 Welcome to module 9 of NPTEL NOC course on point set topology part 2. Today we shall study detraction from Turkex which is closely related to the topic we are studying namely local compact spaces. Let us denote the family of all Hofstor spaces by the symbol H. Throughout this section we shall only deal with apological spaces which are Hofstor and perhaps even not mentioning it specifically. Following the general practice we shall consider various sub-families of H and call them categories. So the word category has a very special and very wonderful meaning in higher mathematics. At this point it is not necessary for us to go into the details about that. For any topological spaces let this kappa x denote the family of all compact subsets of x. So this is a notation for families of all compact subsets of a given topological spaces. Now we define another topological space this is kx. Please do not confuse this kx with kappa x. They are quite different notation. This kappa x to be the topological space its underlying set is x itself but the topology will be different and what is that topology? It will be the co-induced topology from all the inclusion maps from all compact subsets of x. Okay so this is a co-induced topology. Recall that the co-induced topology is nothing but the largest topology on x such that all these functions in this family are continuous. Once you declare a family of continuous functions from arbitrary spaces into x okay fix that family take the largest family such that all the members of this family are continuous that is called the co-induced topology from this family. Since I am talking about the largest family there is only one that is called co-induced topology. Just to recall what we say is from part one where we have studied these things very thoroughly. So I have summed it up here it is equivalent to this definition of co-induced topology. It is 1, 2, 3. What are these three conditions? A subset u in the co-induced topology kx is open if and only if k intersection u is open in k for all compact subsets of x. Now instead of open your same condition is closed. A subset f of kx is closed if and only if f intersection k is closed in e k for every compact subset k of x. The third condition is in terms of the continuous functions. A function from kx to y is continuous where y is any topological state. If and only if for every compact subset k of x the restriction map is the same thing as take eta k and then forward by f okay eta k is the inclusion map. So that map from k to y is continuous where k range is over all compact subsets and eta k is the inclusion map. Okay so these are very easy to verify. If you do not know what is co-induced topology you can take any one of the maps as the definition of one we will use the first one or second one. Sometimes we may have to use this one also. Okay now the following properties of kx are all straightforward. The co-induced topology is always better or finer than the topology which you have started with in this particular case. See x is already topological space and eta k is a inclusion map from compact subsets of k. Compact subsets of x itself right. So before talking compact subsets of x x must be having the topology already. Eta k's are continuous if you take the topology on x the original topology but kx is the one with the maximum number of open sets which is property which is the maximum topology which is the largest topology. So in particular kx is a more open subsets than x. Therefore what happens kx to x is continuous inclusion map is continuous. If kx is half star if x is half star why again kx has more open sets that is why it is soft off. If you take compact subsets of this new topological space kx they are not changed they are same thing as kx kx kx kx kx of kx the same thing as kx kx. So this becomes a surprise a present surprise because if there are more open sets there is a danger that the topology may have such a property namely some compact set may not be compact in the new topology. But here nothing is happening that is the good thing here. The fourth, fifth thing is, the fourth thing is first of all if you take K of KX, KX is a house door space, you can take K of that, you do not get anything new, it will be KX itself. The fifth thing is a continuous function from X to Y, then underlying space is X to Y in both the cases, but you now take KX and KY new topology, F will be continuous in the new topologies also, okay. So set of continuous functions from here to here has not changed either, okay. So finally you may ask why do you need it at all, so that you will be explained so. So KX to X is continuous, I have explained already, okay. Then two follows from one because this has more open sets, every single, if it is open here, there are two points can be separated in this space, so the same open sets will serve here as disjoint open subset separating the given point. The third one needs a little more explanation, take a compact set here under the inclusion map which is continuous, you compact here also, so that is the beauty. So all compact subsets here are here compact find, but suppose something is compact here, why it is compact here, that is what you need to understand, right. So if K is a compact subset in KX, then from one it is compact in X also, that is no problem. On the converse, if K is compact subset of X and U alpha is an open cover for K in KX, what are open subsets of KX, they have the property that when you intersect with any compact set, they are open in that compact set. So K intersection U alpha, these will be open in K, obviously