 Let us begin defining the composition of paths, fix any topological space, take two paths, the first path n set omega 1, omega is the path omega 1, the starting point of the second path tau is same thing as omega 1, so tau 0 is omega 1, in that case we will define this omega star tau, omega followed by tau, this is not the composition of functions, it is not as if x to y, y to z, both, both omega and tau are from i to x, right and what we need is, omega 1 must be equal to tau 0, then we can define omega star tau, this omega star tau is exactly the same way as we have defined in the homotopy, concatenation of two homotopies, but that is precisely what we are going to do, namely in the first half of the interval 0 to t less, 0 less than 2 t less than half, we will define it as omega, but double the speed omega of 2 t, in the second half again we are defining it as tau, double the speed, but the origin has to change, start at half and end at 1, therefore it is tau of 2 t minus 1, when you put t equal to half in the first one it is omega 1, in the second one it is tau 0 and those two points are same, therefore the left hand side will be a continuous function, what is its starting point, it will be omega 0, what is its end point, it will be tau 1, so this is the meaning of composition of two parts, usefulness of this operation is essentially due to its path homotopy invariance, what is the meaning of that let us see, suppose you have omega 1 homotopic to omega 2, this homotopy is always path homotopy and tau 1 homotopic to tau 2, then the compositions corresponding compositions will be also homotopic, path homotopic, remember omega 1 homotopic to omega 2 always means that the end points of omega 1 and omega 2 are the same, omega 1 star tau 1 can be defined means the starting point of tau 1 is the same thing as end point of omega 1, because the same thing should be true for tau 2 also, starting point of tau 2 will be automatically equal to end point of omega 2, because end point of omega 2 is same thing and end point of omega 1. So, the composition omega 1 star omega tau 1 is defined will imply that omega 2 star tau 2 is also defined, not only that this homotopy will imply that the compositions are also homotopic to each other, how does one get this one, by just putting these two homotopies together just like the way we have done it in the case of in the homotopic classes of maps. So, h is a homotopy from omega 1 to omega 2, g is a homotopy from tau 1 to tau 2, in the first half you define it as h, h of 2 t s, don't worry about the second coordinate at all, keep the second coordinate as it is, it is the first coordinate t which defines the functions, which has the paths, the second coordinate keeps changing the paths each time. So, each time you double it 2 t and g of 2 t minus 1 exactly the same way as we have defined the compositions. You can easily verify that this capital F t s will be a path homotopy from omega 1 composite omega 2 to sorry omega 1 composite tau 1 to omega 2 composite tau 2, okay. Next lemma tells us about the basic algebraic properties of path compositions, which I have already summed up in the last module, if you have understood this properly, this path is automatic, whereas it may not be that you have understood it, understood it, so let us go through it. It is actually that remark which I have made in the last module is enough to talk about this associativity identity and inverses, let us see how. So, let us call the first composition omega star tau star lambda, so I have put the bracket on the second ones, let us call it as a gamma 1, suppose that is defined, then the other bracket omega star tau bracket, right, then star lambda that will be also defined and these two will be path homotopy. The associativity is up to path homotopy, obviously gamma 1 and gamma 2 will be completely different because in defining gamma 1, look at what we have to do, the first half of the interval will be occupied by omega, then the second half of the interval will be shared between tau and lambda, whereas in the definition of gamma 2, the first half of the interval will be occupied by omega as well as tau shared between as well as tau, which means only one fourth of the interval will be omega 1. Therefore, it is obvious that gamma 1 as a function is not equal to gamma 2, but a pleasant surprise is that it is path homotopic to gamma 2, okay. Similarly, if you compose with the constant path on this side, whatever path you get is path homotopic to the original path, namely take any path omega starting at say at the point x, constant path at x you have taken, so I can take omega star C B where it is B is the end point, it is homotopic to omega and omega, it is also C A star omega where C A, A is the initial point of omega, A is omega 0 and B is omega 1. So this side is I have composed with the constant path C A on the left hand side, on the right hand side I have the constant path B. So these are homotopy identities for the operation which is the path composition, okay. Moreover, the inverse path, the so called inverse path, namely trace in the same path in the reverse direction. I have denoted by omega underline, okay, omega underline of this omega 1 minus D. If you compose it with omega, it will be constant path at A, if you compose omega bar first and then take omega, then it will be constant path at B. Wherever you start from, go to the other end and come back in the same path, it will be as if you are in the same or in your point, okay. So these are the three statements, all of them can prove, in fact have been proved by one single remark, namely that you can get all of them by different parameterizations, different parameterizations of the same path, that is why they are path homotopy. Left hand side and right hand side are different parameterizations, except the last one, three is somewhat similar but that is not correct, one and two are just different parameterizations of the same path. So let us see why. In other words, all these homotopies have nothing to do with actual omega and actual