 Since we are going to heavily use the notion of Coindistopology and in particular the so called Compactly Generated Topology, let us recall a few fundamental facts about them which will be useful for us throughout especially while dealing with CW complexes. It may be recalled that a special case of Coindistopology was introduced as well as used in part one. So, we assume that you know these things but let us recall these things in somewhat quick fashion. That is one of the reasons why the module 6 is released to you before module 5 that we are going to use is Coindistopology in module 5. So, let us begin with the general theorem here namely start a family of topological spaces. I have denoted them by xi comma tau y because I want to specifically mention these tau y's here. Be a family of topological spaces and look at some family of functions from xi to x. So, which is denoted by curly f. These are just synthetic functions because I have not taken any topology on x right now. Now, you put tau f is equal to all subsets of x such that for every i their inverse image belongs to tau i. So, inverse image must be an open subset in tau i for every i. So, that is a collection and the statement is that tau f is a topology on x with the property that each f i is continuous. Moreover, it is the maximum topology on x with this property. The proof is very straightforward and elementary and straightforward means whatever you are supposed to do you have to there is no trick involved here. So, look at this for example, continuity of f i follows because f i inverse of an open set by definition open set in tau f by definition f i inverse must be tau i in open in the x set. So, it is very clear that if this is a topology then with respect to this one each f i is continuous verifying that this topology and the maximality of this topology etc is easy. Following the above theorem now make a definition. The topology tau f defined in the above theorem is called topology of an x co-induced by the collection f i. To indicate that it has something to do with f we will denote it as x comma tau f. So, we can have a shorter notation by x f. So, it is the coexist topology with respect to this family f. Most of the time we will be considering a special case wherein each f i is an inclusion map of the sets x i to x where x is a larger set and all these x i's are subsets of x with their own topologies. So, that is the situation that we want to apply this one. So, let us see in this special case what are the properties of this topology. Let x be the union of x i's where each x i is a topological space. Let tau equal to tau of eta be the topology on x where this co-induced by the family eta which is eta i from x to x are all inclusion maps. I have also assumed that union is precisely equal to x here. So, that is a special case here. A subset A of x is open or respectively closed in tau if and only if A intersection x i is open or respectively closed in x i for each i. So, this is the first statement of the theorem. The second statement says that if you take the disjoint union of x i's then x being the ordinary union will be automatically a quotient set. It is a quotient map. So, that let us take the quotient map q of x is nothing but the inclusion map whatever x is in some one of the x i's. So, take eta i of x for x belong to x i. Any surjective map or surjective function is a quotient function set theoretically. But now, you can give a co-induced topology tau on x is nothing but the quotient topology coming from the disjoint union topology on x i's. So, when you take disjoint union, the co-induced topology is the same thing as the so-called disjoint union topology which is nothing but a subset is open if and only if you know each part in each A intersection x i is open inside x i. All those open subsets from various things take the union they are the open subsets of disjoint union x i. So, that is the simplest case of co-induced topology. From there, we take an arbitrary union and get this one. Once again, verification of A and B are very straightforward. You have to just look at the meaning of f inverse of A where f y inverse of A is nothing but when f y is eta i is nothing but A intersection x i that is all. So, and then B is an immediate consequence of A because in the quotient topology is also what the definition definition of quotient topology is that something is open here if and only if it is inverse image in each x i is open, inverse image in the total union is open which means intersection with each x i is open. So, you get this one. So, now, we would like to make one more special case in which you can extend the neighborhoods. So, for that, we go back to proposition. Let me just recall proposition 1.1. So, here we had this family of attached cells attached to a y. So, x was obtained by attaching cells to y. Then we had these extensions of neighborhoods. So, similar to this, this is much very much elaborate and much more structured. But there is a very simple extension of this one in the purely topological spaces and that is what you would like to draw to your attention to. So, let us see what is that. So, come here, let y contained at x be a closed subset. That is all I am not, not that x is obtained by attaching cells and so on. Suppose A contains at x is any set and B is an open subset of x such that A inside B. Suppose U is an open set in y such that A