they will cover them, these things will cover K, K itself is a compact subset of, see I started U alpha which are open subsets of KX, but now K intersection U alpha are open subsets of K in the old typology, they are in a typology, therefore they will admit a finite cover, but that is the same cover we will do for K in the old typology also, okay, once you have finite cover that is a theory, so it will cover in whatever typology you take, okay. So K is contained inside, I already want one K intersection U alpha, so it is contained in U alpha, unit of U alpha, okay. So once the compact subsets of this one are same as this one, you use the inclusion maps from here to KX now, okay, and induce a typology, the maps are the same, sets are the same, therefore the coin, the induced typology will be also the same, right, so 5 follows from 4, so the entire family eta K from K to KX is the same, whether you take K to KX or K to X, okay, so finally I will have to verify 5, what is the 5, start the continuous function in two topological spaces from X to Y, now pass on to KX and KY, these are new typologies, what is the relation, you take KX to X, the inclusion map that is continuous, X to Y is continuous, so after composing, I get KX to Y, the same function F, thought of as a function from KX to Y, that is continuous, but I have to show it KX to KY is contained, now KY has obviously more open sets, those open sets, extra open set, why they are in your image, which is same thing as same open set, why they are open subsets inside KX, okay, under F inverse of those, set really same, but new open sets may have occurred here, so why they are open in KX, that is what we have to do, we have to check, right, so given instead of open sets, you can do it with closed sets, so let me do it with closed sets here, given a subset F of Y, which is closed within KY, we have to show that F inverse of F is closed in KX, what is the criteria, take a compact subset K of X, intersect it with F inverse of F, show that that is closed in K, that will do the job, so we have to show that F inverse of K, intersection with F inverse, F intersection with K is closed in F, now F is continuous from X to Y, okay, therefore if you take L, which is F of K, that will be a compact subset of Y, therefore by the criteria for F to be closed in KY, F intersection L is closed in Y, right, hence F inverse of F intersection L, which is F inverse of F intersection F inverse of L, that is closed in X, because F is continuous from X to Y, so from Y inverse may have not taken, they are coming X, this is a closed subset, but now K is already contained in F inverse of L, because L is nothing but F of K, okay, therefore F inverse of F intersection just K, same thing as F inverse of F, intersection F inverse of L, you intersect K further, this set is nothing but K, this is already closed, therefore this intersection with K is closed in K, okay, so that is a closed subset is what we wanted to show, okay, so that completes the proof of this lemma, alright, so we know something about this KX getting X out of, KX out of some, out of X we are getting some other space, what is this space, how to identify it, what are its properties, many properties you have already listed, now we will name it and then start studying it further, the topological space X is called compactly generated if KX is equal to X, so this is a general definition I am making, which is the same thing as saying that remember identity map from KX to X is already continuous, if it is continuous in the other way around also, it is inverse also continuous, then it will be a homomorphism, that is just the same thing as saying KX is equal to X, okay, because already we know that the topology, the underlying set is the same and one topology is contained in the other, the only way you want it is that the two topologies are the same, so inclusion map must be a homomorphism, so that is the same thing as that inclusion map from X to KX is continuous, one way we already know, okay, so such spaces are called compactly generated and remember we are only all the time working inside host door spaces, so far we have not paid much attention to that, okay, but we will keep insisting that we are using host door spaces, then we will have a notation Cg, compactly generated, okay, this is another subcategory now of the whole space, of the topological space, but inside the host door spaces itself, okay, in particular starting with any X, KX is always compactly generated, okay, so you see from general space to you are coming to something special, so set of all compactly generated things is obviously smaller than all host door spaces, right, we now come to the property of KX which we call functoriality, okay, or people also keep referring to canonical property and so on, okay, so this property makes this construction quite dear to topologists, the associations this K from all host door spaces to compactly generated spaces, the I which is inclusion map actually, I do not call it map because this H and Cg are not necessarily sets, so they cannot be domains of functions, okay, but that is why I have used this twiddling arrow here instead of straight arrow, straight arrow to we are using straight arrow for indicating functions, so here I am only calling them association, for all matters of our importance they do behave like functions but there are saturated problems if you insist that H must be a set and so on that is not a set, the collection of all objects, all any all of as soon as world world all is there you have to be very careful, okay, so these two