 Hello, welcome to NPTEL NOC, Points at Apology Part 2. Today we will start studying Manifolds with Boundary. Last time we studied what are called as Manifolds, right? So this Manifold with Boundary, they are also going to Manifolds. Actually we are going to extend the definition of Manifolds to a larger class of topological spaces. So I will touch upon this usage of this terminology a little later. The first thing is that we are already familiar with the word boundary in a different context. Namely, we employed this whenever you have a topological space and a subset of that, the boundary of the subset A is defined as the closure of A minus the interior of A. This manclature for the boundary is totally dependent on the larger topological space X wherein the subset A is sitting inside. Now the term boundary will be used in a much subtle way in a technical sense and it is going to be an invariant of the homomorphism type and it does not depend upon where the manifold is sitting inside. So having said that much, let us come to brass tacks here. The important thing here is even to make this definition, we need the full force of Brauer's invariance of domain, namely remember this one, not the weaker form but the stronger form, namely if you have two non-empty open subsets of Rn which are abstractly as topological spaces, namely take the subspace topology from Rn. Suppose they are homomorphic to each other, if one of them is open then the other one is also open. This is a completely non-trivial statement, a profound statement and this is what we will need. We have used only a weaker version here, namely Rn and Rm cannot be homomorphic if m is not equal to n, that is an easy consequence of this one. So now we will need the full force of this. I will come to that again why we need it. So first of all, our model for manifold itself will change now. Earlier the model was Rn or equivalently the open disk inside Rn, the unit disk inside Rn because both those two are homomorphic to each other. Now we are taking the half space hn, namely the subspace of Rn such that the last coordinate of the points is bigger than or equal to 0. You can take any other coordinate because this is convenient one, only the last coordinate is bigger than or equal to 0. So this is like the half ray from 0 to infinity, half line, 0 to infinity open. When n equal to 1 this will be precisely 0 to infinity, 0 closed infinity open. So we are allowing that boundary point 0, the boundary in the older sense namely this point 0 will be a boundary point of the closed 0 to infinity inside R. That is the starting point, that is our model now. So in hn for example all points wherein the last coordinate is 0 that is subspace of that, that is going to be the boundary of this hn inside Rn. But once we have taken this one you can just forget about Rn and use this model hn to define our manifolds. So let us see how. Let us start with a topological space x, we will call it a manifold width boundary. So I am not defining manifold, I am not defining boundary, I am not defining in the word width here but I am defining manifold width boundary, this entire phrase as one single technical phrase. So a manifold width boundary if it is second countable house door and the local Euclidean part is modified here now. For each point x inside x we have a neighborhood ux and a homomorphism phi from ux to hn. Now you can take this as the whole of hn also no problem on to an open subset of hn that is more convenient that is all. We know that just like ux to Rn I could have taken in the case of ordinary manifolds right or ux to open subset of Rn or ux to just tell the open disc bn all that those three things are equivalent. Here also in the homomorphism could be ux to the whole of Rn or some open subset this open subset allows us little more freedom. Terms such as charts, coordinate function, coordinate neighborhood, atlas etc which we have defined in the definition 12.1 of manifolds remember that in the context of manifolds they all make sense exactly similarly except that now the coordinate functions phi of ux it is not necessarily an open subset of Rn but it is actually an open subset of hn. This is something funny here in fact every open subset of Rn has a copy an open subset of inside hn but not the vice versa. So Rn itself is homomorphic to the strict open upper half space here namely take all points such as last coordinate is strictly bigger than 0 right. So if you have an open subset of Rn that will be homomorphic to some open subset of hn strictly contained inside the interior of hn right. So this half space has more open sets than the full space anyway. So we can have this extra open subset here of hn which may meet the boundary which may meet the plane Rn-1 cross 0 that is the extra thing here that for this is clearly a generalization of the old definition manifolds okay. Let us denote by interior of x the set of all points x having a neighborhood ux homeomorphic to an open subset of interior of hn itself. You see there are two parts one may be just hn but this is a stricter condition interior of h homeomorphic interior of hn okay namely