 Continuing with the classification of one-dimension manifolds, recall that we have represented every manifold as the quotient of a disjoint union of open disks, all of them are, whatever you want to take say the standard Tn. Then we understand that in order to study, in order to understand the topology of the manifold, we need to understand the equivalence relations defined by this quotient map which are given by what are called as transition functions. And we have done a beginning in one of two of the lemmas here with examples, motivating examples. So it takes care of a number of a big chunk of what is happening there. Namely the nature of homomorphism occurring in the gluing data, we can now concentrate on other aspects like the Uij which we have defined as psi i, psi i inverse of psi j of some d1j or whatever. So we would like to know how many components it has and so on. Since Uij is an open subset of this minus 1 plus 1, it is a countable union of open subintervals of minus 1 plus 1. This is one of the biggest advantage of being in one dimension such descriptions are not possible in general if you go to R2 and R3 and so on. Our first claim in this direction is the following. Let X be a host of space and psi 1 psi 2 from d1 to X be any homomorphisms on to open subsets U and U2 of X respectively. Assume that U2 is not a subset of U1. The first claim is that there is no component of psi 1 inverse of psi 2 which is an open interval of the form AB for some a and b strictly inside the interval minus 1 less than a less than b less than 1. In particular this will imply that U1 intersection U2 which is the image of U12 under psi 1 also can have at most two components. This is the statement of course, we will see the proof so on. The second part of the statement of this lemma is that assume for the psi 1 of d1 is not contains of psi 1 of d2. It is more condition now. Here we have assumed that U2 is not contained in U1 which is same as psi 2 of d2 is not contains of psi 1 of d1. So, here this is now U1 is not contained as U2. You also assume that then after changing psi i by a reflection in the region if necessary one of the components of this U12 is as a form b, 1 for some b strictly between minus 1 plus 1 and psi 2 inverse psi 1 from b, 1 to minus 1, c defines an increasing function which is a homeomorphism on to the interval minus 1 plus 1 where once again c is between minus 1 plus 1. So, this is the second part. So, this is a little more elaborate than our earlier lemma which we have seen and this will more or less complete the local analysis of what is happening inside each d1, each copy of d1 that we have that define the given manifold. So, let us go through the proof of this. The first thing is components of U12 are all homeomorphic to open intervals. I am repeating this that is all. The emphasis here is that none of them will be some middle portion of d1. It will not be equal to minus 1 strictly less than a, less than b, less than 1 where a and b are strictly inside that one. It will not be of the form a, b where a and b are strictly in the interval. So, this is the claim here. Now, suppose that is not true. Namely, there is one component of U12 of the form a, b. Now, restrict this psi 2 inverse composite psi 1 from a, b to c, d. c, d is another subset of open interval inside d1, 2 another copy of minus 1 plus 1. Since U2, U2 which is the image of this is not the whole of U1. That means U2 is not contained inside U1. It follows that cd is not the whole of minus 1 plus 1. If cd is whole of plus 1, then image of this will be completely contained inside that. So, cd is not the whole of this. Therefore, c, let us say or d, one of them, which one you have to say, choose say minus 1 is less than c less than 1. If c equal to minus 1 or d equal to minus plus 1, then it would have been the whole thing. That is the only case. So, one of them you have to assume. So, I am assuming that c is strictly inside. Choose that case. The other case will be similar where d is inside 1. So, we will study what happens if c is strictly inside it. We are claiming that it is not. That means we have to get a contradiction from this. Suppose psi 2 inverse psi 1 from a, b to c, d is increasing. So, it is the case again. It follows that psi 2 of c and psi 1 of a, you remember psi 1 and psi 2 are defined on the whole of the interval. I am restricting this to this homeomorphism. So, psi 2 of c makes sense and psi 1 of a also makes sense. They are points of x. What happens if t c is the case? These will be distinct points of x. They are not identified, which cannot be separated by open sets because any neighborhood of c will contain some neighborhood here which will intersect with this one. So, a plus epsilon and c plus some epsilon, they