 Hello, welcome to module 60 of NPTEL NOC, an introductory course on point setup or repark tool. So, today we shall continue our classification of one-dimensional manifolds which we started last time. Beginning with a any manifold with or without boundary, the first thing we did was to reduce the proof to the case of manifolds without boundary. And now, look at a manifold without boundary, take a cover by coordinate neighborhoods. As soon as there is a cover, there will be a countable sub cover because the manifold is second countable. The next thing is we want to understand how the two intervals and two open subsets which are homomorphic intervals, right, they are all coordinate neighborhoods, they are homomorphic intervals, how they intersect in the whole space x. So, we made two list of things which are, which are visual, which are, which we want to happen. And one of them we considered, namely when the two open intervals like this, they actually intersect in a very nice way like this. Then we could, we could get a map to the union, okay, from a open interval which is a homomorphism. So, the union itself was an open interval with a homomorphism. So, this was a nice case. The next case that we want to consider today is that two of them have intersection two components. And they intersect like this properly, not like that or that and so on. So, the two things are coming this way just the way in the first case and the other one also coming nicely like that. So, this is the case we want to understand now and this is one desirable case. So, that is the first one. After that, we will see that these are the only two cases possible. So, that will allow us to complete the classification, okay. So, start with any connected one manifold, psi i from a i, b i to u i, we need to local parameterization such that u 1 intersection u 2 consists of two components. I am just labeling the two components a, b. The first thing is psi 1 inverse of a is a 1, c 1, okay. See the domain of psi 1 is a 1, b 1. So, this psi 1 inverse of a is a 1, c 1 and psi 2 inverse of a is c 2, b 2. The domain of psi 2 is a 2, b 2, so c 2, b 2. So, here it is starting at one end, first end here and other one is ending at the other end. So, that is the hypothesis. The third part is psi 1 inverse of b, exactly the opposite here, is d 1, b 1 and psi 2 inverse of b is a 2, d 2. So, this is the other end, right end, this is the left end and so on, okay. The third condition is, now look at on the intersection psi 2 inverse of psi 1, starting from part of the interval a 1, b 1, go into the even intersection u 2, then take psi 2 inverse, come back to the interval, part of the interval a 2, b 2. So, from interval to interval, this is a homeomorphism that must be order preserving on both the intervals, namely the first portion, second portion a and b correspond to a and b. So, first portion will be a 1, d 1, okay, 2, c 2, b 2, okay, so that is what we want. So, it could be a 1, b 1, a 1, c 1, a 1, c 1 and d 1, d 2, does not matter and the other one will be d 1, b 1, 2, a 2, d 2. So, both of them should be order preserving, then M is homeomorphism is one, conclusion is of course, there are four different conditions I have assumed here, even before that there are only two of them you have to have and we have already assumed that, that none of them covers the other one and so on, remember that, in any case the intersection consists of two components with this part automatically gives you that u 1 and u 2 are not, so contain one contained in the other, okay, so that is not, there is no need to separately state it. So, conclusion is that, the union will be now homeomorphic to a closed interval, closed manifold, namely the circle itself, the entire M is circle, in other words whenever such things happen, there is no other open subsets, open intervals and so on, the whole M will be union, u 1, union, u 2, this is the whole idea, okay, let us see how the proof, the proof is not all that difficult once you have understood the previous one, so here, so just look at the picture, a 1 to b 1, you have one coordinate neighborhood u 1, okay, psi 1 and here a 2 to b 2, you have another psi 2, of course the final picture I have put it nicely, but right now you have to assume that this is some manifold, that is all, right, so this is u 2 part, homeomorphic to that one and that is the u 1 part and the intersection is this a and b, so u 1 comes from here to all the way here, up here and u 2 is from this point to that point, so these two are the intersections, a, b, right, now psi 1 inverse of a is this part, a 1 to c 1, psi 1 inverse of b is d 1 to b 1, okay, similarly psi 1, psi 2 inverse of a is c 2 to b 2 and psi 2 inverse of b is a 2 to d 2, okay, not only that, when you come from, from here, suppose