all xn the nth coordinate bigger than 0. So take such points then they will form automatically a manifold I mean a topological manifold in the old sense right. So why the interior is non-empty as soon as x is non-empty that is also clear because as soon as x is non-empty there is some open subset which is homeomorphic to a non-empty open subset of hn okay. If you delete the boundary part here that will be an open subset in the interior of hn which is obviously non-empty. So everything you have to see inside hn then you get this one okay the complement of interior of x in x is denoted by boundary of x. So this is a notation now I will read it as boundary of x and is called the boundary of x. So we had used this notation earlier for boundary of a where a is a subset of x if x is the topological space in which boundary of x is taken then this is empty. So this has no other meaning there right because the closure is whole of x interior whole of x. So boundary of a boundary of the whole space inside itself is empty all right. So here that is not the case x is the space it is not contained in anything okay this is a topological space on its own. Now boundary of x consists of those points which are not in the interior of x. So interior of x also has a different meaning here okay. So this is in the general topological case interior of x inside x would be whole of x that is not the case here. Here also it can happen that but then x will be a pure manifold not manifold with boundary okay. So I am talking about this one it may happen that boundary is empty just like in the case of manifolds okay which means that x is precisely manifold this is it still known. The points of boundary of x are characterized by the following property. So this is where you have to be you know you have to use that the browse invariance of domain. Namely there is a neighborhood ux of x in x such that we have homeomorphism view from ux question on to an open subset vx of question such that the nth coordinate of this point nth coordinate of vx this itself vanishes. So the Rn minus 1 cross 0 should be hit by this v okay take such a point such that vx is inside the Rn minus 1 cross 0 okay. So those are the boundary points. Now what do you mean by characterized let me tell you that okay so this is where I have to use the browse invariance of domain. So let us see start with such a point now suppose there is another neighborhood ux prime and another homeomorphism side from u prime x to hn similar to that one on to an open subset vx prime of hn such that the nth coordinate vx is now positive either it is positive or it is 0 right it is greater than equal to 0 always. So now it can this happen the question is the moment one this thing happens this other thing cannot happen that is the conclusion that is why this is a characterization. So let us see why this this cannot happen once you have another side like this you can restrict you know this psi to ux prime to a smaller neighborhood we may assume that psi nth coordinate of this psi x psi x of whatever this one is positive for all y inside the smaller subset by continuity of psi. If once it is positive in a smaller neighborhood it will be positive for all the points in that neighborhood that is all I am using and this ux prime is contained inside ux. This will imply that vx prime is an open subset of hn interior of hn now the smaller set and hence any subset of open subset of interior of hn is open inside rn because interior of hn is open inside rn. Now consider f equal to p component psi inverse from this open subset they are all inside our vx prime is inside where it is inside x right inside x. So psi inverse of p come back to f to this one so now what you get is this is inside hn which is a homeomorphism but by by Brauer's invariance of domain it follows that f of vx prime remember vx prime and f of this one these are both subsets of now rn this part is in hn this part is quickly inside interior of hn okay. So f of vx prime must be also open subset right and is contained inside hn but this contains a point vx which has its nth coordinate 0 so that is absurd because an open subset of rn contained inside hn cannot have a boundary point there in rn minus one the moment rn minus one hit if you take any open disk around that point it will flow out there will be nth coordinate will be negative also for some points. So it will go out of hn right so that is absurd okay so that is not possible. So the conclusion is that the definition of boundary of x as well as interior of x is now well defined thanks to Brauer's invariance of domain it follows that u hat which is just a notation take phi inverse of rn minus one cross 0 okay so this may be empty but right now I am assuming that the singleton x is already there so u hat will be a neighborhood of x this is now you know homeomorphic to some open subset here because phi inverse rn minus one cross 0 intersection with whatever neighborhood we have that will be an open subset of rn minus one cross 0. So I am taking phi inverse of that it is a neighborhood of x inside boundary of x because all these points in u hat are coming