X, it is the property of that they are defined on a closed interval. Everything is happening inside the interval, you should understand that one. So this is a very simple idea but it has to be understood that way. So instead of making it like as if some mystery, we will just write it down the formulas for each of them and be done with it, okay. So that is the way many books do it, I also have done it for your sake, okay. So having explained this path, how it works, okay. So I am going to tell you, I already told you that the first path occupies, omega occupies half the interval, first half the interval and the second half the interval is shared again between tau and lambda, okay. So these things we will use and write down the homotopies. So here are clear homotopies I have been written down. First let me tell you how one gets this one. You see the picture, look at the bottom. In the first, there are three squares here, I cross I, I cross I, I cross I, for three different, we have three different statements there. The first one is associativity, omega star tau, then bracketed star lambda. So this is your gamma 2 perhaps in the rotation. At the top, it is omega bracket tau star lambda, okay. So that is the definition, tau star lambda you define first and then compose it with omega on this side. So this is how we get. The first half of the interval will be shared by, you will be omega and the second part will be shared by tau and lambda. So what you do, you join them, this one fourth to half there, this is half two, three fourth there and zero to one like that, join this one. In between homotopies say at t equal to s, at this point, t equal to s, say s equal to s naught, you have to go not at one fourth, here you go up to one fourth more omega, but you go up to this point by omega, then follow by tau and then lambda. By the time you come here, instead of one fourth, you will be up to half, you will be taking omega, then tau and then lambda like this. So this is a picture, what is actually this value etc. is determined by the intersection of these two lines, that's all. This line is given by some s equal to s naught and joining some point t naught here to t1 here, then you have to find out that. So what you get by doing that is the formula here, this one, I explained one here, similarly the two also. So what I want to do is tell you how this switch got, last one, this is omega, omega bar. So what is it, you have path, tell you the matter, how it goes up to half something and comes back. But you know that any path which have same endpoints to any two paths, which have same endpoints, inside i to i, they are homotopy as path homotopies as keeping the endpoints same. So here what happens, if I take omega and come back omega bar, it will have the endpoints, the starting point of omega. If I compose omega bar with omega, it will end point, end point of omega, starting point, the end points, both of them will have same thing. Endpoints of them. Therefore, they are homotopy, though the constant path is at the respective points over. You do not have to write down any formula, because once you have proved, given one single formula, a t, remember a t s was defined as 1 minus s times alpha t plus s times t, showed identity map is homotopic to any map alpha. So any two maps are homotopic the same way. You can write it as a alpha of, sorry, alpha and beta two maps from i to i, end points are the same, whatever end points. So alpha 0 equal to say beta 0 and alpha 1 equal to beta 1 wherever they are in the interval. Then you can just take 1 minus s times alpha t plus s times beta t that will show that alpha and beta path homotopic. So that is the thing that is happening in the last one. If you have difficulties in seeing this way, there are just formulas, you just verify it. How the formulas are obtained? I told you these are the pictures which tell you how to get the formulas. We will have a session in writing down these formulas later on rigorously at another point, not at this point. What we have done here, you can see that s is less than up to s by 2, this is the function. So here what is happening, let us go back, look at this one. S equals s naught, half of this distance, whatever this distance, half of that it will be one thing. After that it will be something else, see up till here. Then it is this part. Then you reach. So this is constant path, this is your omega, only this part. In other words, as the homotopic keeps taking, it will not go all the way at all. In the first phase, you go all the way to the other point and come back. After half the time, you go only half way and come back. At the final point, you know, don't go at all, you stay there all the time. That is what a homotopy is doing. Homotopy of omega composite, omega bar, omega underline is null homotopic, showing that this is how it, all right? Now let me tell you a few things about why interval 01 and so on. It is interesting to note that during these homotopies, the entire action is taking place in the domain itself and so the proofs that the composite extract etc. do not depend upon the actual path is, do not depend upon the SpaceX. Because of property 3 in the above lemma, many authors use the notation omega inverse. After verifying this one, we can also do that but the inverse is somewhat confusing. It is not the same path. It is up to homotopy and that too in the, not as a functional inverse, not 1 by omega and so on, okay? Those things do not make sense. So we shall also use this notation but you have to be careful with this. Some people use even minus omega and that is very much valid because in integration theory, if you integrate on the reverse path, what you get is the negative of the original integration, okay? So many notations are there. They are all, they have their own justification. Intuitively, any continuous map from any interval AB should be called a path. Why restrict to 0 to 1? Why interval, the unit interval should be? Why all the time? So this is what you should have done. Some people try to do. Our definition