associations given by X going to KX, Y going to IY the same, this is just inclusion map, have the following very close relation, nice properties, start with any continuous function from X to I inside this larger space H, inside this larger domain, larger category and that is why I am using category it is not a set and so on, okay, so take a function like this then same underlying set but topology is different, F is continuous, okay, so to distinguish this from F we will just denote it by KF because now we are thinking it off, domain is domain and codomain are both compactly generated, co-induced by the collection of compact sets, okay, so that is why we will denote it by KF, as a function it is F itself, given X belong into H and Y inside CG there is a natural bijection, so again I have used the word natural I will explain it to you but not in the statement of the theorem, okay, what is the natural bijection depends upon Y and X of course, so all continuous functions which I have denoted by apps from IY to X when I write IY, I have to think of this as compactly generated space, okay, but now it is included as if an ordinary topological space doesn't matter because X is an ordinary topological space when I come here this is actually compactly generated space because it's here but now X is converted into KX, so that is compactly generated space, maps here and maps here are related by this side, what is this, the whole thing that we have done, take any F make it KF, so any H make it KH, okay, so when you KH actually you should write here KY, right, KY to KX but we already know that KY is Y, KY is Y that is why I am writing like this, so this PsyYX has a new name, new symbol here it is just the map H going to KH, okay, so insist that this is a bijection, so let us understand that, okay, the first part we have already seen KX to KY we have F of FK, KF from KX to KY, okay, so it is a restatement of our part of the lemma for fifth statement of the previous lemma, the second part all that I have to do is I have to tell you what is the universe image, okay, one part I have to say H going to KH I have to say, okay, take any Y from Y to KX which is continuous, so what is the corresponding map here, same map, say H from Y to X it is also continuous, okay, so how do you see that KX to KY you have a function, right, right, KY to Y the identity map is continuous, so compose what you have got, you have got from KX to KY to KY, KX to Y you have got, okay, but now you can think of continuous from KX to this one, you must see that it is continuous from X to this itself, okay, so by inverse image of this one is given by pre-composing with the identity map from KX to X, okay, so if I have a map from arbitrary space to X, okay, X to Y, KX to continuous KX to X to Y that will be continuous into Y into KY it will automatically continue, so adjective natural what is the meaning of natural isomorphism is the following thing, though the symbol indicates that it has to follow with this domain, the construction of these maps that does not depend upon what domain you have, all the time start the map and apply K of that, so that is why this is natural that is the meaning of this one, but this natural property technically we have to express like this, namely given Y1 and Y2 inside CG X1 and X2 inside H, maps G from Y1 to Y2, F from X1 to X2, these are totally arbitrary, okay, then we have such a commutative diagram, Psi Y2 X1, Psi Y1 X2, these are defined already, what is their relation, no problem, look at wherever you have maps F and G, you pre-compose and post-compose, start with the H here, okay, look at F composite, H composite G, that will be from Y1 to X2, starting with a map from Y2 to X1, so I wrote down where I have written compositions correctly, Y2 to X1 is alpha, okay, then okay, Y2 to X1 you have your H, not alpha, all right, alpha is defined here, alpha of H is F composite, H composite G, G starts from Y1, goes to Y2, H from Y2 to X1 and then F from X1 to X2, that is what I have written, F composite, H composite G, similarly here you see it is beta H is KF composite, H composite G, so that is all you have to make, this is something here, don't make K of alpha, okay, essentially it is that, but better write completely what this will mean, that is what I have done here, instead of K alpha, beta is K alpha here, okay, so what I have done, start with start with any H from here to here, Y2 to KX1 this time, see H once here, you composite this one, you come here, so now H is here, namely Y2 to KX1, okay, G is from Y1 to Y2, H is from Y2 to KX1, F is from X1 to X2, so what you take KF, that is from KX1 to KX1, so that is why this map is from Y1 to KX2, so whether you first compose here and then apply K, then beta or first compose it and then apply K, you get the same thing, which requires no proof because we have already seen all these things, the fact that K of KX is KX that allows us to call K as a retraction, retraction, it is as if you know CG is sitting inside H and we are getting back a map which is identity on CG, so that is the meaning of this, if it is X is already CG, then K of X will be X itself, so in any case K of KX is KX, okay, so that is R squares, R square is R, that is like retraction, okay, property 1 attracts the name Functor, namely not only the objects have been associated, the functions are also associated corresponding, okay, so with certain properties which we do not want to go into detail, so that is the name, Functor is given, retraction of Functor is what I call it, property 2 makes it what is called as adjointness, the two