it is from here, you go to here inverse image that is psi 1 inverse, okay and then sorry, so if she first comes psi 1 here and then take psi 2 inverse here, so you get a map from a homeomorphism from a 1 c 1 to c 2 b 2, that must be order preserving, okay, similarly you start from here d 1 b 1, come here and then go by inverse image, you get d 1 b 1 to a 2 d 2, that must be also order preserving, so these are the assumptions that have, both are order preserving, then the conclusion is that this u 1, u 1, u 2 is circle, not only that, once you have that one, there is nothing else, m is a connected manifold, so it has to be whole of this one, this is what we have to see, okay, so pick up any point t 1, s 1 as shown here, namely a 1 less than t 1 less than c 1 and d 1 less than s 1 less than b 1, then there exist unique points t 2 and s 2, what are they, look at the image of s 1 here and that is image of something here, because this is homeomorphism another way, so there is a unique s 2 here, which comes to that one, these are the points, t 1 psi 1 of t 1 is this point, psi 1 of s 1 is this point, they are also equal to psi 1 of psi 2 of s 2 and psi 2 of t 2 respectively, okay, so you have started picking up this point, so clearly these t 2 and s 2 will be obviously inside that a 2 to d 2 and c 2 to b to that soft, okay, after that what you do, now you just define the map lambda from 0 to 2 pi, closer interval to m by the formula in the first part it is psi 1, in the second part it is psi 2, the only thing is you have to adjust the whole thing by re-parameterizing the intervals, okay and where do you take, up to 0 to pi I am going to take psi 1, pi to 2 pi I am going to take psi 2, here the picture is clear, so from this point I will map up to this point using psi 1, okay, from there, so parameter is pi here now, the pi to 2 pi I will use this map, so s 2 to t 2 I will ignore the rest of this overlapping part, s 2 to t 2 I will re-parameterize from pi to 2 pi, similarly t 1 to s 1 I am parameterizing 0 to pi, so map will be like this, okay, so this part is coming here, this part is coming here, so the arrow is here, this arrow is here this way, so you have to understand, so this is counter clockwise since I have taken here, so I have to when I map this point t 1 as 0 here, okay, e power 2 pi as 0 or cos theta cos 0 plus sin, comma sin 0 that is the point here, go all the way up till here and then pick up psi 2 here, so the idea is clear and formula is also clear, what I have to do, put some a t plus b so that t 1 to t 1 to s 1 with this interval goes to 0 to pi, so that is all idea, so this is the map, similarly t 2 to s 2 to t 2 goes to pi to 2 pi, okay, so s 1 minus t 1 by pi times t plus t 1, when t is 0 this is t 1, when t equal to pi pi pi cancels out, t 1 t 1 cancels out it will be s 1, similarly when t equal to pi here this will be 0, so this is s 2, when t equal to 2 pi this will become pi pi, so a t 2 minus s 2 plus s 2 that will be t 2, okay, so you take psi of this psi of that except when these 2 points are the same namely there are 2 formulas you have to see they are same, why they are same because psi 1 of this psi 1 of this point psi 2 of that point are the same that is all, similarly when it is 2 pi also at the end they are also same you see, so that is what you have to see, okay, so what happens is first of all 0 to 2 pi it is continuous okay because on the interval psi on the common point they agree, so it is a continuous function in each interval 0 to pi pi 2 pi it is given by a homomorphism, so it is injective, this injective because they are mapped into different components, different parts of the things here you see and here comes the important thing that they are order preserving, so there is a common portion here namely I have taken this part, so this part is covered by psi 2 also, I have covered this part, this part of this, this is covered by psi 2 also on the intersection they are order preserving therefore the left out parts are you know they are never mapped by psi 2 on to this part, so that is the whole idea, so this is similar to what we have verified in the first case, okay, so the entire thing is injective except 0 and 2 pi are mapped to the same point, that is relevant grouping that is precisely what we wanted here therefore this 0 to 2 pi will factor down to what to s1, see 0 and 2 pi are mapped to same point right, therefore lambda factor down to continuous bijection from s1 to u1, what is the projection map here from 0 to 2 pi to s1 capital s1, t equal to e raised to pi i, e raised to t just e raised to pi i, e raised to i t because 2 pi I have taken, so t equal to e raised to i t or just cos t plus i t, so that is the plan, so under that this will give you a homeomorphism now, continuous bijection