuous function from the entire disk to circle g takes values in the circle its domain is the entire disk and on the circle it is identity and that is a contradiction to the first corollary here there is no such function there is no retract out we will do. So, why we get a contradiction we started with the assumption that assumption must be wrong fx is not equal to x for any x what is the conclusion there must be a point x such that fx is equal to x ok. So, that is the conclusion next time we will prove another method to compute the fundamental groups both these methods will be later on strengthened far beyond what you see in the beginning in the course later on ok. So, next time we will prove what is called as van Kampas theorem a starting very simple version of that there are many other versions later on all right yeah thank you.