here in the rn minus one cross 0 so they are qualified to be inside boundary of x okay. So what I am talking starting with a phi like this okay ux and phi where x has gone to a point inside rn minus one cross 0 the point phi n of x is 0 so with that you take phi u hat equal to phi inverse that will be a neighborhood of x inside boundary of x and homeomorphic to by the same phi will give you homeomorphic to open subset of rn which just means that the boundary of x itself is a manifold pure manifold in the older definition of dimension one less namely n minus one okay this is if it is non-empty that is all otherwise of course empty is allowed okay and this boundary of x itself will not have any boundary because now it is pure manifold this is this is not half space now here it is rn minus one okay and easy consequence of this observation is that boundary of a manifold with boundary is a topological invariant what is the meaning of that if you have a homeomorphism f from x to y x is a manifold with boundary then y will be a manifold with boundary and f of the boundary of x will be equal to boundary of y you think about it this is not difficult at all okay so boundary always is mapped on to boundary of y by the homeomorphism so restricted boundary of x it will be homeomorphism okay now I come to again this nomenclature about manifolds somewhat apologetically strictly speaking our first definition of manifold should have been manifold without boundary right and the second one here should have been named manifold with or without boundary the only problem is right from beginning before defining the manifold I have to define what is the with boundary and without boundary so one does not like that one the second point is there is a standard convention of using the smallest words to represent the most commonly used concepts so we will be studying manifolds and quite often without boundary so that should be just called as smallest so we have called that one as manifolds okay so that is all explanation for why we are making this one in any case if there is an initial confusion about this what is a manifold what is a manifold with boundary what is a manifold without boundary having spent sufficiently enough time now explanation I hope this confusion if at all has disappeared now okay let us have some examples any closed dick in our n is a manifold with boundary so we have to we have to prove that one since any two of them are homomorphic to each other it suffices to show that the standard unit disc dn inside rn is an n manifold with boundary in fact boundary itself will be the n-1 dimension the sphere all x at norm x equal to 1 okay we will prove that let us look at first the case n equal to 1 then this closed disc is nothing but minus 1 to plus 1 the interval right you can write it as minus 1 to 1 open union minus 1 open comma 1 okay I have written it as two open subsets of this closure interval each of them is clearly homomorphic to 0 to infinity 0 closed infinity open and that is h1 so I have given a atlas there are two charts here okay so I have given an atlas over in the general case what I will do instead of you know subtracting this one and adding this one and so on there are two of them I will do this one dn minus the top point the north pole whatever 0 0 0 1 and dn minus the bottom part 0 0 0 minus 1 so this is similar to that but in n dimensional version okay and we claim that these two open subsets of dn are homomorphic to the entire of hn just like this one okay carbon copy of that one n dimensional version that is all so let us denote shorter notation n equal to 0 0 1 and this point will become minus n let us consider one of them u equal to dn minus this capital N okay this one the other one is homomorphic to this one by just taking the nth coordinate x going to all other coordinates same nth coordinate whether it is a reflection all right so it is enough to show that this is homomorphic to hn all right so we have ready made map here actually only thing is we have to see how to use it all right so first what I do I will identify this disk dn with a subspace of rn plus 1 indeed a subspace of sn itself see sn is sitting rn plus 1 this is inside rn so I need one more coordinate I will put it in the first one here or the 0th coordinate the rest of the coordinates are exactly as x1 x2 xn okay the 0th coordinate is 1 minus summation xi square and take the square root why I have done this if you take now squares of v all these that will be equal to 1 so I will be inside the unit sphere here in sn plus in rn plus 1 okay that is all clearly this is a continuous map is actually the graph of this function okay this function x1 x2 xn go into that to take this f is just like x comma this one so it is like a graph okay so you can rewrite it as y1 y2 yn plus 1 put fn okay equal to 0001 okay that is n prime that this n prime is a notation f of capital N is nothing but 0001 I do not I am not defining this one this n prime is just another notation it follows that f defines a homeomorphism of dn minus n the whole thing is defined the whole if you throw away that it will be a subset of b