from the circle to m, s1 is compared, m is Hausdorff therefore this is a subjective map, so it is not subjective, we do not know that one, it is subjective on to u1, u1, u2, so whatever it is, it is homeomorphism on to its image and the image is u1, u1, u2 that much is covered, but then u1, u1, u2 is open as well as being compact it is closed also, therefore it must be the whole space because m is assumed to be connected, so every bit is used here, so we have completed the proof that m is actually homeomorphic to the circle in this case. So two important cases which will produce the two different final conclusion have been covered, so now the claim is that there is no other case, these are the only two cases that is the whole idea, so of course in the second case we have to complete the proof also, there is also that part, having taken care of these two favourable situations, we now claim that we are always in one of these two cases, so what is the meaning of that, let me elaborate, start with a house door space, psi 1 and psi 2 from minus 1 to plus 1, this does not matter, you could have taken any open interval minus 1 to plus 1 to x be homeomorphisms on to some open subsets u and u2 of x respectively, neither of them contained in the other, assume that intersection is non-empty, then these are the only things that can happen, no component of psi 1 inverse u1 intersection u2 which I have assumed it is non-empty will be an open interval of the form a, b for some minus 1 less than a less than b less than 1, in other words it is this open, every component is an open interval right because they are all subsets of now minus 1 plus 1, so connected components of an open subset are open interval, those open intervals none of them will be able to avoid both the end points away from the end point, that is the meaning of minus 1 less than a less than b less than 1, then what can have, what else can happen, it can happen only if one of them, one of the end points must be there in the interval, so I assume minus 1, if my plus 1 is also there it is the whole space, it cannot be whole space, so only minus 1 to b it can be or it can be a to 1, so there are only two possibilities, therefore the conclusion is in particular u1 intersection u2 has at most two connected components, see once you have proved this one, this in particular this is obvious because connected components of u1 intersection u2 are in one one correspondence with the connected components of psi 1 inverse of u1 and u2 inside the interval, see this u1 u2 is happening inside the topological space X, but here we have come to the open interval, so there you can see there are only these possibilities, so there are only two possibilities at the most you have two components, so this is the strongest thing first we have to observe, next if u1 intersection u2 is connected, that means it is only one component, then u1 union u2 is only more fit to an open interval, this is our first case, so that is what I have to verify, if u1 intersection u2 has two components, then u1 union u2 is only more fit to s1, this is the second one, so this is precise in the meaning of saying that there is no other possibilities, let us verify this one, so we have to just verify this one, these two I have just elaborately tell you how this one compares all, you have already seen it, but let us see, so why this happen, indeed the argument for this is already used, so I will again elaborate on this one, first of all note that psi i inverse of u1 u2 is a proper open subset of minus 1 plus 1, it cannot be the whole space because neither u1 nor u2 is contained in one, one contained in the other, so u1 intersection u2 cannot be the whole of u1, similarly cannot be the whole of u2, moreover its components are all homeomorphic to open intervals and they are in one-one correspondence with the components of u1 intersection u2 because psi i's are homeomorphic, the emphasis here is that none of them will be some middle portion of minus 1 plus 1, minus 1 less than a less than b less than 1, so that is not possible, assume on the contrary that one of the component is of the form ab with minus 1 less than a less than b less than 1, suppose it has happened, we want to say it does not happen, that means we have a middle portion here, then look at psi 2 inverse of psi 1 of ab that is some other cd that is contained inside again minus 1 plus 1, here I do not know what happened, psi 1 something bad has happened, here I do not know I do not care, something has happened here, where cd, c equal to d, c equal to minus 1, d equal to plus 1, all places are allowed, only thing you know that here also that cd is not the whole of minus 1 plus 1, okay, let us see that may be useful or that may not be useful, so let now alpha from minus 1 plus 