 So, continuing with the classification of one-dimensional manifolds, let me restate our earlier theorem that we wanted to prove. So, let X be a connected one-dimensional half star and second countable that is a under assumption abstract topological manifold. Then X is homomorphic to one of the following, the open interval zero one, half closed interval zero one, closed interval zero one or the circle S1. Recall that we also reduce the k, reduce this proving this theorem to the case of proving when boundary of X is empty. Then we have to prove that is either it is open interval zero one or the circle this is what we want to prove. So, we had already reduced that case to that one. So, where do we start off? We start off with the definition and of course, later on we will say look at this is connected and it is half star which is second countable etcetera we keep on using those. So, the definition gives you immediately the following lemma namely take more generally, take any n manifold not necessarily one-dimensional manifold. Then X is the quotient of a countable disjoint union of copies of open disk of the n. It is a quotient space of some disjoint union of open disk. All manifolds are like that. By second countability argument, there is a countable atlas every manifold has a atlas. Here we have a countable atlas. So, let us call it ui fi i belonging to n this i belonging to n for X where fi i from ui to dn are homeomorphisms. Now, you take m to be disjoint union of dn as many as you know countably many whatever it is finite than finite many of this family disjoint union of dn i. What is dn i? Each dn i is a copy of dn open unit disk but disjoint union abstractly. From there I can define a quotient map now namely q from m to X be the map given by on each copy dn i take it as the inverse image of fi i. Remember what are fi i? Fi i's are from maps from ui to dn. So, fi i inverse will be from dn to X going into ui. So, they are all maps into X but they go into ui. So, q restricted dn i is fi i inverse since dn i's are disjoint copies. So, this definition makes it completely defined on q and it will be continuous because this is an atlas that means union of ui is in the whole of X this q will be surjective. So, it is a quotient as a as a fact theoretically it is a quotient map but it is also an open mapping on each of these open subsets dn i's it is a homomorphism. Therefore, it is an open mapping any surjective open mapping is a quotient map. So, it is more than that it is very stronger than it is a open mapping. So, the theorem is over this lemma is over we have displayed it as a countable union of this one. Of course, when I say I at the most I it could be finite also here. So, that is also allowed. So, next we would like to make some bargaining here. We do not want all those you know some of them may be useless and so on what are the meaning of useless. So, this is what we want to do by deleting those open sets which are already covered by the union of previous members. See we have already listed we have already made a listing u1, u2, u3 and so on. If u2 is already covered by u1 what is the point you just deleted u2 there is no need for u2 go to u3 go to u4. Suppose u1, u2, u3, u4 cover u5 then what is the use of u5 delete that one and so on. So, delete those members which are already covered by all the previous members union of the previous members this is what I want to do. So, in the listing ui, fi starting with n0 equal to w0 equal to un0 u1 that is u1 this is our notation u1 is un0 and that I call it as w0. Let n1 be the least number of for which un1 is not contained inside w0. Keep going maybe u2 is not contained then u2 will be called as un2. There is some sequence I am going to a2 will be n2. If 2 is already there u2 is already there then I will go to u3 like this. The first one the smallest one it is contained not containing so that will be n1 un1. So, call here just n1 be that one. So, you call w2 equal to un0 so that is second one w2. So, like this you put wk plus 1 define it as wk union un plus 1k. What is the property of un plus 1k? It is not contained inside wk because wk is the union of the previous members. So, I think this is very clear. Of course, if the list is finite you will stop at some place otherwise it can go go on you do not know. Even if the original list is infinite you may end up with a finite list like if the compact space you may be somebody may be a might have union infinite covering. But because it is a compact space you can always get a get away with a finite covering. So, take the modified covering with the property that the nth open set coming is not contained the union of the first n minus 1 of them that is the property I want that is all. So, some simplification. In what follows we shall assume that ui phi i is a countable and minimal at last in the previous sense. So, all the extra extra n s ui s have been deleted that is it. Now, take phi i equal to psi i equal to phi i inverse from minus 1 plus 1 to x. Now, it is a special case namely n equal to 1 now we are into the n equal to 1 case now. Phi i inverse of minus 1 plus 1 d 1 minus d 1 is minus 1 plus 1 to x. If you may think of x as a quotient of a distant union of countably many copies of minus 1 plus 1 that is the lemma and q is the quotient map d 1 i to x where q restricted to d 1 i is psi i the inverse of the these coordinate charts. So, they are called actually parameterizations. Therefore, to understand x what our manifold is we need to unravel what kind of identifications are taking place under this quotient map. Any quotient map is same thing as introducing equivalence relation. So, what are these equivalence classes there which points are identified with where that is what we have to understand all right. Therefore, we start. So, this is the picture that I have this is some one-dimensional manifold possibly these are all intervals of minus 1 to plus 1 and there are these parameterizations this part will cover something that part will cover something and so on. So, it is quotient of distant union of minus 1 plus 