will be identified. So, that will contain the portion of it. So, they will never be, you can never be separating them. This is the point which we have used this argument. So, this is what happens here. On the other hand, suppose now this is decreasing. Decreasing means what now is a, b would have become 1 to, a would have 1 to d and b would have 1 to c. That is the one. Then it follows that psi 1 of b will have a problem, b and psi 2 of c. So, they will be very, very nearer, but they will be distinct. So, they cannot be separated by open sets. In either case, we get a contradiction to the horse-drawnness of x. So, exactly same thing will happen if instead of this I assume d is less than 1. So, in either case, something has to happen because the whole interval cannot behave the whole thing. So, we have a full contradiction. Contradiction to what? Contradiction to the fact that the one of the component is of the type a, b. So, this completes the proof of 1. So, once that is the case, what is the meaning of this? That means components must be running from minus 1 to some interval or from minus 1 to some point or some point to the other end plus 1. So, such intervals are only two of them, only two possibilities. So, there are no other possibilities out of which you may have only one component or you may have two components wherein this is, this will be the picture. So, u 1, 2 is completely understood. The picture is that either it is one interval or this end or that end or both of them. Now, assume the same hypothesis on the other side, namely u 1 is not contained inside u 2. We can assume that the same thing for u 2 1 also. The intervals inside components of u 2 1 also will have the same problem, argument will be the same. So, once you have that one, now assume that one of the component of u 1, 2 is of the form b comma 1. It follows that by composing side 2 with a reflection if necessary, this b comma 1 should have been mapped to either minus 1c or some d 2 1. If d 2 1, you change by reflection and make it minus 1c. So, I can always assume that b comma 1 is mapped to minus 1c by side 2 inverse side 1. The picture for the components in both d 1 as well as d 2 is the same. So, we have to make whether this interval goes to the other end or the same end, the initial end or the terminal end. So, that is what we have to make cases. Finally, having established this, there are two cases. Either this map, this homeomorphism is increasing or it is decreasing, which the same thing is order preserving or order reversing. Suppose this is decreasing, this would imply that if it is decreasing means what now? b goes to c and 1 goes to minus 1. So, 1 goes to minus 1, there is no problem, but b going to c, there will be a problem. So, side 1 of b and side 2 of c will be distinct inside x, but cannot be separated by open sides. Therefore, this must be increasing b to 1 minus 1 to c. So, this must be increasing. So, this is what we wanted to establish. So, that completes the proof of this lemma. So, now, we can consolidate the whatever observation we have done in another lemma. Let x be a one connected manifold. So, all these we assume that nothing about x exists that exists host of. Now, I want to make a clear picture. Now, let it be one connected manifold having an atlas consisting of only two members. Then I will write to complete what happens here, complete the classification itself. Suppose only two memory units such that one is not contained or even is not contained side unit. Then the intersection is non-empty and has utmost two components. This is what we have seen already. If u1 intersection u2 has only one component, then x is homeomorphic to an interval, of course an open interval because we are assuming all the time in the case when x is without boundary. So, this is what our first lemma ensures of yesterday. If u1 intersection u2 has two components, then x is homeomorphic to s1. This is the second lemma that we had. So, this is nothing new here, but let us see. So, how we have wanted u1 intersection u2 non-empty has to be there first of all. Why? Because if there are only two open sides, if they are disjoint that means x is disconnected. So, it must be non-empty. And then it can have only one component or two components. So, now claim one follows from the previous lemma, u1 intersection u2 non-empty and has the second part. The second part is that it is exactly same as lemma 6.4 and 6.5. 