1 to minus 1 plus 1 denote the function t equal to minus t, the reflection, by replacing psi 2, I do not want to change the psi 1, psi 1 there is something bad has happened, replacing psi 2 by psi 2 composite alpha if necessary, I could have taken any other homeomorphism, right, so take psi 2 composite alpha instead of psi, we may assume that psi 2 inverse of psi 1 from ab to cd is increasing that is order preserving, if it is order reversing I will do this one, if it is preserving I do not want to do any, I will not change it at all, okay, if necessary do not change it unnecessarily that is all, okay, when if it is increasing do not do anything, if it is not increasing that may be increasing because there are only two possibilities for a homeomorphism of intervals to intervals, okay, therefore you can change the second one by reflecting, so now it will be increasing, so once you assume that cd is a proper subset of minus 1 plus 1, right, so it follows that minus 1 less than c less than 1, okay that is my assumption or minus 1 is less than d less than 1, okay, so I am assuming I am not assuming anything on ab, ab is a bad, assume to bad here, what cd can be there are only two possibilities, alright, so I am using that fact now here that cd is not the whole of minus 1 to 1, so one of them at least c already must be strictly inside the interval and that is going to cause a problem namely in the former case suppose c is the, c is in the middle of the interval, in the second interval, okay, what happens psi 2 of c and psi 1 of a they are coming very near but they are not identified, they are in the open part, right, so what happens is psi 2 of c and psi 2 of d are distinct points of x which cannot be separated by open set, in the latter case the same thing happens with psi 1 of b and psi 2 of d which cannot be separated by open set, so in either case you have got a contradiction to the house dorseness of x, okay, so here is the picture you know fully explaining what is happening, this is your u1, okay and that is the portion of u2, other portion I have drawn I do not care what is happening, okay, so psi 1 a is coming like this, this psi 1 part is coming like this, okay a is some in between minus 1 to a is coming here and psi 2 of c just comes here, this portion and this portion will be in every neighborhood of this a as well as b, as well as c, so a and c are not, a and c are distinct points, okay, sorry, not this portion, this the other portion because they are, they have to continue afterwards, right, this is not the end points of the interval, this portion will be common, so for every open neighborhood of psi 1 of a inside the u1, there will be some portion every common with psi 2 of a, this is the intersection part on the left hand side, similarly here what happens, there will be on the left hand side there will be intersection part, these portions are distinct points, but as soon as you see a psi 1 of a and psi 1 of b, okay, so they are the image, they are there inside the, inside our x, right, but they are distinct points, they are in different ones, they are not assumed to be in the intersection, right, they are open intervals, so these two points, you know contradict the first portion of the interval, so some psi c prime to c, psi on this part, okay, will be coming there, that is all, so I do not know how many I have drawn, drawn rest of them, some d prime to d that will be distinct, that is the whole idea, so they are in the, they are in this part of this, so rest of the part here they are, they will be common, so here all of them will be common, so you cannot separate them by distinct open subset, okay, so that completes the proof of this first claim that no interval can be of this form, once you have that, you have only two components, at the most two components for the intersection, now I have to say that all the hypothesis of the first case is covered or all the hypothesis for the second case covered, that is all, then these two and three are the only possibilities, okay, that is the second part and third part here, so second part is the first part from whatever we have seen, it follows that for each i equal to 1 and 2, psi i inverse of u and u2 is of the form minus 1 to ai or bi to 1, this end or the other, this end, okay, therefore in both of them because I can apply the same thing to psi, once I observed psi 1 by symmetry the same thing should be true for psi 2 also, okay, the conclusion once you have done for psi 1, there is a completely symmetry, 1 and 2 you can change, that is all, so here a1 minus a1 to, minus 1 to a1 and 1 b1 to 1 and then other case, there also minus 1 to say a2 and b2 to a1, so these are 4 cases are possible, you know, you can take 4 cases to be considered but they are all symmetrical, so for