intersection y, U is an open subset of y. So, A intersection y is contained inside U because U is an open subset of y, but U itself is contained inside is V intersection is V intersection y. Okay. Then there exists an open subset V prime in acts such that A is contained inside V prime, contained inside V and V prime intersection y is precisely U. So, what I have done is that for the part of A inside y, U was a neighborhood inside y that U gets extended to a neighborhood of A, extended neighborhood of A inside the whole of x, but contained inside the original neighborhood V. So, that is the meaning of this. The V prime is contained inside V, V prime intersection y is U. So, this is what I mean by extending neighborhoods. Okay. You can extend the neighborhood, but at the same time you can control it so that it is not too large. So, it is contained inside and already it is a neighborhood. Okay. So, this is an ordinary statement for subsets from a closed subset from y to x which we can now generalize to a large extent. So, that is what we are doing there. So, the proof of this one is very straightforward though it looks somewhat complicated and all that. So, here I have drawn a picture. So, in this picture, so whole area is you can take it as the whole plane as x and this square is y. Okay. Then this hexagon represents A. Okay. A intersection is part y. Okay. This part y that is contained inside this U intersection y. Okay. This dotted thing is U intersection y. I deliberately return now like this because U may be as an open subset could be anything that is you do not know where it is going. Okay. But it must be intersection which you contain inside V. V is this ellipse. The area close by ellipse. Okay. So, under this situation what I am doing here is the proof. Let W contained as x be an open subset in x such that W intersection of y is U that is the meaning of that U is open inside y. So, y is a subspace topology. So, there is an open subset like this. After that you just take V prime equal to V minus y. So, this part, open part here. Okay. Union W intersection V only the portions, you know W intersection V part you have to take. So, this part for example is outside V. You should not take. Okay. So, once I have written down this one, all that you have to verify that this V prime is open and A intersection y is containing A, V prime intersection y is exactly equal to do and after that is very clear. V prime is containing that V is very clear. Okay. You verify that one. Yeah. So, that is just one set at a time. Now we want to generalize this to a larger extent. So, here again take x to be the union of all the x i's, each x i is a closed subset of x i plus 1 in the next one. Okay. So, they are subspaces in some sense union, I have taken subspace is closed subspaces. Okay. And x is given the core induced topology with respect to the family of inclusion maps from x i to x. x itself may have no may not have topology to begin with. Each x i is topological space and each x i is a subspace of x i plus 1 and it is a closed subset of x i. So, this is the hypothesis. Okay. And then we will give the topology x eta to tau eta to x. Okay. That is the core induced topology. Now suppose B is an arbitrary subset of x contained inside W. Okay. And W is such that intersection with each x i is a neighborhood of B intersection x i. If B is a subset of x, B intersection i will be a subset of x i. This W intersection i is a neighborhood of B intersection x i and x i. Then W itself is a neighborhood of B in x. Recall that a neighborhood means what? Okay. Something is W is a neighborhood of A means what? There is an open substitute such that A is contained inside, U contained inside W. This is what we have to prove finally. Then W is a neighborhood of this one. I have to find an open subset. Now here neighborhood, W intersection x i is a neighborhood. We are not given any open set. There is some open set. So, each time you take open subset, they may be different. When you intersect with previous x i, they may not coincide with the old one and so on. That is the problem here. Otherwise, suppose I change the neighborhood to open neighborhood here and open neighborhood here, then this is obvious from the definition of coidist policy. W itself is a open set. So, point is that open neighborhood has been replaced by arbitrary neighborhood here. That is why you have to prove this state. This comes from step by step by the previous lemma. Okay. So, we have to the previous lemma with each time take x equal to x i plus 1 and y equal to x i. Then x i, then y is a close subset of x i, x and A equal to x i intersection B. B equal to x i intersection W for each time. We then have, this is a neighborhood now. Then inductively, you can choose open sets u i in x i such that B intersection x i is contained inside u i, contained inside W intersection x i and u i gets extended to u i plus 1. So, u i plus 1 intersection x i is u i. This is the statement of the lemma. This is the lemma gives you that one. So, this you do for each i, then you take u equal to union of u i's. When you take intersection of this one with x i, it will be precisely equal to u i. Because all the larger ones, u i plus 1 intersection x i is u i. u i plus 2 intersection x i plus 1 itself is u i plus 1. Intersection with that one will be again u i and so on. So, they all will be equal to u i. Therefore, what