of a path as a map from the closed interval causes a minor irritation namely if you restrict a path omega from 0 1 to x to a closed sub-interval, it is no longer a path in our definition, right? Restriction should be also, they say 0 to, you have traced a path. Half the path is also a path in that way but in our definition, we have to re-parameterize it, make it a function, look like as if from 0 1 to x. So people object, this is not a good definition, okay? When you say when people object, you give me a better definition, you just ask, right? So it is true that if you restrict to 0 1, start with omega from 0 1 to x, just take the restricted function 0 to half. Is it a path? That function is not a path in our definition, you see. Notice that composition law had to be defined after re-parameterization even if we adopt the more general definition, okay? So if you take AB arbitrary and CD arbitrary, so omega of B is equal to tau of C, then what is the interval for omega star tau you take? Do you take AB or CD or AB union CD? AB union CD may not be an interval. Even if they are, there may be more overlapping than just the end points, right? So arbitrary intervals again have to be converted into some standard intervals and then only you can compose them. So that is a point. I mean it is not as if 0 1 we have taken and this is too restricted and so on. Strict associativity law always fails. Similarly, the constant path is not a strict unit. Indeed, there are a few different ways to avoid some of these difficulties, but some other difficulties you will acquire. We shall discuss such a thing. I have given you an example in a form of exercise later on. So what I want to say is no matter what kind of definition you take, there will be some problem. For example, I will right now tell you instead of 0 1, etc., you can take the entire inter, entire real line, but then you want the path should end somewhere. Otherwise, you know non-compact thing will not be called a path. It should have a starting point and an ending point. Therefore, what we have to take the entire inter, entire real line, but take only those functions which are constant after a, after some point. After say 1 and before 0. So everything less than 0 will be omega 0. Everything bigger than 1, omega 1. Omega 1 is equal to omega of every t, t bigger than 1. So you put that condition. Okay. You understand instead of 0 1, you put any two points A, B and there it is function is just continuous everywhere, but for all t less than A, it is omega A. For all t bigger than B, it is omega B. If you take this definition, then restrictions, etc., have no problems. So such things can be slightly modified, but whatever you do, there will be some other problem. This is what I want to tell you. Okay. Therefore, the definition with 0 1 is found the best among all other definitions. Okay. So how to deal with this when you cut, take a path. Okay. Now we cut it into two parts. Then I would like to think of this as composition of the original path as a composition of these two paths. This is called subdivision. The subdivision should be allowed and it is easy to adopt it in our definition and we have a beautiful theorem there which will take care of all objections of this type. So let us come to that one. This is precisely called as invariance under subdivision of the homotopy of up to homotopy what you call as the concatenation or the composition of these things. Namely, take any path 0 1 to x. Okay. Now you divide it, this interval, some into n parts 0 less than t 1 less than t n less than 1. So this is subdivision. Now you restrict the original path to say gamma 0 1, 0 t 1, t 1 to t 2, t 2 to t 3 and so on, gamma 1, gamma 2, gamma n. Take the composition in our sense. You will not get the original path gamma but what you will get is homotopic to, path homotopic to gamma. To prove this one, you have to do it only one single division namely 0 1, put a t 1 in between, cut it down, re-parameterize as these two paths as s, original as 0 1 to 0 1, both of them. Reparameterize and define the composition, what you get is the original path up to homotopy, absolutely. Okay. Therefore, you can safely say that all objections of this kind, you know, half the path, it is a path fine. Only thing is you think of this as again taking place between 0 and 1, that is all. So here is a proof there. I prove it for n equal to 1, namely only t 1, then by repetition it will follow. Okay. By induction it will follow for all n. It is enough to prove this for n equal to 1. So with t 1, I will call it as a 0 less than a less than 1. By repeated application, we will get the general case. So let us prove this for this one. Consider the following re-parameterization of the unit interval from alpha 0 1 to 0 1 given by alpha t is equal to twice a t, 0 less than t less than equal to half, 2 t minus 1 plus 2 t twice 2 minus 2 times a half less than 2 t less than 1. Okay. So what I have done, the half, first half of the interval and t equal to half, this will go where it will go to a. When t is 0, it will go to 0. So half has gone to a and 0 goes to 0, half goes to a and 1 goes to 1. Put t equal to 1, what do you get here? t equal to 1, this is 0, this is 2 t minus 1, this is 1. So this is a path. So this is a re-parameterization of, suppose 0 1 itself, you think of this identity path and I am cutting it instead of half, I am cutting it at a. These two are homotopic, path homotopic is what this alpha t says. Therefore, once you take this one, gamma restricted to 0 to a, re-parameterized by this method, that is the gamma hat I have put, star gamma a to 1 hat is nothing but gamma composite this alpha. Therefore, it is homotopic to path homotopic to gamma itself because alpha is a re-parameterization. So this is the proof of this general theorem, invariance under some division. After this, you must be, you know, you can do all the algebra of composition of this path is smoothly. So now we shall specialize to the case when the end points are the same. That we will do in the next module. Thank you.