 So, today we shall continue the study of manifolds with boundary, the basic thing here is instead of the model Rn we shall use the half space model hn, this h upper n denotes all points of Rn where in the nth coordinate is greater than equal to 0, so if n is 1 this is just the closed interval the whole ray from 0 closed to infinity open, so it is like that the half space at the topological space x is called manifold with boundary, if it is second countable harsh darkness those things are there as usual, but now the at last the charts are taking place at each point x we have a neighborhood ux and a homomorphism homomorphism of an open subset ux to hn namely inside an open subset of hn, so that is the only difference, so if you understand what kind of open subset of hn come then as compared to open subsets of Rn then you will understand the meaning of this definition as compared to our old definition of a manifold, namely you have to, being a subspace of Rn a subset of hn is open if it only if it is the intersection of an open subset in Rn with hn, so therefore suppose I take a open disk for example in Rn and when I intersect with hn it may be like a portion of that open disk, cut off by the plane hyper plane namely xn equal to 0 everything xn less than 0 xn less than 0 will be thrown out, so on the other hand the intersection with Rn minus 1 itself will be there ok, so those points should be there, those points if they are present then that such a set will not be an open subset of Rn itself hn itself is not an open subset of Rn ok, so that is the point that you have to pay attention to, so denote by interior of x for such a manifold such a definition I mean whatever you have defined the set of all those points x having a neighborhood ux, homeomorphic to an open subset of interior of hn wherein n is strictly positive since this interior of hn is an open subset of Rn, this will be an open subset of ux will be a homeomorphic to an open subset of Rn itself, such points will constitute our manifold with the old definition ok, so I am going to denote interior of x to be all those points having neighborhood ux, homeomorphic to an open subset of interior of hn which will be same as open subset of Rn, I do not have to change, I do not have to go to Rn within interior of hn it is fine work, clearly this forms an open subset of x ok, once neighborhood has that point every point inside that neighborhood will be also belong to that therefore, it is an open subset of x and it is a topological n manifold in the old sense now because the open subsets which are not in the interior they may cause problems, so those things are not taken here ok, so can you see why interior of x is non-empty if x is non-empty, why it should be that interior should be non-empty for the moment something is an open subset of hn intersects with interior of hn it has to be non-empty, therefore that part of the neighborhood will constitute to the interior of it goes inside the interior of x, so it is interior of x is non-empty, the complement of interior of x in x is denoted by boundary of x, so this you should not confuse with the definition of boundary of a subset of a topological space, so in topology the boundary of a set is defined by the old definition of this boundary, the boundary of for example a disk is a circle and so on no matter where they are contained in this is a manifold boundary and this was sacrosanct the definition topological definition of a boundary of a subset is adopted from the properties of the boundary of x rather than the other way around, so this is called the boundary of x, so we will have this notation, we will not use this notation for an arbitrary subset a of x and then boundary of x, we will not use that, we will use a bar minus a interior a dot or some such thing, so clearly it is closed subset of x because it is a complement of the interior which is an open subset, it may happen that boundary of x is empty, we cannot say that because all the charts may actually land up inside interior of h n, in which case the entire of x will be in the interior of x, so this interior of x in ordinary topological space, if x is a topological space interior of x is itself but here it is not the case, so this is some other definition interior boundary but it can happen that interior of x is the whole of x in which case x will be a manifold in the old sense, so in the new definition we have not thrown away the old definition, we have adopted that one, so this new definition is actually an extension of the old definition, we could have just called these are manifolds, the only possibility is that they can have, so what are called as boundary points, the points of boundary of x, now comes the crucial thing pay attention to this one are characterized by the following property, what is it, there is a neighborhood u x of x and a homomorphism of field from u x such that the nth coordinate of phi x, nth coordinate of phi x actually is 0, I can always arrange this, this is the characterization of the points on the boundary, you have a field such that the nth coordinate is not 0, then it will be automatically in the interior of h therefore it will belong to interior of x, remember our definition of interior of x and boundary of x are completely complementary, one cannot be the other, there is no intersection there, therefore so all such phi which are all such points x must have such a homomorphism and once it has a one single such neighborhood and a homomorphism like coordinate like this, it will have to be in the boundary of course it is again a simple consequence of topological invariance of the main, I repeat this one namely if under say phi 1 the nth coordinate is 0 and phi 2 the nth coordinate is not 0 for the same x, this cannot happen because when you take phi 1 into phi 1 inverse phi 2 then you will get a half or half open interval half open neighborhood namely open set inside h n which is not open in R n, you will