definiteness you just cover one of them, argument will be same in other cases, okay, so consider the case when these intervals of the form minus 1 to a1 and minus 1 to a2, so a2 for psi 2, this is for psi 1, we claim that psi 2 inverse of psi 1 minus 1 a1, I am assuming that it is only one component, in part 2 we have only one component, so one component this is what is happening, if you look at the again psi 2 inverse psi 1 minus 1 a1 to minus 1 a2 has to be decreasing, if it is increasing what happens, similarly psi 1 of a1 and psi 2 of a2, they will be coming very close to each other but they are different, so the rest or other part will always be, they will have common part in every open neighborhood, there will be open intervals common to both of them, so they will not be selected, that is the meaning of this one, so if they come like this there is a problem, so it must be decreasing, it should be the other way around, so lately it should be like that, not like this, this is not allowed, so that is the meaning of that picture, this kind of identification is not allowed, they have to be like this, which means this if you this way the other function should be like that, so this means that they must be here, so this must be decreasing, otherwise it follows that every neighborhood of psi 1 of a1 and every neighborhood of psi 2 of a2 will intersect each other constantly in the half Gaussian, therefore this psi 2, psi 1, psi 2 inverse psi 1 from minus 1 a1 to minus 1 a2 is decreasing, now consider psi 1 prime, namely composite with alpha, psi 1 prime t equal to psi 1 of minus t, then the two homomorphisms psi 1 prime psi 2 prime fit the hypothesis of Lemma, 12.39, now they will be exactly like this can join them, no need to worry about what is happening here, so we are done, so that is case 2, case 3 also similarly, but I have to show that we are inside the second case correctly, now we are assuming that we have two components for intersection u1 intersection u2, it follows that phi i of u1 intersection u2 must be again end point a minus 1 to a i disjoint union b i to 1, i equal to 1 and 2, this psi i is i here, these two there are two cases to be considered again here, namely psi 2 inverse psi 1, this component a minus 1 a1 may be mapped to minus 1 a2 there or minus what b2 to a2, which way they are mapped that is what I would do, so this psi 2 composite psi 1 inverse may be from minus 1 a1 to minus 1 a2 or it may be minus 1 a1 to b2 to 1, as soon as this happens the other one namely b1 to b1 should be the other component that is fixed, you have freedom only to choose where one of the component goes, the other component has to go to the remaining component, there is no choice, so I have to only these two cases here, okay, the first component here goes to the first component there or the first component goes to the second component there, so these are two cases, again by symmetry I have to just see what happens to the one of the cases, so case a let us take, so let us look at the case a, so same reason as in 2, we conclude that this minus 1 a1 to minus 1 a2 has to be decreasing, okay, so that the thing is coming again and again, so let us consider change psi 1 by a reflection, then we now claim that now you have the whole thing changed, right, so as soon as you change the sign the increasing, decreasing will change on both the components, here also it will component, the point is you can change the thing only on one of them, automatically you cannot change the other one because it will change there because the whole thing single interval only, right, so the intersections have two different components, okay, as soon as you adjusted the first one correctly, you may or may not, okay, but here it is decreasing you have to change, okay, so that is the whole idea in this one, so that is why I have written in this picture I have shown this portion coming here, this portion coming here, okay, so now if you change this interval so that the both of them are like this, then you are in a nice shape that is all, alright, whether you want to do it or not it just depends upon you, but what happens once you do that the other one, okay, so this one we have seen already, yeah, here, so once you change psi 1 like this we now claim that the homomorphism psi 1 prime psi 2 fit the hypothesis of lemma 12.40 completely, clearly psi 2 inverse of psi 1 prime minus 1 to b1 now, see all the things have changed because t has been replaced by minus tf, minus 1 b1 to b2 to 1 and therefore psi 2 inverse of psi 1 of minus 1 1 2 will be equal to minus 1 a2, okay, 1 to a1 was minus 1 to a1 was there now I have changed the sign, so this will be the minus 1, 1 to minus 1 to a2, okay, the far end here left, right end is coming into the left hand end here also it follows that