 we would like to compute the fundamental group of the circle. The basic tool here is, as I have told you, the exponential map r to s1 namely theta 1 to e power 2 pi i theta. The fact that r is contractible, therefore if you take any base point little r, pi 1 of r r is going to be a single point. Is that easy to see? Any loop based at a single point we have seen is the longotopic inside any interval. We have seen that. So in particular it is true in r also. So this fact will come to us very, very much useful. But when we go down to s1 under e power i theta, something strange happens, but not too much strange things. So it gives you complete control over what is happening in pi 1 of s1 namely the exponential map that gives you the control. So let us first concentrate on what is the big feature of this exponential map. It is a surjective function and exponential of t1 plus t2 is what is exponential of t1 into exponential of t2. So the addition inside r goes to multiplication inside s1. So it is a group homomorphism. What is the kernel? Kernel is determined by 2 pi, multiples of 2 pi. Exponential of t1 is equal to exponential of t2. If t1 minus t2 is an integer because now I have put theta going to equal 2 pi i, the 2 pi i has been divided. So it is just an integer. All the integers go to the same point including the 0. Wherever 0 goes to namely 1, so they all go to the same point. In fact difference is an integer, the image will be the same. So this is the meaning of exponential of t1 equal to exponential of t2. If and only if t1 minus t2 is an integer. So these things I have sketched in this diagram. So all the integers I have put a bullet kind of slightly larger points 0, 1, 2 here minus 1, minus 2 here and so on. All of them are going to this bullet here in the circle. I have taken interval say 1 4th to 3 4th, 1 4th to 3 4th and 1 plus 1 4th to 1 plus 3 4th and so on. What will they make? It will be from this i you have to trace all the way up to minus i. This is angle pi by 2, this is 3 pi by 2 under 1 4th will go to pi by 2 under this map. That is the meaning of this because I have multiplied by 2 pi, 2 pi i and the real number. So this is the exponential function. It is injective restricted to any open interval of length less than 1. In particular 0 to 1 open interval it is injective. 1 and 0 go to the same point. You take any interval which is of length less than 1 then exponential map is injective because difference of any two members there is not going to be an integer that is all. So this is the property of this function which would be exploited to the brim. So this lemma says the following. For every point z in S1 the open set exponential inverse of S1 minus z throw away one point. How does the inverse look like? It is a disjoint union of intervals and if you take exponential function restricted to each of these open intervals it is a homeomorphism on to the this S1 minus a single point. No matter what point it throw away. So what happens? The inverse image of this one single point which you have thrown out it will be all various points look at one single point say r naught. Then the next point will be r naught plus 2 pi sorry r naught plus 1 because I have divided by 2 pi I keep saying 2 pi and previous point will be r naught minus 1 and r naught minus 2 r naught minus 3 and so on. The difference will be always an integer where r naught e power 2 pi f r naught is your z. In between these two intervals from r naught to r naught plus 1 it is an injective mapping on to S1 minus z. Okay what is this inverse? Inverse is precisely what you call a logarithm function chosen inside S1 minus z. Log function is not defined on the whole of S1. If you throw just one point it is defined. The complex logarithm of any unit vector has the real part zero. So it is purely imaginary and you divide by 2 pi I then what you get is the inverse of this one. So this is all a little bit of complex analysis that is all that I am recalling here. But this property is going to be very very fundamental for us. An inverse of exponential defined on any sub arc of S1. Any sub arc means at least one point should be missing then you take any subset of that which is connected that is a sub arc. Okay it is called a branch of the logarithm if you want to use this this terminology. Maximal sub arcs on which a branch of a logarithm is defined may be defined are S1 minus z. As soon as you include the full thing it is not defined. If you throw a one point it is defined. In what follows we will use branches of logarithm defined on open arcs two of them either throwing one or throwing minus one. If I throw one I get one arc okay it is a very big arc except one point it is whole circle right and I take another arc like this throwing away minus one. So these two things are important first we will use them okay. So the branch is why I am saying branch is if you choose say minus one to zero that is one branch open branch. It is same log exponential find that will not work when you take zero to one that is a different branch. But once you define once you choose the interval of maximum length minus one maximum length one okay open interval there it is a one one mapping so it has inverse and that inverse is branch of the logarithm alright. So let us do something the first thing is start with any connected space actually you would like to path connected but connectedness is enough