associations K and inclusion, they are adjoint of each other, okay, this is a wonderful property. Now finally, I want to come to one important, very very important use of this idea, namely this Functor K, namely compactly generatedness, in the study of algebraic topology, one of the central thing that we do is study maps from compact spaces into some space, especially from the spheres, you know all the spheres are compact, set of all maps from the sphere into X, all maps from sphere cross interval, close interval into X, these are of prime importance in algebraic topology. Now if X is compactly generated, then whether you take, whether you take, sorry, you start with any X, then if you look at maps from S n to X, it is the same thing as map from S n to K X, why? Because S n is already compact, therefore it is compactly generated, therefore K of S n is S n itself. So, map studying from S n to any algebraic space is the same thing as studying from S n to K X, that means what, without loss of generality, we can assume that K X itself, X itself is compact, for X itself is compactly generated. So, that will reduce lots of problems, namely constructing such maps, etc., becomes easier that you have to do things only on compactness. This is not just this one, I have just indicated it will help you in the study of homology function also. So, here is an example which you must notice that given, giving a hostile space which is not compactly generated is not all that easy. That is good because that just means that lot of hostile spaces are compactly generated already. So, that is a happy situation. However, we will give you an example which is not all that difficult, but it is not all that easy also, so pay attention. So, here is an example of a hostile space which is not CG. Start with n cross n, where n is given the discrete homology. So, n cross n is also discrete space. Take one extra point, like we do it in Sierpinski space, disjoint union with extra point, I will denote it by infinity. With the following, now I have defined the homology on the whole space X. So, a subset U of this X, X is what? n cross n union infinity. X, a subset of X is open, means what? It is in tau. So, I have defined this tau. If filled only if f is a subset of n cross n, that is one condition. If f is a subset of n cross n, any subset, it is open inside tau or second condition is there are finite subsets f1, n cross n, f2 of n, such that X minus U, now I am putting a condition on X minus U, is looks like a finite subset of n cross n union with, you can call it n cross capital, little n cross capital N as vertical line. Vertical lines, n inside f2 just means that finitely many vertical lines. Take a finite set of n cross n and finitely many vertical lines. Take the union. If X minus U looks like this, then U is open. Or there is one more condition. X minus U intersects n cross n, namely the, now I am again looking at vertical line. A vertical line intersection with X minus U is a finite set for all n. So, there are three different types of open sets. I will repeat, the first one is all open subsets of n cross n, all subsets of n cross n because they are all open subsets because he is a discrete space. Or the complement should satisfy either two or three. What are these two? The third one is easy to remember. Interception with every vertical line must be finite. It may be empty that certain lines may not be intersecting. But all whatever is intersect is finite, that is okay. Such a set will be taken and complement of a set will be taken as open subsets inside, inside X. The second condition you have to understand carefully. Here I am allowing in the complement full lines here, full vertical lines. But I am allowing only finitely many vertical lines in the complement along with some more finite subsets of n cross. So, two and three are especially designed for what? Neighborhoods of infinity. If a subset U does not contain infinity, then this condition what tells you what you have to do? What you have to do? Nothing. Any set which does not contain infinities are dramatically open. Two and three will give you neighborhoods of infinity. So, this is somewhat unusual. So, I am going through carefully here. So, do you understand now that the description of this topology, it is very easy to check that this is a topology. All that you have to do is the neighborhoods are correct. Neighborhoods of infinity are correct because this part is already at topology. This is described there. So, why? Suppose you take two of them here, two of them here, U1 and U2. What will be their intersection? Compliments will be unions of such two things by De Morgan law. Union of finite subset F1 and F2, union of F1 and F1, prime F2 and F2, prime income. So, such unions will be alright. Or it may be like this. Again, intersection with each N is finite. If union of two such finite, finite, finite. Suppose you have one here, one here, then what happens? You are allowed from finite subset. Some finitely many things are there, fine. So, rest of them are all, what? Rest of them are all finite here. So, this condition, to take care of this one, N cross N for these things are empty. So, fine. We did not intersect. Compliments did not intersect. But if it contains some of them, it is fully existent. So, take a topology like this. So, that is going to give you something. What is that? So, how is the topology on X? That is fine. The subspace topology on N cross N is discrete. That is also fine because we started with N cross N as discrete space. But X tau itself is not discrete. Not all subsets are open. Once