psi 2 inverse of psi 1 prime is from minus 1 b1 initial segment goes into far end b2 to 1, so both of them will be inquiry, finally it follows that psi 2 inverse composite psi 1 of minus 1 1 to minus 1 a2 is also inquiry, the other part is also inquiry, okay, for otherwise you will have psi 1 prime minus 1 a1, minus a1 and psi 2 from a2 will be violating the hypothesis, okay, so that is why we are in nice situation of second lemma there, therefore the conclusion is that in this case the entire manifold has to be s1, okay, we do not need that right now, we just want because I have not assumed that x is connected here, okay, so I say even in unit is s1 that is fine, alright, this is the lemma in which which just says that whatever we desired only that will happen that is all, now let us complete the proof of the theorem, recall that we started with a connected one dimensional manifold by second countability we get a countable cover ui of x by open sets all of them homeomorphic to say let us say fi from ui to minus 1 plus 1, inductively we define a finite or infinite we do not know but countable, okay, increasing sequence that is why sequence means countable wk of open subsets of x such that each wk is connected union of wk is the whole space that is all first we have to do this way but we will do it in much more elaborate way so I will describe you as follows, so how do we do that start with w1 equal to u1, so you have got a countable cover so you have indexed it somehow never mind but that indexing may not be very good so we are going to do some changes here, so start with w1 equal to u1 having defined wk, okay, so what I am going to do look at all those ui's which are such that they are not contained inside wk and those which intersect wk, so in the first case whatever so w1 is defined what is this s1, s1 is all those indices i in n such that ui is not contained inside u1 and ui intersection u1 is non-empty, now suppose there is no or none of those things will intersect it at all that will contradict the hypothesis that x is connected, okay, therefore there must be some ui's which intersect, right, now suppose all of them are contained inside ui then you do not have to take them at all you can drop them out, so if everything contain everything intersecting this one is contained inside u1 then again you will have a problem there will not be anything so u1 is the last thing, so you have finished, so u1 equal to whole of x, right, so you are done no problem, so that is why they are this set sk is non-empty, non-empty subset of n, so there is a minimal element, so take the first one that is the minimum, so sk is non-empty, okay, if sk is empty you are finished, wk has to be x there is nothing that, okay, so and x is connected that is what we have to use we have used that one, right, because otherwise it will disjoint union, okay, in that case we have achieved our goal there is nothing to do otherwise take the minimum nk that depends upon k the minimum take the minimum, so u nk you take put wk plus 1 equal to wk union u n, u nk, so in the first case it is u1, w2 will be u1 union u n1 then next one will be u n2 and so on, so u ns are selected from this collection several of them may be left out may be all of them will come but how they will come they will come only the way they will intersect the previous thing whatever has been constructed, okay, the collective thing, so this is the union it is not one single open cell set, collective thing it has to intersect one of them intersect the whole thing something, alright, but should not be contained inside, okay, because this wk plus 1 is larger than w, so we keep we keep increasing otherwise we get stagnated that is all, now let us see inductively we claim that each wk is homeomorphic to an open interval or it is s1, there are two cases at each stage as soon as it is s1 we know that we have come to an end, so what is the other case the other case is each time you get that it is an interval, there we have not yet completed the proof, each time it is an interval if you have stopped there it is okay it is an interval you have completed the proof, but it may not stop it may be infinitely it keeps going on, so in that case you have to write a small proof there that is all it is a means, okay, so let us see why this happens clearly this is the case where k equal to 1 because w1 is u1 there is nothing to prove, u1 is an open interval for k equal to 2 there are two cases to be considered as what are they by the previous lemma we have in the situation of lemma 1 12.39 or 2 of lemma 12.40, alright, accordingly we have the above two conclusions see any interval u1 is already an interval okay or in the wk is an interval interval means what homeomorphic interval another one is another interval and they