happens is this u difference opens up a sector. So, B is contained inside u because P intersection x i is contained inside u i and all of them are going to double intersection. So, finally, it will be contained. Now, we shall study what is the relation if at all. It is not a complete answer, but what is the relation between the topology on x and the cohesion topology on x. Suppose you start with topology on x and these x i's are subspaces. Now, you can come back and give the cohesion topology on x. Is there some relation? This is what we want to study, but we are not going to complete answer here. But some sufficient conditions on which this happens is our, is this theorem now. So, start with x equal to u n of x i, where each x i is given a topology tau i. Assume that the following compatibility conditions are satisfied. Namely, this one is for each pair i j, the two topologies on x i intersection x j tau j, induced from x i tau i and x j tau j are the same. So, this is subspace topologies on the intersections coming from the two different sets must be the same, that is the first condition. Second condition is a more generous condition. For each pair i j, intersection x i x j is an open subset of both x i and x j. You can replace open by respectively closed subsets. So, that is a separate statement. For all of i j's, you have to do that. So, either you take all of them open or all of them closed. So, that is cc2. With these two conditions, now if tau is the induced or coinduced topology on x, with respect to the family eta i, that namely inclusion maps, then the subspace topologies tau x i, you see tau i's are already given topologies. Now, subspace topology tau x i is coming from tau eta. For each x i coincides with tau i, hold a topology. Moreover, each x i is open or closed in x according to condition cc2. So, this is the beginning of our attempt to understand what happens to the x itself on x i's. So, what we did? We have started with some topologies, gave the coinduced topology on x, then look at the subspace topology. Will those be the same? That is the question. Whatever topology you have on x, if these are subspace topologies on each x i, then you go to intersection. There will be two different ways of doing this. Namely, first come to x i and then go to x i intersection x j or secondly, come to x j and then take the intersection. Both of them will be subspace of the original x. Therefore, they must coincide. Therefore, condition cc1 is a must. So, this is not, this is a necessary condition. We have to put this one, no matter. But this condition is not necessary. This is a generous condition. This will actually imply whatever we want. But whatever we want may be there without this strong condition. Okay? So, we have to be somewhat apologetic about this one. That is such an affirmative answer. Conditions cc1 is a must. It is a necessary condition. Second condition is not necessary. But it is sufficient. It is, it is a stronger condition. Okay? The proof is very straightforward. We need to check that the original topology tau i is the same thing as tau eta restricted to x i's. Okay? For each i. Of course, by the definition of tau, what is the definition of this tau, this tau eta in coin topology, each eta i is a continuous function from where to where, from the original tau i to tau restricted to x i. Therefore, tau little x i is contained inside tau i. Open subsets of this one will have to be open here. Okay? The inclusion map itself is continuous. Therefore, this must, one way is clear. Now, let us prove that tau i is contained inside tau restriction x i. Okay? So, first consider the case when x i intersection x j is open in both x i and x j. So, we are using the second condition also now. And second condition has two, two different cases. One is all of them open, one all of them close, second one. So, let us take the case when x i intersects x j open in both x i and x j. To begin with, this condition itself will imply that each x i is open in tau. Right? Because fix x i, when it is open in tau in the coin superpower by very definition, intersection with each x j must be open. And that is what it is. So, each x i is open in tau. But then if a is open in x i, it will be open in x also. Right? Open subset of an open subset is open. Okay? If a is an open subset of x i, that will be open in x i also. Therefore, tau i whatever you take tau i, okay? tau i means what some open subset in x i must be open inside tau. So, it means x i is contained inside tau restricted to x i. Okay? The same way now you can use for closeness also. Okay? The second case is similar. x i intersect in x j is closed in x j for every j implies x i itself is closed in x tau. Hence, each closed subset of x i, closed subset of a closed subset is closed. Okay? So, it is similar. All right? So, there are nice theories here about the first question that I raised here. But we will not need them. So, I am skipping them. So, let us now go to another special case of this coinduced topology namely compactly generated topology. Okay? So, we make a notation here. Take any topological space to start with. Okay? Then look at all the compact subsets with their induced topology subspace topologies. So, there is a family of subsets, right? Topological spaces. And look at the inclusion maps. So, those maps will coinduce a topology on x. That topology is denoted by x w. This is