6.4. So, this is the lemma. It is homeomorphic to an open interval 0, 1. So, let us go back. In the third case, we may assume that one of the components psi1 to compose a b12 minus 1c. You can always assume this one because if the initial component goes to the initial component by changing the sign of psi2, we can make it the terminal component. You can always assume this one. And it is increasing, but this is a part of the previous lemma actually. I do not have to repeat it. It follows that the other component, we have two components psi1 inverse of u1 intersection u2 is of the form. It means initial component minus 1 to a and the other component of psi2 inverse of u2 is of the form d, 1. So, minus 1, a is identified with d, 1. Again, for the same reason that exists Hausdorff, it follows that minus 1 a to d, 1 must be again increasing. If it is not increasing, a and d will collide and they will come closer, but cannot be separated. Therefore, we are in the situation of lemma 6.5, the next lemma here. The conclusion follows. What was the next lemma here? This is what we had proved earlier. Then it is homeomorphic, psi1. So, this is the situation here. b12 minus 1c and minus 1a to d1. What are these alpha, betas now? In our case, this is psi2 inverse of psi1 and this is psi2 inverse of psi1 restricted to two different components. That is all. So, now we can continue with the proof of the theorem. Earlier, we had arranged the sets u1, u2, u3, etc., such that ui is not contained ui plus 1 and so on. Actually, all that was not necessary, but now something stronger than that is necessary. That is what I am going to do. By second countability, we want to, we have a countable family of uj's, homeomorphic to the open disks and covering the whole of x. Step one, we want to arrange them in the following way. What is this? There exists a countable family of uj's such that each uj homeomorphic interval, this part is okay. Second one is the kth member is not contained in the union of earlier k minus 1 elements. Which I am denoting by wk minus 1. This is just, wk minus 1 is just definition. The k minus 1 component intersects the kth component. The union of up to k minus 1 components, wk minus 1, intersection uk is non-empty. k greater than 1. I want to arrange it so that. Finally, the fourth condition is if wk is not the whole of x, then it is homeomorphic to an interval. So this is what I am going to claim. So this is my first step. Let us see how we prove this one. To construct such a family, we start off with any one member from the family uj, any one member, call it u1 equal to w1. w1 is the union of all these. So it is w1. Having picked up u1, u2, uk, so as to satisfy the above requirement inductively. So they satisfy 1, 2, 3, 4. Then I want to pick up uk plus 1, the next one. Having picked up u, u2, uk, so as to satisfy requirement, we check first of all that the wk which is the union of uj up to k is the whole space or not. If it is the whole space, there is nothing more. Even if there are other uk, they will already contain these others. So we do not have to do anything. If so, we stop. Otherwise, it means that there are members uj not contained in wk. If none of them intersect wk, then what happens? x will be the disjoint union of two open sets, at least two open sets, disjoint union. One is wk and rest of the other uj. That would mean that x is disconnected. So that cannot happen. So at least one member which we have not taken yet must be intersecting wk. So you take that. It would mean that x is, so it would mean that there is some such thing. Therefore, there exists a member. Label that one as uk plus 1. This uk plus 1 is not contained in wk, but it intersects it. So 1, 2, 3 you have done. Now you have to look at wk. I have, by induction, I have proved that wk is an interval. I want to show that wk plus 1 is also an interval provided x is not the whole space. So this is what I have to get to prove. Essentially, there are two cases to be considered. Depending on whether the number of components of wk intersection uk plus 1 is equal to 1 or 2. So this is prior previous lemma, this is the only possibility. And these two components have to be end components, not in the middle. That does not occur. All these we have already seen. Accordingly, it follows that if there is only one component, wk plus 1 by previous lemma is union of these two intervals, patched along a common subinterval. So it will be an interval. If there are two members, it will be a circle, full circle. If it is a full circle, what happens? We have already seen that one earlier. It will be both open as well as closed in x. But x is connected, so it must be the whole space. So problem is over. But if I assume that, see the fourth condition is if wk plus 1 is not equal to the whole space, then it cannot be a circle, so it must be an interval. So therefore, inductively all these things have been satisfied for k equal to k plus 1 now. That means the construction is over, step 1 is over. Let come to the proof of step 1. So let us finally complete step 2. Step 2 is to only analyze what