 Welcome to module 27 of point set topology part 1 course. Last time we have studied the union of open sets, union of topological spaces, coherent topology and such things. So those things are much more used when you go to manifolds or more generally simple shell complexes and CW complexes and so on. One of the problems there again is that you have some neighbourhoods of a point in the smaller space, you want to extend it to a larger space and then again larger space and again larger and larger and so on. So the first step of that we will study here which is purely you know elementary point set topology. If you are stuck with this one there then that will be more difficult, okay here is a lemma. Take Y contained in ZX a closed subset, okay X is the topological space, Y is a closed subset with its topology coming from X now. So as a subspace also you can think of this one. Now take A contained inside X any set let V be an open set in X such that A is contained inside V, okay so V is an open subset of X. Now suppose U is an open subset of Y, open subset of Y, okay such that A intersection Y is contained inside U and the whole thing is contained inside V, therefore they are contained inside V intersection Y also. The question is whether you can find an open set V prime in X such that A is contained inside V prime and contained the open contents are V, I mean I am not going out of V but now this V prime is such that when you intersect with Y it is exactly equivalent. So Y part of A this A intersection Y, okay was contained inside so U is a neighbor of A intersection Y inside Y. Now I want to extend it, extension means what when you take intersection it should come back to you, right. So this extended namely V prime to a neighborhood of X to an open subset of X such that this becomes a neighborhood of A in X, okay. If A is a single point this will be easy to do, okay but if A is not a single point and not contained inside of Y and so on then this is somewhat complicated that is why I have put it here. Actually I feel like you know having explained this one feel like giving it to you in SSI but this kind of point set topology somewhat new and many people have difficulty in this one so I am going to explain this one later. Let us understand the problem correctly, you start with a topological space X and this Y is a closed subspace, okay it is subspace and as a subset it is closed also inside X. Now here an arbitrary subset of A here, A is an arbitrary subset of X contained in open subset V of X, okay. When you take the intersection of A with Y that is contained in open subset U of Y and whole thing does not go out of V which means that when you take intersect with Y it contains U, okay. What I want is if this were equality then I am happy, right you do not know this equality so what I want is may be something smaller here V prime, subset V prime intersection Y will be equal to U, so can you bring it down subset smaller one, subset equality whole thing is a question, okay the assertion is there is, okay question is like that this question mark here but assertion is that it is true, so this is the thing, so you can find a V prime here open contained inside V containing A when you take intersection with Y it will be exactly equal to U that is why this is an extension of U, okay. You do not call V an extension because when you intersect you do not get U but you get something larger maybe that is all, okay. So what is V prime, how to get it, so I have put it directly here but here is a picture what all kinds of things can go wrong, so this is your X, this is your, this square is your Y let us say in the picture, okay the picture cannot be too bad after all, this is your V which is an open subset of X, it is not contained in your Y then everything will be trivial, this is your A this X are gone which is also not contained inside X, contained inside Y, okay, part of it is inside X part only part is inside Y, okay. Now what is U once again, so see U is an open subset of Y and contains A intersection Y, so this is A intersection Y, okay and this dot dot dot up till here is an open subset of Y and it contains A intersection Y, okay. Can you extend it to an own neighborhood which will contain the whole of A like this, when you intersect it why it should be only this part, okay. This U is open in Y, U will be equal to something like W the full thing dot dot dot intersect with Y that is the definition of opens of sets of Y, okay but this W has gone out of B and it does not contain A either, so this W the choice of W here by the definition of that U is open inside Y is not doing the job, it neither contains A nor contains inside W, I want both of them, so what should I do, this is the point, so I have to use that is Y is a close subset of X, okay I do not want to I have to use that, I do not want to go out of V, okay V is a opens of set Y is closed, so V minus Y will be an opens of set, okay so this take this opens of set, where is it open, is Y is closed inside X, so complement