minus n prime where b set of all y is in sn the first coordinate y1 this part is greater than equal to 0 now this is a positive square so this is not the full sn it is only half the sphere right given by the first coordinate greater than equal to 0 so this is the preparation I have made now now what I do I take the stereographic projection stereography projection is defined from sn minus this n prime the north pole to entire of rn so that is a homeomorphism we have studied this one carefully restricted to the half disk it will go into the half disk h here half space h here namely y1 y2 yn plus 1 okay in the rn plus 1 the y1 coordinate is always greater than equal to 0 but the last coordinate yn plus 1 is 0 because stereographic projection so this is clearly homeomorphism to hn the only thing instead of last coordinate yn greater than equal to 0 I put y1 greater than equal to 0 so you have to interchange those two coordinates so look at these methods I mean these are important methods in just inside you know handling various subsets of rn that's all so more generally any convex polyhedron in rn being homeomorphic to dn is an n-manifold with boundary details are left to you as an exercise why I don't want to get into what is the definition of convex polyhedron and so on if you don't know you will have to learn it somewhere else that's all that is not central to the theme of this course note that the manifold boundary of dn and its topological boundary as a subset of r in coincide namely they are the unit sphere right in both the case indeed this explains why the term boundary is used in general topology you know the classically these manifolds were defined and studied even without definition perhaps even before the point set topology was conceived forget about the boundary of a subset and so on so boundary of a subset that that concept is a copy from examples like this okay unfortunately we are learning it the other way around for the same reason as for manifolds without boundary manifolds with boundary are also matrizable and hence para compact why because we have assumed second countable host dorm and whether they are inside you know homeomorphic to open subsets of rn or hm they will be locally compact okay so they are t3 second counter t3 space they are matrizable we should often use the word manifold to mean manifold without boundary the old definition okay so that is why this short term manifold often the results that we state for them are valid for manifolds without boundary as well though we cannot take them for granted there are some results which are not true at all and if they are true you have to prove them separately sometimes it requires quite a lot of effort as compared to manifolds without boundary okay whenever things are not true at all we will try to mention them separately okay so here is an easy consequence of existence of partition of unity because they are we have shown that they are host dorm and para compact okay we shall now obtain the so-called collar neighborhood theorem which is a very useful result on its own so that is an important result about manifolds with boundary let me make a definition first let x be a manifold boundary of a non-empty now okay otherwise the rest of the definition is actually true it does not hold at all so start with the manifold with boundary non-empty by a collar neighborhood or simply you can say just a collar of boundary of x inside x we mean an open set you okay open subset of x okay with a homeomorphism p from u to the boundary of x product with zero infinity this is a topological product boundary of subset subspace of x okay you must have a homeomorphism with one extra property namely for all x on the boundary phi of x should go to x comma 0 okay there are bound u is a neighborhood of boundary of x so boundary of x is sitting inside here right so for those points phi of that must be just x comma 0 so such a open subsets open neighborhood with a homeomorphism will be called a collar neighborhood now there is nothing special about the choice of this this half open interval 0 to infinity you could have taken 0 to epsilon because we know 0 epsilon is also homeomorphism infinity that is all indeed once u is a collar neighborhood as above you look at these subspaces namely phi inverse of boundary of x cross 0 to r take the inverse measure that will be another open subset which will contain boundary of x and it will be again homeomorphism boundary of x cross 0 r obviously which you can you know you can compose with another homeomorphism make it 0 infinity to satisfy this definition for h n the entire h n itself is a collar for boundary of h n obviously h n is a manifold with boundary right so what happens in the case of h n its boundary you know has a neighborhood which is the whole of h n that h n is homeomorph to r n minus 1 cross 0 mainly boundary of h n cross 0 cross r n minus 1 okay so this is the neighborhood the whole h n is a neighborhood similarly if you take the d n minus a concentric ball ball is center r center 0 radius r r smaller than 1 of course okay take a ball of radius r closure ball subtract it that is