 Hello, welcome to module 60 of NPTEL NOC, an introductory course on point set up all as we park tour. So, today we shall continue our classification of one dimension 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 a 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 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 you 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 comma b the first thing is psi 1 inverse of a is a 1 comma 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, see 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 these are all hypothesis, the second third part is psi 1 inverse of b exactly the opposite here d 1 b 1 psi 2 inverse of b 8 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 a part of the interval a 1 b 1 go into the u 1 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 a and b correspond to a and b, so first portion will be a 1 d 1 okay to 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 are 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 contain in the other okay, so that is not there is no need to separately stated, so conclusion is that the union will be now homeomorphism 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, so 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 homeomorphism 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 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 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 union 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 so there is a unique s 2 here which comes to that one these are the points t 1 psi 1 of t 1 which is point psi 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 such 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 2 that is all 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 from 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 this part is coming here this part 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 I 0 or cos theta cos 0 plus sin the comma sin 0 that is the point here and 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 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 it will be s 1 similarly when t equal to pi here this will be 0 so this will 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 take psi of this psi of that except when these two points are the same namely there are two 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 why similarly when it is 2 pi also at the end they are also same 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 on the interval psi on the common point they agree so it is a continuous function in each each in each interval 0 to pi pi 2 pi it is given by homomorphism so it is injective this injective because they are mapped into different components different parts of the things here you see and here 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 covered by psi 2 also on the intersection their order preserving therefore the left out parts are you know they are never mapped by psi 2 onto this part so that is the whole idea so this is similar to what we have verified in the first case okay so entire thing is injective so 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 a continuous bijection from s1 to u1 u2 what is the projection map here from 0 to pi to s1 capital s1 t equal to e raised to pi e raised to t just e raised to pi i e raised to i t because 2 pi I have taken so t goes to e raised to i t or just cos t plus i t so that is the map 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 surjective map so it is it is it is not surjective we do not know that one it is surjective 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 in 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 okay so two important cases which will produce the two different final final conclusion have been covered so now the claim is that there is no other case okay these are the only two cases that is the whole idea alright so of course in the second case we have yet to prove we have to complete the proof also there is also that part having taken care of these two favorable situations we now claim that we are always in one of these two cases what is the meaning of that let me elaborate start with a half-door space psi 1 and psi 2 from minus 1 to plus 1 this doesn't matter you could have taken any open interval minus 1 to plus 1 to x the homeomorphisms on to some open subsets u1 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 was I have assumed it is non-empty will be an open interval of the form a comma b for some minus 1 less than a less than b less than 1 in other words it is this open every every component is an open interval right because they are all subsets of no 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 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 okay see once you have proved this one this in particular this 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 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 this possibility 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 homeomorphic 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 homeomorphic to s1 this is the second one so this is precise in the meaning of saying that there is no other possibilities okay 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 happened 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 or 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 is happening psi 1 something has something bad has happened here I do not know I do not care something has happened here where a cd c equal to d c equal to minus 1 d equal to plus 1 any all places are allowed okay 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 but 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 the psi 1 there is something bad has happened replacing psi 2 by psi 2 composite alpha if necessary I could have taken any other 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 might 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 the 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 all right 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 us problem namely in the former case suppose c is the in 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 path right so what happens is psi 2 of c and psi 2 are distinct points of x which cannot be separated by open set 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 not 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 some in between minus 1 to 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 a and c are not a or c are distinct point okay sorry not this portion this the other portion because they are they have to continue for afterwards right this is not the end points of the interval this portion will be common so for every open interval open neighborhood of psi 1 of a inside the u1 there will be some portion every common with psi 2 of this is the intersection part on the left hand side similarly here what happens there will be on the on the left hand side there will be intersection part these portions are distinct fine but as soon as you hit it a psi 1 of a and psi 1 of b okay so they are the