of Y is open inside X, so V intersection that will be open inside X itself, so take the intersection with this one whatever you have, this portion this portion will get out that will be inside, but still it is not covering A right, so how to cover A to think properly and that will give you an answer, how to cover A, so W is there I can intersect with V minus Y, okay that will give you an opens of set contained inside V, but I should finally make it containing A also, right, so the picture will tell you what you have to do, here I have an answer here anyway, so what I am putting it is start with W open in X such that W intersection Y is U, I repeat how do you get it, because Y is subspace barrage and U is opening Y, now put V prime equal to V minus Y, Y is closed, so V minus Y will be open because V is open, union with W intersection V, okay this W is there, okay intersect it to V this part, okay union W intersect with V, V minus Y this part was not there right, so you take this part union of this part, this part V part will cover that one and W intersection part will be whatever your reluctance will cover that one, okay, so V minus Y union W intersection V, okay, so this is my V prime, W is open inside X, V is open inside X, intersection is open, this is union of two open sets that is open, the only thing you have to verify is suppose you intersect with Y, it must be equal to U, when you intersect with Y this part does not contribute anything because it is complement V minus Y, so it will be only this, W intersection Y is U, okay, but W intersection V is already part of that one, V is already inside you, so this is also, the V is already larger subset than U, therefore when you intersect it with Y this will be just W intersection Y and that is U, okay, so finally you have to verify that A is contained inside F prime, I will review it to you, A is contained inside E prime is what you have to find the proof, so that much I will leave it, so once you have just one single step thing here, this lemma, now I will extend it to, now I will extend this result, not the neighborhood to the general case here, namely suppose you have X written as union of Xi where each Xi is the close of space of Xi plus 1, so this is union of increasing sequence of closed subspaces, okay and X is given the coherent topology with respect to the family, Xi to X, that means what X is a topology, these are subspaces and then you have the, this Xi inclusion means these are subspaces and then something is closed or open if and only if intersection with each Xi is closed or open, all those things are, that is one wing over a topology, coherent topology. Now suppose you take B contained inside W, contained inside X, okay, these are such that for each I, W intersection Xi is a neighborhood of B intersection Xi in Xi, okay, when you localize at each Xi, you have this W intersection, it is a neighborhood, neighborhood inside Xi, okay, then W is a neighborhood of B in X, this looks very strange because by the very coherent topology this seems to be correct, watch out here, I have just sent a neighborhood, suppose I replace it by open neighborhood then this is obvious, W intersection Xi is open for each I inside Xi, then W will be open inside X over, it already contains B, okay, so the whole point is that these only neighborhoods, they may not be open, yet W will be a neighborhood of B, I am not claiming B, W is open, okay, okay, so we appeal to this lemma which we have done just here, this lemma, one by one we keep getting open subsets, neighborhoods contain open subsets, right, there is not single one W round, so that is why you have to extend it correctly each step, then you will get it, that is the whole idea, so appeal to lemma 2.49 or whatever, okay, start with X equal to Xi plus 1, Y equal to Xi, so Xi is a close aspect of Xi plus 1, so in general we had lemma was, Y is a close subset of X, so I am applying that, take A to be, Xi intersection B, the V to be extension W, these are X, Y, A and V were coming inside this lemma, so I have set up what they are, then inductively you choose, suppose you have already chosen UI, such that UI is an open subset of Xi and contains this Xi intersection B, okay, then this lemma says that, okay, so that is what I have, so the B intersection Xi is contained as UI, contained as W intersection Xi, then there will be UI plus 1, when you intersect with Xi it is precisely equal to UI and of course this UI plus 1 will be contained inside W intersection Xi plus 1 and so on, now you put W equal to union of all these UIs, intersection with each Xi will be precisely equal to UI which is open subset of Xi, therefore it is open, intersection with Xi contains B intersection Xi, therefore this will contain B, the whole thing is contained inside W because intersection with each Xi is contained inside W intersection Xi, that is all, okay, so one step extension, I said this lemma will give you this whole extension here for an inductive family which is, this is an increasing