a collar neighborhood for boundary of d n that is namely the sphere s n minus 1 okay for an interval a to b okay closure interval a to b a to t union s to b is a collar neighborhood of boundary of a b what is boundary of a b where a b is an interval close interval little a comma little b right just the two elements okay that has a neighborhood like this a to t comma s to b where t and s are like this namely a less than t less than s less than b so that is the example now here is the final theorem so pay attention that x be a manifold with non-empty boundary and w contained in that x be a proper open subset in x such that boundary of w is contained inside w I do not want this w to be the whole of x okay then there exists a collar neighborhood u of boundary of x such that this u is contained in u bar contained inside w and the complement x minus u is homeomorphic to x again so throw away the collar neighborhood whatever left out is again homeomorphic to the manifold itself okay the collar is always like boundary cross an interval right so you you remove the open half open interval that will be again homeomorphic to x that is the that is the meaning of this one here and such neighborhoods can be chosen as small as to please inside an open set this w itself is a neighborhood okay then you can take u inside that just a u bar inside that all right so the proof is likely longer but not difficult if you understand what is going on the two examples three examples etc whatever I have given they are the guiding principles here let us look at what happens in that place of zero to say one okay open so zero is the interval zero is the boundary it has a neighborhood no matter how small you take zero to some epsilon which is homeomorphic to zero to infinity if you remove zero to epsilon whatever is left is again a closed interval proper closed interval not a sing routine so it will be again homeomorphic to the whole of zero one right okay similar to that that is what is happening but we have to work a little harder here we need to use partition of unity also because we are not assuming compactness if you assume compactness you can probably write down a much shorter proof over to you after after you learn this proof that will be left as an exercise to okay so idea here is a very very beautiful idea here what we do is okay we attach an external caller to x I am going to explain what the meaning of this one namely you know you are taking a larger space you are creating a space larger space containing x okay so the rest of the extra space is boundary of x cross in interval minus one plus one minus one zero so that is what I am doing so I am taking y as a quotient of disjoint union of x and boundary of x cross minus one zero if you take disjoint union that is not not much fun what I do I identify boundary of x cross zero this copy with the boundary of x here okay by identify x comma x belong to here with x comma zero here okay the rest of them is floating outside okay observe that y is also an n manifold which is boundary homeomorphic to boundary of x cross minus one this time so this portion becomes the boundary now the point the all the points x comma zero they have become interior points right the idea is to define a homeomorphism f from x to y which is identity outside this w prime and then take u equal to f inverse of boundary of x cross minus one to zero clearly then u bar will be contained inside w and x from here clear is a homeomorphism okay so what is this w prime I will explain this is a small subspace of w chosen right in the beginning like this you can choose an open subset w prime so that boundary of x is contained inside w prime contained inside the closure contained inside w so here I am just using regularity and this one is a closed subset closed subset contained in open subset in between you can introduce this another open subset right so now we begin with a countable partition of unity theta i how did you get countability because the whole space is second countable why where I am working on working on boundary of x which is the manifold so that for each i the support of theta i is compact so this is one of the remark which we made because because of the local compactness okay you can assume that the support of theta i is compact okay and is contained in a coordinate open subset ui the coordinate subset can be assumed to be such that their closures are relatively compact closures are compact right and if this is this support of theta i is contained inside this one it will be automatically being a closed subset will be compact also okay so so how do I do that start with a coordinate cover namely at last go to a locally finite cover because of para compactness after that go to a countable finite countable cover because second countability use the partition of unity which exists because of para compactness automatically you will get these ones are compact okay so these are coordinate neighborhoods i have phi i from ui cross 0 1 to vi these are coordinate neighborhoods for for points inside boundary of x okay now I am going inside the manifold x itself by using that these are all