go to an open subset of R n and that will contradict what invariance of domain, if you have two subsets of R n which are homomorphic, if one is open the other one must be open inside R n, so here it will happen open inside h n only so that is not possible okay, so once a point belongs to the boundary with this characterization with one chart namely the nth coordinate is 0 then for any other chart at that point the nth coordinate has to be 0 for that point okay, so let us go ahead it follows that if you take u hat which is equal to phi inverse of R n minus R n minus 1 cross 0 just take the restriction R n minus 1 cross 0 of the image and then take the inverse that is phi inverse of R n minus 1 cross 0 so that will be a subset of actually a closed subset of u or u hat u x this will be a neighborhood of x because x belongs to the nth coordinate function of x itself is 0 okay but all these points now must be inside because the nth coordinate is 0 here must be inside boundary effects okay, so this will give you a chart for x inside the boundary of x so what we have just proved is that the boundary of x itself is a manifold of dimension 1 less namely R n minus 1 now, hostersness and second countability automatically are satisfied therefore the boundary of x if it is non-empty is itself a topological n minus 1 dimension manifold okay and that manifold boundary effects will not have any boundary because now everything is inside open R n minus 1, R n minus 1 cross 0 is homomorphic to R n minus okay not h n not h n minus 1 okay so they will all be in the interior or this manifold so boundary of a boundary is empty always we shall most often use the word manifold to mean a manifold without boundary okay only for that reason we define the manifold with model inside R n instead of h n okay often the results are stated okay for manifolds so called manifold without boundary but they are valid for manifolds with boundary also however sometimes what we have to take a you know slightly different version and it proves to me also be slightly different but because of time constraint and to emphasize on the concepts involved we will prove many of these results only for manifolds without boundary which we keep calling only manifolds no without boundary manifolds meet automatically without boundary if there is a boundary we will specifically mention it because those cases are less okay so we shall now introduce a notion which comes handy in proving a number of results which are valid for manifolds for manifolds okay with boundary also once you prove some result for a manifold okay then automatically it will prove for manifolds without boundary okay so that you do not have to mention that one or maybe let us with a little more effort so there is one but not all of them will do but many of them will do so that is why we are proving this so that is the concept of double of a manifold start with any manifold in which boundary of x is non-empty that is important we may assume that x and boundary of x are connected also to get a picture what exactly is going on okay think of a arc for example half arc or a whatever inside a circle closed circle closed arc now take the same another copy of it and along the boundary points you identify them a point corresponding point on the other one because I am taking same copy right identify them what you get you would let a circle that is what I mean by a double of a manifold so we will should define this one more carefully here okay let ui vi be an atlas for x with phi i taking values in h n as usual let x plus minus denote any two copies of x okay that plus minus denotes the same x with you know one copy x plus and that is what for x plus we take the same atlas as for x for x minus we will the atlas will be denoted by phi i minus which is phi i but I follow the reflection namely the nth coordinate I will change it to minus the phi i r composite phi i where r from r into r n is given by x 1, x 2, x n x going to x 1, x 2, x n minus 1 minus x n okay all right now let eta denote boundary of x plus to boundary of x minus the identity function because it is the it is the the same copy after all okay that dx denote the quotient of x plus determinant in x minus where x belong to boundary of x will be identified with eta x belong to boundary of eta boundary of x minus so same copy same point will be identified only along the boundary okay corresponding points all right that is my quotient space on the quotient space we can define a chart if point belongs to the interior of x whether it is in x plus or x minus there is no problem you can take the same phi plus or phi minus so we get a chart inside r n now okay maybe it is plus part or minus part okay because after taking reflection the nth coordinate might have become we will become minus does not matter the problem is only at points wherein the point belongs to the boundary of x okay boundary of x plus and similarly it will be a copy of that one it it is the quotient of these two it is identification so the same xx okay representing that that has become a single point so what do we do there there also it is very easy what we do take the image of take a take a chart phi at that point phi i look at the image okay just to reflect it right take the urine of those two that will be homomorphic to the two of these phi plus and phi minus on x plus x plus union x minus modular the relation okay so that is that is what the chart would look like for this double of x therefore double x is a manifold without boundary no boundary at all now the original boundary for x has become an interior point now okay and it contains two copies of x which are closed subspaces of the x and boundary of x is their common boundary in dx okay so you have two closed subspaces they are manifolds with boundary okay their set theoretic boundary is precisely equal to boundary of x now you see the relation between the manifold boundary and set theory boundary so this double