family of closed subspaces, this kind of thing occurs in CW complexes, study of CW complexes, okay, so that was the part of union, the next topic was about quotient ecology, I will have to do that one for the rest of the time here, though that is not exactly the theme of the today's talk as such, okay, they all fit into the theme of constructing new spaces, so today I will not do anything about spaces but just about sets, okay, the next topic is quotient spaces but today we should just recall the synthetic aspects of a quotient map, hope you know it but let us not assume that you know exactly what I want you to know, so I will tell you such few things, consider the following three concepts in set theory, start with X any set, okay, first thing is there is a surjective function from X to another set Y, okay, the second thing is X is written as a mutually disjoint union of some of its non-empty subsets indexed by a set Y, what do I mean by this, X is disjoint union of AY, AY 3Y inside Y that is indexed inside, okay, AYs are non-empty, non-empty subsets, AY intersection AY prime is empty, Y not equal to Y prime, it is mutually disjoint, alright, so there is no repetition and so on here, if there are two repetitions then they will be equal not empty, so Y not equal to Y prime, intersections must be empty, so this is the meaning of mutually disjoint open sets, sometimes I just disjoint union that is what I started with, right, so here also I am looking at disjoint union, that is the second condition, so such a disjoint union often is called partition, partition of a set, some people call it decomposition of a set also, so I will use both the terms so that you will get familiar with both the terms, that is all, the third one is we are given an equivalence relation on X, so I have told you three different things, surjective function, a decomposition and an equivalence relation, what they have to do with each other, the claim is that these concepts are equivalent to each other giving one of them is same thing as giving the other two, is the same thing, what is the meaning of same thing I have to explain to you, okay, so let us see how, what is the meaning of that these things are same, to start with suppose X is said and so what is the surjective function, that is the first one, right, it is a here, take a surjective function, I want to say that it will immediately produce a partition, what is that, the fibers, the fibers of Q, what do you mean by fibers of Q, Q inverse of singleton Y, right, that is the fiber over Y of Q, as Y varies over Y, okay, that will cover the whole of X because this is a surjective function, okay, that means what union of A Y is equal to X, because the function Q inverse Y intersection Q inverse of Y prime will be empty, okay, again because surjective each Q inverse of Y is not empty, so this is a partition or a decomposition, so A implies B, implies B means what, a surjective function gives rise to a partition, all right, let us go further now, suppose you have given a partition indexed by a set Y, I want to define an equivalence relation now, relation I am defining X is equivalent to Y, X1 is equivalent to X2, if both X1 and X2 are in the same subset AY, remember X is union of AY, so given any X it must be inside one Y, AY, but if given two of them, both of them are in the same AY, then I said they are equivalent, they are related, why this equivalence relation, that is again easy to see because if X, Y, Z are such that X, Y is in same thing and Y, Z is in same thing, X and Z must be in the same same thing, that is all, right, same set AY, both of them will be, so that is a transitivity, reflexivity is obvious, symmetry is obvious because by definition it is symmetric, X and Y both belong to means what, so it is already symmetric relation, so this is an equivalence relation, start with a function, subjective function gives you a partition, partition gives you a equivalence relation, you can go back whichever way you like, I can directly go to now the function, suppose you have a equivalence relation, okay, you want to construct a function from X to some set Y, Y is not given now which is a subjective function, so how do you do that, not through some arbitrary, that must be something to do with the relation which you have been given, equivalence relation, okay, so take the equivalence classes that is you are set Y, okay, take the equivalence relation is there, so deliberately I have written it is a different notation here, okay, now you have some equivalence relation, not necessarily coming from a function, alright, some equivalence relation, I do not know what function is, I am just concentrating equivalence relation, take the equivalence class, take all equivalence class that is your Y and take Q of X equal to the equivalence class of X, that is a function, each equivalence class contains some point, after this equivalence class, therefore this function is surjective over, so you have got a surjective function, the point is that