happens when we keep on running this one infinitely. That is the only case. If it stops at anywhere level, there are only two cases. Either wk is a circle or it is an interval. The problem is over. So we have classified the all manifolds. If it keeps on going on infinitely, then what happens? So that is the case. So to analyze the case when the sequence wk is infinite. For a fixed homeomorphism f1 from w1 to minus 1 plus 1, fix some homeomorphism because w1 is nothing but u1. There is already a chart. You do not have to worry about any other thing. You call this as f1 now, a new notation that is all. From the lemma 6.5 which follows that f1 of this w1 intersection u2 is nothing but f1 of u1 intersection u2 is of the form minus 1a or b, 1. Accordingly, we can extend f1 to a homeomorphism f2 from either on the left side minus 2 to 1 or on the right side minus 1 to 2. See minus 1 plus 1, you extend it by 1 on the left side or 1 on the right side. Which way you do not know? Only one side. The next one may be the other side. The next one may be the same side but other side and so on. You do not know which side. So both ways you have to keep extending as and when ui, u1, u2, u3 keep coming. Suppose you have constructed the homeomorphism fk from wk to ikjk. Remember wk is a subspace of x. It is not a subspace of r. Now we are constructing homeomorphism of this into open intervals inside r. Suppose you have constructed this one. The next one namely uk plus 1 may be coming and intersecting the left side of the interval or the right side of the interval. Accordingly, you have to extend that which is an extension of fk minus 1 from here to here. So all this I am assuming inductivity where ik will be smaller than ik minus 1. If ik minus 1 is 5, minus 5 this will be minus 6 or if it is 5 or itself this will be I am going for a minus and plus 1. So it will be all these things will be negative or this jk will be positive. jk minus 1 is bigger than jk. jk minus 1 is smaller than jk. So ik is smaller than ik minus. So and they are integers with the property that the difference is just 1. jk minus jk minus jk minus this is again this is a mistake nobody pointed. What I wanted is jk minus jk minus 1 here and ik minus ik minus 1 that difference is 1. Therefore, once you have extended this way from fk to fk plus 1 and wk plus 1 to ik plus 1 to jk plus 1 we keep extending like this just the way we have extended f1 to f2 either way. A homeomorphism if you keep changing each time there is no way each time you have to extend the old homeomorphism on w1 it is f1 on w2 f1 gets extended to f2 and so on. So that is important. Once you have that take a to be the limit of all these ik's it may stop at some point this interval these sequences are either constant after some stay there is no further intervals coming or if they are keep coming then it will be ik's will be minus infinity similarly jk will be either some number or plus infinity. So a and b could be plus infinity minus infinity plus infinity that possibility is there. It follows that a is minus infinity or b is plus infinity one of them must be true because we have infinite sequence of the wk's each time we are adding one extra to the length of the interval minus 1 plus 1 is length is 2 then at the second level it will be 3 4 and so on the length definitely increases. Therefore the one of the points either the left end or the right end must go to infinity minus infinity plus infinity. So whatever it is this a and b are not may not be number they include infinity or plus infinity minus infinity define f from x to a b to be fx equal to fk of x where x itself is in wk once it is in wk it may be in wk plus 1 but fk plus 1 will agree with fk there so it is well defined. So f is well defined on each interval it is a homeomorphism already therefore it is a homeomorphism we can now compose f with a homeomorphism a b see once a b I do not know it may be minus infinity plus infinity there is a homeomorphism a b to 0 1 because any open interval or open ray is homeomorphic to 0 1 so conclude that x is homeomorphic to the interval okay. So finally what we have done is starting with manifold x half star what second countable one-dimensional manifold we have shown that it is either homeomorph to 0 1 or the circle okay. Now final thing is that we have already done in the right in the beginning I recall if it is a boundary then how did we complete the proof if it has boundary this 0 r is the interior of that therefore the boundary points can be at most two either 0 or 1 or both okay so in either case it will be one of the two things 0 1 0 close 1 is the homeomorphic to 0 open and 1 close so we get two more possibilities and that completes the proof of this the entire classification so thank you.