something contains infinity, it cannot be arbitrary set. It has to satisfy the discontinuation of this condition. Okay. So, it is not discrete. That is obvious. The third thing is here, X tau is actually a P4 space. It is half star and it is normal. Once again, because of N cross N is already discrete space, we have to verify it for only one of them contains infinity. For example, any point here can be separated from any point here because you can take this point itself as an open subset. And this compliment is also an open subset because it is a closed subset also. So, that will contain infinity. So, that is over. So, host darkness is over. Similarly, take a closed subset here and a closed subset here which are disjoint. You must be able to show that it is normal. I will leave it to you. You must show it is an open subset. I mean, there are disjoint open subset. I will leave it to you to verify the normality. Okay. So, it is a P4 space. The third, fourth thing is for each N, look at the vertical line union infinity. This itself is a discrete closed subset of X. All points of N cross N, they are open. That is fine. Okay. So, if I show that singleton infinity is open here, then this will be discrete. Okay. Why singleton infinity is open inside this subspace? Why? Because you take singleton infinity union all other lines N cross N, little is this N cross and you leave it out. By property 3, that will be an open subset in the in the in X. It is intersection with this one will be just singleton infinity. So, therefore, the singleton infinity is open here. Okay. So, that proves that this is a discrete space. Why this is an open subset? Its complement is some subset of N cross N. So, every subset of N cross N is open. So, this is a closed or discrete subset. Final thing is very important thing here. Every compact subset of X is finite. Every compact subset of X is finite. Okay. So, let me, I have already included A, B, C, D. Let me prove E completely here. That can be any compact subset of X. If infinity is not there, then K is a compact subset of the discrete space N cross N. In a discrete space, we know that every compact subset is finite. So, we are done. But if infinity is there, suppose assume that infinity is inside K, then we shall show that infinity itself is an isolated point of K. Therefore, if you remove that, that is an open subset, isolated singleton points are open subsets of the spaces. So, if you remove it, the remaining set will be also compact because that will be a closed subset of the compact. But then that will be inside N cross N. So, that is finite. Therefore, K will be finite. So, let us see why infinity is an isolated subset inside a compact subset of K. The proof is similar to what we did here, but little more thing we have just examined the definition carefully. Okay. From D, Y N which is N, singleton N cross N union infinity is a closed subset of X. That is what we have seen. Hence, K intersection Y N is compact because K is compact. Okay. Intersection with Y N will be a closed subset of K. So, it will be compact subset of D, compact subset of K itself. Again by D, K intersection Y N is finite. Okay. K intersection Y N is finite. Why? Because we have just now we have seen that this is a discrete space. Okay. Now, take Y N prime as Y minus, this Y N minus infinity. Okay. Throw away infinity. So, now we are inside N cross N. But then this U which is equal to X minus all the Y N primes. Okay. That is a open neighborhood of infinity. Okay. Therefore, moreover K intersection U is just infinity. See, what I have did? I took K compact set intersected with each Y N, that is a finite set. Okay. Take Y N prime as Y minus, Y N minus infinity. Okay. So, throw away infinity. Right? Then we still have these things are open subset or closed subset now because singleton infinity was a open subset of Y N. So, these are open subset. These are, these are closed subset. So, X minus union of Y Ns, that is an open subset. It's a neighborhood of infinity, infinity is still there because I have thrown away Y infinity from here. So, infinity is not in any of them. So, when I throw X minus that infinity will be there. Okay. So, U this is an open subset because I have thrown away a closed subset. Right? So, K intersection U is singleton infinity. So, that is open subset of K. So, this verifies 4 and this verifies 5. So, it follows that K X is a discrete space. Why? Because we have proved that every compact subset is finite. Take an open subset, intersect it with the compact subset. What do you get? You must have those things are open. Right? Automatically, if the finite subset is discrete space, they are open. Okay? And conversely. So, take any set open. Take any set. Okay? Intersect it with the K. K is finite already. Right? So, all finite subsets are open. In particular, all singletons are open. So, that means that everything is open. So, K X will be a discrete space. X is not discrete. Therefore, X is not equal to K X. Okay? Just by adding one single point, compactly generatedness has gone away. Okay? You may wonder. You see, N cross N is a discrete space. Is it compactly generated? Yes, of course. Any discrete space is compactly generated. Okay? Singleton sets are compact. There it's fine. Finite sets are compact. That's fine. Okay? But this CG is not compact. CG is not compactly generated. There are other close subsets, other open subsets, which are not discrete. Okay? All right. So, next time, we will study compactly generated spaces a little deeper. Okay? I have already told you why they are important and long. Thank you.