x dx is called the double of x okay this could be useful in several cases okay so as an example of our claim in the previous paragraph let us derive that any manifold with boundary is compact see we prove it for manifolds without boundary but now it has a boundary then you do not have to prove it again all the way afresh here is how you want to do this okay all that we do is take dx which contains x as a closed subspace dx is a manifold without boundary and our theorem appraise there that is paracompact and it is a closed subspace every closed subspace a paracompact space is a paracompact okay so you are done as an immediate consequence of this existence of partition affinity we shall now obtain a slightly better picture for this for this boundaries namely the caller neighborhood theorem is a very useful result on its own it takes me a manifold with boundary non-empty what is the meaning of a caller caller of boundary of x we mean an open substitute of x which is homeomorphic to what boundary of x cross zero infinity open zero close infinity open so that zero close infinity open this is the caller it is like a slight extension of a boundary of x itself such that this homeomorphism should be such that whenever it is in boundary of x it is px is x px is identity okay so that is the meaning of so this u must be an open subset of x itself okay now you can see that if you take a disc okay round disc the boundary you know is the circle what will be a caller neighborhood all points which are of the norm say bigger than some one minus epsilon okay so that will be a caller for s1 inside d2 or more generally sn minus 1 inside dn all right once you have some epsilon you can use the whole you can stretch it to the whole of infinity also so in the definition I could have put zero epsilon here okay but then you may ask what is epsilon and so on so instead of that I have zero infinity okay so once there is some epsilon neighborhood like this boundary of x cross zero epsilon here open is also homeomorphic to boundary of x cross zero infinity so there is no problem so in every manifold x boundary of x has a caller neighborhood okay a caller is a neighborhood which is a very special kind of neighborhood all right moreover for any caller neighborhood q of boundary of x if you look at x minus u throw away that caller whatever is left out is homeomorphic to x if you throw away the epsilon neighborhood of the circle from the disk provided that epsilon is what less than one okay if you want less than epsilon means something less than you have to you have to less than one throw away that right again you get a small disk a closed disk again which is homeomorphic to the whole disk so this is easy for the disk because you have taken a nice object but this is true for all manifold all topological okay so that is what we want to prove okay so theorem is that every manifold x boundary of x is a caller neighborhood then when you remove that it is homeomorphic to x in fact the second part will be part of the first part was proof itself so you will see that we do not have to waste so much of time for proving the second part again so let y be the space obtained by attaching an external caller to x so instead of working inside i am going to extend x beyond like we have did the double don't do the all the double but slightly you have your h n okay that line r n minus one cross zero you bring it slightly below try it at r n minus one cross x minus epsilon okay so what it will be it will be it will contain a neighborhood of our original boundary namely r n minus one cross zero and that neighborhood will be r n minus one cross zero cross zero to epsilon so this is the picture which happens in the model and we want to say that the same thing will happen for the manifold itself okay so we start with constructing this y the space obtained by attaching an external caller to x via y is a quotient of disjoint union of x in the double what we did another copy of x we did don't big the whole copy but take boundary of x cross minus one zero okay take boundary of x cross minus one zero and identify x this boundary of x with x cross zero in boundary of x cross minus one zero the minus one part is identified with the original boundary right so from minus one to zero you have this extra thing okay observe that y is also a manifold with its boundary homomorphic to now boundary of x cross minus one so so r n r n minus one cross zero is standing there but I have come up to minus one below minus one to zero I have taken this extra thing so now the new boundary will be boundary of x cross minus one so this part I am not going to prove that this y which is obtained as a union of old x at boundary of x cross minus one along identified along this one is a manifold that is proof is same as in the proof of bound double double of a manifold okay observe that y is also manifold with boundary homomorphism the idea is to define a homomorphism from x to y and then take u equal to f inverse of boundary of x for minus one zero so this y which is slightly larger manifold is again homomorphic to x in such a way that the boundary of x cross minus one this boundary will go to the boundary of x under a homomorphism by the way boundary has to go to boundary invariance of domain okay so it will go to boundary so if we have a homomorphism like this you will have this one then what happens to the minus one to zero power it would have a copy inside x now the image of minus one to zero boundary of minus one to zero it is f inverse of that that will be my my u that is the idea okay so how to get this one how to get a homomorphism like this okay begin with a countable partition of infinity theta i on boundary of x so that support of theta i is contained in a coordinate open set ui of boundary of x together with a homomorphism vi from ui cross zero one on to an open subset vi of x okay so instead of choosing arbitrary neighborhood of of a point on the boundary inside