intersect, so the connected number of connected component is at most 2 if it is 1 the union will be homeomorphic to again interval if it is 2 the union will be s1, so that is these are the two cases okay, so this way from u1 was 1 to 2 we pass suppose now we have come up to wk some k, wk plus 1 is homeomorphic to s1 the step stop the sequence stops otherwise it keeps going what is it keeps going each time you are concluding that all the wk's are homeomorphic to open interval okay and we have what an infinite sequence one bigger than the other sequentially right, so in that case why the entire union is a you know is homeomorphic to open interval this is what we have so these things are not happening inside r of course right that is the whole conclusion I mean finally you have to so right now they are abstract manifolds one dimension manifolds you want to show that the entire thing is homeomorphic to an open interval which is same thing as showing that homeomorphic to r alright, so that is the last part here it remains to consider the case when wk is infinite in which case each wk is a proper open subset of a homeomorphic to an open interval proper means what it is not the whole space this whole space sequence stops no, so the infinite case is the one we have that is all starting with the homeomorphism f1 psi 1, f1 was psi 1 right you could remember I am re-indexing psi 1 from minus 1 plus 1 which is w1 okay, apply proposition 12.30 with this a hat less than a less than a prime less than b less than b remember this is proposition being equal to minus 2 I am choosing them now minus 2 less than minus 1 less than minus half less than half less than 1 less than 2 respectively and take g from alpha beta to w here I do not know what it is okay, any homeomorphism we get a homeomorphism f2 from minus 2 to this must be taken as u2 the next one minus f2 minus 2 plus 2 w2 is the union such that f2 restricted to the small interval minus 1 minus half plus half is your f1 okay, f1 is defined minus 1 plus 1 but on and the whole larger one we do not know only in the smaller open interval it is f1 and that is extended okay inductively having got a homeomorphism from minus k see this is f this is the starting point f1 I have instead of psi 1 I am using the notation f1 in the inductive hypothesis then I say f2 f2 is for minus 2 to 2 f3 will be minus 3 to 3 and so on okay, so fk is for minus k2 plus k to wk similar to the above step we get a homeomorphism k plus 1 from minus k minus 1 to k plus 1 to wk plus 1 such that restricted to a smaller interval of minus k plus k only minus k plus half comma k minus half so check away half half from both sides on that it is fk the old map okay, so now you define f from r to x by the rule ft could fk t whenever t belongs to this interval given any t inside r it must be in one of these intervals okay, once it is in this interval even if you take k larger fk t the value of fk t on this interval is the same thing fk plus 1 fk plus 2 they are all equal to fk t here therefore ft is very defined alright it is straight forward to check that f is a homeomorphism okay all that you have to do is continuous bijection and an open mapping directly okay to show the open mapping you can show that any small open interval like minus epsilon plus epsilon image is open you do not have to show the whole image is open any small interval every small interval is open image is open then the whole thing will be open mapping so I will leave that one to you but learn this method how to patch up homeomorphism if things are arbitrary homeomorphism not agreeing with the other one then you will have a problem so when you have inductive steps like this it is possible to patch it up to get a homeomorphism the entire thing why it is subjective here because union of all wk's is the whole of x okay so here are a few exercises which will help you to understand the next topic so take some time to think about them even if you do not solve them completely alright with those words I will just leave it to you to you know keep looking at them so that you have some time to think about this so here are two of one is on quotient spaces then this is our old friend about you know what is happening to homeomorphism from what kind of homeomorphism from one interval to another interval be there okay so this is our old topic which you have been discussing several times here so there is some new concept here called isotope it just like homotope human so I have defined it carefully we will go through that then we have this the transitive action so we have got psychic use so can you see that they are also isotopic to each other this is what one has to think about okay so next time we will study a little bit about surfaces so two more lectures on that alright thank you