the classical notation is referred to as weak topology for some reason. Okay? So, that is why x w usually one calls it weak topology. Topology coinduced by a set of compact subsets of x. Okay? So, what is the relation between x and x tau, x and x w? That is what we would like to understand. This is going to be very, very important. So, pay your attention. So, take start with any topological space. The very first thing to notice that is identity map is continuous. What is the meaning of this? Every set which is open in x is already open in x w also. So, in other words, there are more open sets here. So, x w is final than x. Though it is called weak topology, don't be under the assumption that it is weaker than the original topology x. It is not that. It is finer. Okay? Identity defines a bijection of compact sets in the two topologies. Though there are more open sets here, the set of compact subsets is the same as set of compact subsets of x. That is the beauty of this topology. So, that is part b here. The third condition is that, third statement is that if f from x to y is continuous, then the same map f, when you change the topologies on both sides to weak topology x w to y w, it is also continuous. Okay? So, let us see how a and b work out. Then I will leave c to u to work it out. All right? Recall that a subset S of x is closed in x w. I am just recalling the definition of weak topology here. If and only x intersection k, where k is a compact subset, it is closed in k. This should happen for every k. That is the meaning of that x, something is closed in x w. That this topology is compactly generated or co-induced by the family of compact sets. This will be the case if S itself is closed in x. If something is closed in the whole space, intersection with each subspace will be closed in that subspace. So, in particular, intersection with each compact set will be closed. Okay? So therefore, this proves a, namely something is closed in x, then it is closed in x w. By taking the complement, it is open in x, it will be open in x w. Okay? Now the second one, if k is compact in x w, you have to show that it is compact in x. But identity map is continuous. Image of a compact sets is compact. Therefore, k is compact in x also. The converse is important. Take k, which is a compact subset of x. Why it is in compact in x w? Okay? That is what we have to show. Let k be a compact subset in x. Let f be a family of subsets of k closed in x w with finite intersection property. Now I want you to remember that to show something is compact is the following. Namely, take a family of closed subsets, which is finite intersection property. Then show that intersection of all these is non-empty. Okay? By De Morgan law, this is the same thing as saying that if you take an open cover, then there is a finite sub cover. Okay? Just by taking De Morgan law, these two are equivalent. So, I just wanted to recall that. So, I am going to use that property here. Let f be a family of subsets of k closed in x w with finite intersection property. By the very definition of x w, for each f in f, these are closed subsets, we have f equal to f intersection k. Okay? Just of k after all is closed in k. Okay? See, they were closed subsets of x w means now f is a closed subset of k. Since k is compact, in the ordinary interpolation, we have passed through the ordinary interpolation here, an original interpolation. k is compact, follows that intersection is non-empty. So, this is set, intersection of all f is non-empty. It is what we wanted to prove. All right? So, we will use part three, how to show this is continuous to you. Okay? However, let us make now this definition. A space is said to be compactly generated if and only this identity map, which we have seen that it is continuous, must be a homeomorphism, which means that identity map the other way around is also continuous, which just means that the original open subsets remain open and nothing more comes. The topologies are the same. Okay? So, this x w is compactly generated always. If x is equal to x w, then x itself is called compactly generated. Okay? So, here is a remark. If you use this one, everything will be easy for you. So, what are these? Take a compact subset k, then a function f from k to x is continuous. If filled only if k to x w is continuous. Okay? k can be any compact space and f could be any function. Suppose now x is compactly generated, then a function from x to y is continuous. If filled only if every compact subset k of x, the restricted function f restricted to k from k to y is continuous. This fact b is going to be the key to understand compactly generated topology and its use in constructing continuous functions. Okay? So, this will be used again and again in the CW complex. I have emphasized this one several times. In the situation of this theorem, theorem 1.6, I want to write this, let us say. So, this is an exercise. Theorem 1.6, assume that x i is half star. Then prove that x is half star. Let me just recall what is this exercise? What is this theorem? The theorem was that union of x i is one contained in the other and so on. And then extended neighborhood you have to. So, you may have to use this to prove that exercise. Okay? It is not very difficult but you have to work out. So, that is roughly what we wanted to tell about coinduced apologies and compactly generated apologies. Okay? Thank you.