you can come back and you will give you the same thing, whichever we like, you have three cycle here, you know cycle means consisting of three points, either you can go from here to this what I have started surjectivity, I came to a partition and I came to equivalence relation, then I came into surjective function, you can come back directly also or you can go here and come back, you will get the same thing which you started with, let us just test it just for this one, starting with an equivalence relation, I have constructed this map Q, okay, so how do I get back this one, all that I have to do is define this equivalence relation whenever they are in the same fiber, they are mapped to the same point, the same point is what, equivalence relation, equivalence class, so I am getting back the same equivalence relation, so you can check the other three also from here to here, here to here, so this is what you mean by, you know there is a bijective correspondence here, it is canonical, you can go from here, you come back you get the old thing, okay, so these three different pictures of a quotient function has to be kept in mind, the simplest being a surjective function over, but the other two descriptive pictures are also very helpful many times, okay, so keep up this one, okay, so there are these words, the map Q is also called identification map, quotient map, quotient function so far, map I use only usually for continuous function, identification space is what, the set of equivalence relations, the classes of equivalence, sorry equivalence classes corresponding to equivalence relation which is, which you can say that same thing as fibers equivalence classes, okay, so these only set theory so far, okay, I want to tell you that it has some properties, so set theory function, set theory property, what is it, take a surjective function Q just like what we are done here, okay, now take any function from x to z to some other set, okay, I want to have a function f hat from y to z such that this f hat, f total composite Q is f, this will happen if filled only if this condition is satisfied, 2x1 equal to qx2 if filled only if implies fx1 equal to x2, okay, implies fx1 equal to x2, then there will be a map like this, so this is the lemma, this is the set theory, so Q is from x to y, it is surjective, f is some map here, suppose you can fit a map here, function here from y to z such that this diagram is commutative, what does it mean there, Q hat, now Q composite f hat, f hat first you should writing f hat composite Q must be equal to f, when does it happen, suppose there is such a map, first some function is there, okay, then I have to verify that perhaps you know if filled only if I have to verify this, suppose Qx1 equal to Qx2 used to coming here, then what happens, take a fact of that it will go to same point, but what is a fact of this Q composite Q is f, so fx1 will be fx2, so this condition is easily satisfied if there is one fact, conversely suppose this is true, then I want to define its effect, how, take a point here, how do I take a point here, you see the equivalence class or it is Q of something here, Q of some x, take f of that and define f hat will be equal to f of that point x which comes to this y, so define f hat of y equal to fx, what is x, Q of x is equal to y, so that is the way you have to choose, okay, when you have chosen you have problem, why this is well defined suppose I choose some other x prime here, okay, but that is the condition here, if there are two points x1 and x2 that is a Q of x1 equal to Qx2, then fx1 is equal to fx2, so therefore f hat of this y or y prime whatever, you have y you have chosen to find, but x and x prime differently, f of those two points will be the same thing, so f is, f hat is well defined, but then what is f hat of Qx, if this y is already Qx, by the very definition it is fx, if y is already Qx then f hat of that is by definition fx, okay, so whenever two points here go to the same point here, f should also map them to the same point, so this is used everywhere, everywhere whenever there is a quotient map like group theory, you know function analysis when you want to construct quotient vector spaces, linear algebra, linear algebra you must have done it already, so all these things there are, this is function theory, this is just saturated, there you may have f, f is linear map, this is also linear map, automatically this will be linear map and so on, if this is a normal subgroup and this is quotient here, where this is a subgroup, then this becomes a group, then this will be automatically become a group homomorphism, okay, so that is the relation, this is the set theory, so different disciplines use that in a correct way, so now we are going to use this topology, so today I am not going to give you topology, okay, so that we will do next time, so this much elementary points of topology you must remember, so tomorrow we will discuss the topology of this one and also some examples before going further, thank you.