1. Now, put u 1 to u 2 1 u 3 1 u 4 1 or u 2 i and so on. So, u i j more generally equal to psi i inverse of psi j of d n 1 take a take an interval go to its image through one of the j's whatever psi j and then take psi i inverse of that. So, psi i inverse of that may not be the whole interval because that may be covering something else right. So, but it will be a subset of when you take psi i inverse it will be subset of u i. What is u i or d i I mean d 1 i right. So, it is a subset of d 1 i I have been called that d 1 i is u i also more generally okay. It is coming from u j. So, therefore, I am calling u i j this is notation psi i inverse of psi j of d 1 j okay. So, each of them is an open subset of d 1 i the first one it is subset of d 1 i. To begin with we need to know what are the possibilities for each component u i j components of u i j inside this u i is minus 1 plus 1 u i or d 1 i is minus 1 plus 1 okay. I would prefer to write it as u i and remember that they are all minus 1 plus 1. Remember u i j is some open subset of this minus 1 plus 1 okay. What are these components and how the maps are related how what kind of maps we do we get this what we have to understand okay that it is possible to understand these things is surprising because it is one dimension it is possible okay. So, let me illustrate this with an example okay. This example will tell you most of the story actually what is happening. Suppose that class consists of just two charts only concentrate on one u i and u j and psi i there and psi j there just concentrate on that part what is happening to begin with okay. So, we can assume that m itself is d 1 1 union d 1 2 instead of i and j and so on. Then let us consider some simple cases here namely the following cases the first case a is u 1 2 is open interval 1 by 2 to 1 contained is a u 1 u 2 1 is open interval minus 1 to minus 1 by 2 okay that is a possibility because e 1 u 1 2 is subset of u 1 and u 2 1 is subset of u 2 the other one. So, here I have 1 by 2 to 1 that is minus 1 to minus 1 by 2 okay. Now if I go from here psi i and then psi 2 inverse psi 1 and psi 2 inverse I will get from here to here from this interval to that interval a homeomorphism composite of two homeomorphisms to homeomorphism. So, what could be let us let us say that it is something like t going to t plus t plus 3 by 2 I have to be careful if I add 3 by 2 half will go to so from here minus 1 will go to half and minus 1 by 2 will go to 1 okay. So, from psi 2 to psi 1 inverse psi 1. So, from u 2 1 to u 1 2 this is the homeomorphism t going to t plus I am just assuming I am not proving anything one of the homeomorphisms. So, take the easiest one or it may be the other way around namely what is this one just minus t this one going to minus t even simpler okay. So, what happens under these two cases let us understand okay in the first case we can take alpha from m m is what m is disjoint union of these two this is notation d 1 and d 2 like this okay. So, from m I will define minus 1 to 5 by 2 by restricting on the first copy of minus 1 plus 1 it is just t on the second copy I will just shift it by 3 by 2 like this one and take t plus 3 by 2 okay. So, that this alpha will agree with these identifications okay. So, that is the whole this t goes t, t goes 3 by 2 and then goes t. So, that is why you take alpha t here and on the this part you take t plus 3 by 2. It is easily checked that this map which is defined on the disjoint union factors down to x x is a quotient space remember okay to alpha hat from minus 1 to 5 by 2. You can see that this is a surjective and it is a 1-1 mapping okay it is it is open it is actually homomorphism on each part can be verified. So, this will be homomorphism okay. So, here is the picture for that. So, this is 1 by 2 to 1 this is your u and 2 here and this is u to 1 and this is your x now I do not know what x is but I have actually got it this way okay the quotient space has to be this gets identified with this one by shifting the t going to 3 by 2 this goes to t here but this goes to 3 by 2 times this one. So, they agree a point here is identified the point here but both sides will agree on that part. So, therefore these two patch up together to give you a map from x to this whole interval from minus 1 to what 7 by 2 or whatever 5 by 2. So, in this case what we have proved is the manifold x is again an interval open interval its length may be larger do not worry we are not interested in much in the length okay. Let us see what happens in the second case which is this looks like simpler namely t going to minus t t going to minus t you are just reverse in the direction. So, in the picture I can just write this way. So, this is minus 1 to plus 1 this is minus 1 to plus 1 the other way around okay here right this minus 1 to plus 1 the other way around all right. Now a point here see 1 by 2 to this one goes to actually 1 by 2 and 1 by 2 1 by 2 and minus 1 by 2 would be the same thing 1 by 2 goes to minus 1 by 2 and that is going here. So, this one goes to minus 1 to minus t and that goes to here that is why I have written like this okay drawing picture is this way is here okay. So, what happens is this portion they are identified but this portion and this portion are different. So, you have to draw pictures like this what happens to the minus 1 by 2 and 1 by 2 they are not identified only open interval are identified this point this black point here it is not a part of the manifold because I have taken open intervals remember that. So, on this side I have not shown them at all but here I have shown it because it is ignored but here I have put a landmark here what happens these points are not identified the open interval is identified what happens to those two points they will be in the closure of both both this both of them will be closure in the this open interval part here. Therefore, half-sterveness of x is violated here conclusion is that we do not want to accept that half-sterveness is not violated but we want to say that our own assumption is wrong namely we cannot have the second case at all