here okay take two functions x to r okay such that when you compose it exponential they are the same okay. Then there exist an integer n such that f one of x minus f two of x is equal to this integer for all x the difference is given by one single integer okay look at this one exponential function from r to s one this is stress looks like our fundamental problem in the lifting problem this is the corresponding p from e to b is r and b is s one x is an arbitrary space we are studying the lifting problems here this is the first case of that what this theorem say what this lemma says is up to an additive constant all the lifts are the same take any two lifts f one and f two they differ by a one single integer additive integer if f one minus f two is a constant function n is statement clear once the statement is clear the proof will be as easy as it is okay how do you prove look at this difference function g f one minus f two may make sense because they are they are taking the real value so difference makes sense the difference is also continuous f and g are continuous if you will continue this has the property today exponential of g remember exponential is a homomorphism from additive group to automotive group therefore exponential of g is exponential of f one divided by exponential of f two but they are the same so it is equal to one and this is true for all x therefore exponential of g x is contained in the set of integers because exponential of anything is equal to one means it must be an integer this is what we have seen right therefore g from x to r is a map which takes only integral values but x is connected if you have connected space image of a connected space under continuous map must be connected so what is the connected subset of z integers it has to be a single point and that point is n some n so for all x g x must be one single n so what we have proved now is that lifts of functions taking values in s one two functions taking values in r via the exponential function they are unique up to additive on a strand now let us look at this proposition one by one whatever let f be a function from i to s one that is a loop now I could have denoted by omega okay but I would like to loop function theoretic notation here take any map from i to s one let us take say f 0 is one this is just to standardize okay this is not a very essential thing so it is starting at one then there exists a unique map g from i to r such that the starting point zero g of zero is zero sitting over one zero this zero is inside r the one is in in our s one okay so the first one one this one this one is a value in s one it is a unit complex number zero is zero and exponential of g is f so this says that every function from i to s one can be lifted not only that it can be lifted you can choose the starting point whichever integer you want I have taken it as f f zero is equal to one suppose I have what a lift like this then what happens to other lifts I know by the previous lemma I have to add an integer and I get it so whatever integer I add f zero will be corresponding integer suppose I subtract by one then f zero sorry g zero g zero may can be made into minus n or plus n or any number therefore along with this proposition it says that any function any smooth function from i to s one can be lifted and there are so many lifts namely infinitely many lifts one at each point one at each integer so this will be the meaning of this proposition so we have to do it only for one namely g zero equal to zero okay then we are done is that clear let us begin how we we are going to do this but complete proof we will do next time okay so how are we going to do this so the previous lemma the uniqueness follows this what I just told you okay there is a unique map suppose there is one then another one will be differing by this one by an integer but I have fixed a g zero equal to zero so that additive integer n must be zero that means g minus g1 minus g2 is zero this means g1 is going to get that is the uniqueness okay so we have to show only the existence so what is the idea of doing existence what we do is we will use the connectivity of i okay essentially but we will do it in a little more economic way look at all points set inside i such that there is a g with the property namely g zero is zero and exponential of g is f okay you know that there is only one such g if at all so suppose g is defined up to t zero to t then you put that t inside z so z is the subspace of the interval obviously zero itself is in z right because I can take g zero equal to zero and that's all e power zero e of exponential of zero is one we know e power two pi i zero is just one so this is this set is non-empty okay in the usual parlor what you would like to do is that we will show that z is open and closed suppose we do that then because i is connected and z is non-empty z must be the whole of i if z is the whole of i what happens zero to t that t must be taken can be taken as one so g is defined on the entire of zero one so that is the solution so this is the scheme of proof this connectivity is used to prove lots of existence theorems like this in topology more generally not for just i any connected space that's why we did the first lemma for just connected spaces but lifting cannot be done all the time okay for that you have to use some special property of i that we will come to next time all right thank you