neighborhoods of this whole thing is a neighborhood of uh this w prime is a neighborhood of boundary of x and ui are compact ui bar are compact so I can extend this neighborhood to 0 comma 1 right phi i from ui cross 0 comes some see this is the first thing here because of the compactness you can extend it to a small neighborhood 0 to 1 to vi is vales theorem okay homeomorphism phi i to some rn is there now rn minus 1 is there now you can take an open subset in rn okay open subset vi these vi's are inside w prime of x everything is working inside of such that the phi i cross 0 0th coordinate x comma 0 is x inside vi prime for all x so these ui phi i's I am actually writing as a parameterization they are you know parameterizations for points inside the boundary but the parameterization is occurring for neighborhoods inside x itself okay so you choose such homeomorphism there is no problem now comes the construction so this is all preparation how to choose these things etc now put eta not equal to 0 I am going to do some uh what is it inductive construction so starting point no you do not do anything in or not equal to 0 eta k is the sum of theta i is up to k i rank to 1 to k remember these are just non-negative real value functions so it can just somewhat sum it up finite sum I am taken and for k greater than 1 that is a definition next put this z k is a space x comma t such that x is inside u k the u and u k's are open subsets of boundary effects x is inside u k the t varies between minus of eta k minus 1 to 1 after all all these eta k are taking values between 0 and 1 okay so I take minus of this less than equal to t less than equal to 1 up to there I so this is the interval strictly contained inside minus 1 to plus 1 this part is plus 1 on the positive side the negative side how much you can go at the most minus 1 okay so these are subsets of now u k cross minus 1 to plus 1 okay but much smaller so x must be only here now z k prime is all x t such that the same thing but use eta k here eta instead of eta k plus 1 so since eta k is slightly you know bigger because one more function is as been added here this will be slightly larger interval than this one only on u k so these are defined only on u k all right now take alpha k from z k to z k prime be the homomorphism which linearly stretches the segment singleton x cross this interval to this interval so linearly stretch it means what 1 goes to 1 eta k minus 1 x goes to minus eta k of x you know we have linear homeomorphism mapping any interval a b a to a prime b prime right so take those homeomorphisms for each x you do that automatically because you can write down the formula in terms of eta k minus 1 and eta k here so it will be automatically a homomorphism on the whole space okay so put y k see again I am defining these things inductively right put y k put x union this x comma t where eta k is taken up to 0 only okay so portion of z k is taken z k prime z k prime is taken only up to t less than equal to 0 okay that is your y k namely these are extra spaces you see that is what okay but take it with union text okay so they are all subspaces of y now that beta k from z k prime to y k be the embedding by given by beta k of xt is phi k of xt where t is positive remember this phi is ui cross 0 1 so the second coordinate is here non-negative so that is what you have to do so non-negative you take phi k when it is negative from 0 minus 1 to 0 you take it as xt itself okay identity map itself so that makes sense because they are also say we are working inside y now so it will go inside y k because this beta k is defined only up to minus eta k for t negative not less than equal to 0 you define this when t equal to 0 the two definitions coincide okay so we have got these embeddings of z k prime inside y k we define g k on the image of beta k intersection with y k minus 1 to y k by the formula g k of beta k of xt put beta k complete alpha k this alpha k is stretching k the smaller space to larger space okay so first stretch it and then take beta k every member here is this is the image so it is beta k of xt okay so take that xt stretch it and then again take beta k so it is like first you take beta k inverse instead of writing inverse I have just written g k of every element here is after all unique element like this beta k of xt if you write it as some y then this will be this xt will be beta k inverse of that so that is more interesting than writing so this write like this so g k is defined like that okay if xt is not in the support of theta k then eta k will be 0 right so if eta k is 0 what is this summation it will be eta k minus 1 so that point will be inside that k not in that k prime it will be here though I put this one okay for points wherein x is not in the support of this theta k eta k is eta k minus 1 therefore alpha k of xt will be just xt therefore g k of beta k of xt will be beta k of xt throughout outside for negative things it is just xt all both of them xt but now because of this one g k of beta k of xt will be just beta k of xt because this is identity so the stretching are extending one from the next one that