image they are there inside the inside of our x right but they are distinct points they are in a 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 false darkness of the interval so some psi c prime to c on this part okay will be coming there that's all so I don't know how many I have drawn drawn arrest of them some d prime to d that will be that is the whole idea so they are in the they are in this part of the rest of the part here they are they will be common so here all of them will be common so you can't separate them by distinct open success 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 a second case covered that's 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 from whatever you 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 once I observed the 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 this complete symmetry 1 and 2 you can change that's all so here a1 minus a1 to minus 1 to a1 and 1 b1 to 1 and then other cases there are also minus 1 to say a2 and b2 to a1 so these are four cases are possible you know you can take four cases to be considered but they are all symmetrical so for definiteness you just cover one of them argument will same in other cases okay so consider the case when these intervals of the of the form minus 1 to a1 and minus 1 to a2 see 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 where I assume only one component so one component is this what is happening if you look at the gain psi 2 inverse psi 1 minus 1 a1 to minus 1 a2 has to be decreasing if it is increasing what happens similarly the a1 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 is if you this way the other functions should be like that okay 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 contradictory the half-doubtsness 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 could 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 no need to worry about what is happening here okay so we are done so that is case case 2 the 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 endpoints a minus 1 to a i disjoint union v i to 1 i equal to 1 and 2 okay this psi i this is i is i here these two there are two cases to be considered again here namely psi 2 inverse psi 1 you know this component a minus 1 a1 may be mapped to minus 1 a2 there or minus what b1 b2 to a2 which way they are mapped that is what I would 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 that is fixed side okay you have freedom only to choose where one of the component goes the other the 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 you know one of the cases so case a let us take so let us look at the case a for the same reason as in two we conclude that this minus 1 a1 to minus 1 a2 has to be decreasing okay so that that thing is coming again and again so let us consider change the change psi 1 by you know by a reflection then we now claim that now you have the whole thing changed right so as soon as we 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 can't change the other one because the whole thing is one 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 that is the whole idea in this 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 all right 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 homeomorphism psi 1 prime psi 2 fit the hypothesis of lemma 12.40 completely clearly psi 1 psi 2 inverse of psi 1 prime minus 1 to b1 now see the all the things have changed because t has been replaced by minus t here 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 comma 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 incredibly finally it follows that psi 2 inverse composite psi 1 of minus 1 1 to minus 1 a2 is also incredibly the other part is also incredibly 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 a situation of second lemma there therefore the conclusion is that in this case the entire manifold has to be s1 okay we do not we do not need that right now we just want we have 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 homomorphic to say let us say pi 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 okay wk of open subsets of x such that each wk is connected union of wk is the whole space that is all first we are 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 you have got a countable cover so you have index 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 which are 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 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 we have finished so u1 equal to whole of x right so you are done no problem so that is why there this set sk is non-empty non-empty subset of when 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 of 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 in that case we have achieved our goal there is nothing to go otherwise take the minimum nk that depends upon k the minimum take the minimum so unk you take put wk plus 1 equal to wk union unk so in the first case is u1 w2 will be u1 union un1 then next one will be un2 and so on so un's are selected from this collection several of them may be left out maybe 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 subset collected thing it has to intersect one of them intersect the whole whole thing something all right but should not be contained in there was 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 we 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 right accordingly we have the above two conclusions see any any interval u1 is already an interval okay or inductively 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 the 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 kf wk plus 1 is homeomorphic to s1 the the step stop the sequence stops otherwise it keeps going well what is it keeps going each time you are concluding that all the wks 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 what we have to show 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 we are abstract manifolds one dimensional 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 place this whole space sequence stops no so the infinite case is the one middle that is all starting with the homeomorphism f1 psi 1 f1 was psi 1 right you could have I am reindexing 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 this 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 being any homeomorphism we get a homeomorphism f2 from minus 2 to this must be 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 smallest 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 the 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 equal to 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 fk plus 1 fk plus 2 they are all equal to fk t here therefore ft is very defined all right 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 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 all right 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 homotopy you meant so I have defined it carefully we go through that then we have this the transitive action so we have got psi pq's 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 all right thank you