x okay there are coordinates I am choosing a product neighborhood we choose a neighborhood first vi and then vi cross zero open one one one one two one sorry zero two one open okay half open interval all right for each point neighborhood we can do that one in a round thing you can always have a square open subset it's like that ui cross zero one you can have okay this after all here rn rn is rn minus one cross r it is a product of poly there on to no one set vi inside x once you have this one for each ui okay and you have partition of infinity what I do I start defining take eta not to be zero and eta k to be island who want to k theta i theta one theta two theta three some ordering is there it is a countable partition of infinity okay so eta only theta one eta two is theta one plus theta two and so on okay i ring to one to k theta i all right let us take z k to be all x comma t x is in uk what is uk you can open subset of boundary effects okay they cover the whole of boundary effects minus of eta k minus one of x less than equal to t less than equal to one so t ranges from in this this interval okay so this is definition of this is subspace of boundary effects cross of zero one because entire sum total is what is sum total here it is always one at the most right so I am taking half partial say here so this will always be less than one so minus of that will be bigger than minus okay minus of something less than equal to one nine okay so z k prime as for all those x t belong you to the next one u x u of k uk eta of the next one minus eta k less than equal to less than equal to one so it is slightly bigger name it is because sum is larger here that alpha k from z k to z k prime be the homomorphism which linearly stretches the segment this segment only t part from minus eta k plus one to eta k okay you see they may be equal also then there is nothing true because the theta k at some point may be zero okay inside uk it may be zero no problem but if it is not because these are they will slide some more will be added up after all so you just stretch it this part remains same one the other end remains the same these are homomorphism keep stretching that okay so homomorphically any two intervals half this this is half closed okay these these are close intervals sorry so they are close intervals so they are homomorphic to each other okay so homomorph on to the segment minus eta k to one for each x you do that okay it depends upon x right okay so this is what you have to do that if we if the end points are are parameterized by x a continuous parameterization then the homomorphism it is parameterized so this is a non-trigger step okay but what you can do you take a linear map linear maps are there which goes to one to one and this one to this one that will give you a formula so that will be automatically a homomorphism okay continuously depends continuously on x put y k equal to now x union all those x comma t set minus of eta k of x is less than equal to zero t okay this is now kth stage construction is over that beta k b from z k prime to y k be the embedding given by beta k of x t is equal to phi k of x t t greater than or zero and beta k of x t equal to x t for t less than equal to zero okay so you have to see that these these things agree at t equal to zero namely it will be x itself there the point x I don't know okay so we define homomorphism f k from y k minus one to y k so inductively as follows f k of z will be just identity if it is inside it's outside phi k of u k cross zero one okay at one there is nothing changed that's why this is this is okay it will be x t okay where result itself is x t but x is not in u k if x is in u k you check it to be beta k times the alpha k of x t okay is alpha k of x t okay so you have to keep keep constructing this inductively like this finally you put f equal to to composite f one and then followed by f two the other way around okay f k depending upon where k is where x is x will be at a final stage u k one of the u k after that no matter what you take f k plus one f k plus two and so on it is identity it doesn't change okay if it is outside it is just that okay so all f k's if x is already in u k and not in u k plus one and so on okay then it will be identity so this is this infinite composition though it looks like infinite composition so first of all note that on the complement v which is equal to union of phi i phi i of u i cross zero one f is identity see for each boundary point we have taken these neighborhoods okay so everything is happening inside that smaller neighborhood even the pushing everything so what we have done is we constructed from minus one to plus one okay and then pushed the minus one to zero inside without bothering the plus one part plus one part is already there okay so all that you can do entire thing is just that one only you take some r n or any x and take x cross minus one to plus one okay the only thing is you don't have uniform one all the way because it's infinite set in a compact set this would have been much simpler okay then you could have have a uniform epsilon okay on v itself this is union of this one f makes sense since given any point x belongs to boundary of x there are only finitely many i for which x belongs to u i and f k of x t is x t if x is not in u k that's what i had already told you indeed in a neighborhood of x all f k right in t except those k for which theta k of x is not zero okay for this reason f is also a proper mapping since summation theta k is one it follows that f is subjective since f k are embedding f is an injective therefore f is only one so the partition of unity i have been used here and eta k is how much i have to go that will depend upon partition of unit so eta k so add up to theta one theta two up to k that's called eta k we will stop here next time we will consider that euclidean topological spaces can topological manifolds can always be embedded inside r n so sometimes thank you