the psi 1 inverse of psi 2 will not be minus 3 this is acceptable to gives you a long line longer interval. So, the identifications if you have an interval like this an interval like this identification like this is allowed identification like this this is not allowed that is the meaning of if you do like that there will be a point here and a point here which will come close enough but remain distinct that means what an open part open interval has both of them is in the closure. So, there will be no interval around them which are disjoint which means that half-sterveness is violated. So, that is the gist of this one example I will tell you one more example which is much more interesting one. So, this is done so b so here I am assuming u12 is minus 1, 2 minus 1, 3 and 1, 3 to 1. So, u12 has consists of two connected components in the case a it was connected here it is not connected but it has two components u12 and u21 is also minus 1 to 1, 3 and 1, 3 to 1 this is just an assumption this is a simpler case okay. Here suppose psi 1 inverse of psi 2 is again I have to make a case t plus 4 by 3 on the first component mainly minus 1 to minus 1 less than t less than 1 third minus 1 third t plus 4 3 and on the other part it is t minus 4 third what does it mean see you have two things here one interval I did not make a picture of this one because this was much much easier actually. So, I have I will let you make a picture for yourself. So, if using this picture itself for example what is happening is there are another interval here this interval is shifted to this interval another interval here and this interval is shifted to this interval okay it is just a shifting homeomorphism okay that is the case I am I am looking at okay in which case you define beta from the disjoint union of these two copies of minus 1 plus 1 by these formulas namely take t going to e power pi i 3 t by 4 see minus 1 to plus 1 if you take pi i t it will cover the entire circle exactly once right but here I am covering a only three fourth of it so 3 t by 4 okay the first one covers that much of some other part is left out beta is the second part is covering it d d 2 is covering it namely 3 t plus 4 by 4 which means rotate it by pi rotate it by through an angle pi whatever you whatever you have taken so those two will cover the whole thing okay. So this beta is defined from m to s 1 okay there are these two are disjoint copies you can define whichever you like so the beta is a subjective mapping all right what are the points which you go to the same point they will be precisely given by this so these are the relations here therefore what happens is that beta factors down from m through the quotient map q from x to s 1 and that map beta hat from x to s 1 will be homeomorphism okay so what you have done is taking two arcs like this and then forming this kind of operation identification portion here overlapping portion here overlapping that will give you a circle so that is what is happening here okay so I think these two examples explain the whole thing whatever I wanted to tell you the the the key here is though these are look like very special examples they will take care of most of the cases so I have to make a general statement here okay there is a general fact indicated by these examples we formalize this in the following two lemmas the proof will be completely obvious however I implore you to write down it on your own so take it as an assignment okay so this is a statement let a less than b less than c d less than e less than f be real numbers now I am not even using that they are minus 1 plus 1 and all that okay so this is the whole idea because for any open interval is homeomorphism to any other open interval so that is why I am writing arbitrary a less than b less than c d less than e be real number m be the disjoint union of the two intervals a to c and d to f okay let alpha from b c to d e the b c is one end of the first interval and d e is other end lower first interval in this this part initial part of the second interval okay d e so alpha from b c be an order preserving homeomorphism like a shifting adding in the other case which which we violated what was it it was order represent t going to minus t that was not a good one so you have to take order preserving homeomorphism then the identification space namely m is the disjoint union you identify t with alpha t where t is inside b c open interval b c this you call it as m alpha why alpha because this alpha is the identification map so it is just even m alpha this will be homeomorphism to the open interval 0 1 okay now I can't write it is 3 by 4 4 to 1 by 4 1 to 1 5 and so on okay I don't care it either therefore I am writing it as some open interval namely it is homeomorphism to normally 0 1 so this is the lemma okay similarly you can have the second one which will give you a circle okay exactly similarly so this this this is the first case this is a 1 a 2 was a case which is not occurring so b case b is discovered by this one okay in the example given now I am taking minus 1 plus 1 itself to be more accurate given minus 1 less than a less than b less than 1 minus 1 less than c less than d less than 1 let alpha from this end b to 1 you see b 1 to this end minus 1 plus 1 okay the end of the first one with the initial part of the second one or the other beta is minus 1 to a to d1 so this end and that end both of them be order preserving homeomorphism again let x be the quotient of the disjoint union of the two copies of minus 1 plus 1 by the identification t is identified with alpha t on this part where t is between b and 1 and here it is using beta t is identified with beta t on the second part minus 1 to a then what you get is s1 homeomorphism to s1 the quotient space is homeomorphism to s1 okay proof is you have to correctly copy this one without using these 4 by 3 etc that is the challenge okay try so we are going to use this one so further things we will take next time so let these two examples sink properly in you next time we will discuss further proof of the classification thank you