is important okay now we define f k from y k minus 1 to y k see on subspace is this beta k so we have defined okay now for the whole full y k to be g k on image of beta k and identity outside image of beta k okay so this is very important here now because these things are just stretching things right so on this on whenever it is minus equal to plus 1 it is identity already there okay so on the press when it is identity so also when theta k is 0 for for points which are outside the support these two are the same also right so therefore what happens is that is an important the point is that once you are outside this image of beta k okay on the on the boundary of beta k itself which is 0 therefore on the outside of this one so you can just define it as identity so extend its identity then it will follow that f k is continuous indeed since each alpha k is a homeomorphism it easily follows that f k is also a homeomorphism finally there are infinitely many of them I am putting f equal to the compositions of these f i's in that correct order okay the reverse order so what is the meaning of this how do you explain that look at any point okay f 1 is defined f 2 is defined after the return stage it will become identity why because this f k once the point is outside theta k outside the support of theta k this f k will be just identity so you are taking only the composition up to here stretching stretching stretching along the these intervals vertical intervals x will be fixed x comma t goes to x comma t prime that is the kind of homeomorphism we have here okay so this makes sense because at each point only finitely many functions are are not identity after that it is identity so this is like infinite product when you take most of them equal to one infinite product of elements I mean real numbers that similar to that okay so note that compliment on the compliment of v which is p i union p i of u i cross 0 1 take the union and therefore outside w prime f is identity on v itself f makes sense since any point x belongs to boundary of x there are only finitely many i for which x belongs to u i and this is like local coordinate locally finiteness and so on okay and f k of x t is equal to x t if x is not in u k x is not in the support of theta k then already this f k is identity but these things may be they are all inside once it is beyond u k that is all theta k's are so theta on theta theta t are all 0 that is the whole idea indeed in a neighborhood of x all f k are identity except those k for which theta k is not 0 right so that is the whole idea so for this reason the map f is also a proper map proper map why it is outside you take any point here then the neighborhood it is the compact neighborhood and outside that it will be identity so this goes on like that okay so only at each you have to take some compact neighborhood of boundary of x okay then the rest of them you know these points where you can use this formula it will be identity so the changes are occurring only inside a compact neighborhood for each compact subset you will get a compact subset of for each compact subset of boundary of k you will get a compact neighborhood something 0 1 minus 1 to plus 1 inside that everything is working here outside that it will be identity okay so for this reason f is a proper mapping summation theta k is 1 it follows that f will be surjective summation theta k is 1 if you have anything less than 1 it will be hit in some k then y k to that y k plus 1 will contain that okay so it will be a surjective mapping okay stretching after all 1 will be also hit there from some point given any x some summation finitely many some this summation though it is infinite some given any x there will be finitely many coordinates theta i for which summation theta i is equal to 1 you look at maximum of this theta i at that point the point x comma 1 will be hit so that is why it is surjective since each f k is an embedding f is surjective also therefore f is a memory a proper injective okay surjective map that is why it is and it is an embedding for each embedding means what open map or closed map you can use there is an open mapping okay to see open mapping you have to restrict take an open open take a point take in our neighborhood compact neighborhood okay inside that only finitely many compositions are there they are all open the composite is open so all these study so this means that f is a memory okay yeah so that completes the proof that we have caller neighborhoods for the boundary of x the entire boundary of x whenever boundary of x is non-empty of course okay here is a easy exercise show that a manifold with boundary is connected if and only if its interior is connected this is a slightly tricky thing you have to think about this interior is connected of course interior is connected but what about the points in the boundary why the whole thing is connected that is what you have to show okay but that will make you think what exactly should I use here all right so next time we will do some more properties of manifolds mostly we will now deal with only manifolds that is manifolds without boundary thank you