 Hello, welcome to module 53 of NPTEL NOC on point set topology part 2. The title of today's talk is an application to portion maps of part of the exponential correspondence that we have studied last time. A natural question that arises with respect to the quotient topology is the following. Suppose you have two quotient maps qi from xi to zi i equal to 1 and 2. Is the product map q and cross q2 from x1 cross x2 to z1 cross z2 a quotient map? So, you may be surprised that in general this is not true. See, subjectivity will be still there, continuity is still there. If q1 or q2 is a q1 and q2 are open maps, the product is also an open map. But in general, we have to study other quotient maps. Not all quotient maps are open maps. So, here is the satisfactory answer. In any case, this question was actually raised in part 1 itself when we were studying quotient maps. So, today we will have a satisfactory answer. q1 cross q2 can be written as q1 cross identity composite identity cross q2. This is just a set theoretic fact for any function of a product sets. So, if I have shown that each one on the right-hand side is a quotient map. Composite of quotient map is a quotient map. Therefore, q1 cross q2 is also a quotient map. Therefore, question is reduced to the special case when out of q1 and q2, we assume one of them is an identity map. By symmetry, whichever one you assume, it is the same thing. q1 cross identity or identity cross q2. If you show for all q2, this is true. That is fine. It will also imply it is true for all q1. So, what we shall do is that we shall show that if x is locally compact regular space, then for any quotient map q from y to z, q cross identity of x is a quotient map. So, this is the condition locally compact or regular. Locally compact or absorbed implies locally compact regular. So, we will take this step in. We will take this one locally compact regular. Let us recall a fact about quotient topologies. A continuous surjective map of topological spaces is a quotient map if filled only if it satisfies the following property. Given alpha from a to b, that is a continuous surjection is given. When is it a quotient map or when the topology on b will be a quotient topology? This is a condition. For every topological space w, every function g from b to w, some synthetic function, if the composite is continuous, then g must be continuous. G is continuous, then the composite is continuous is obvious. Because composite of continuous functions is continuous. Here it is the other way around. If this is satisfied for every g, then alpha will be a quotient map. So, I am not going to prove this one. This has been proved and used several times. I am going to use it now. Starting with q cross identity, g surjective and continuous under the hypothesis that the topological space x is locally compact and regular, we are going to prove that this thing is a quotient map. What we need to prove? We need to prove the following condition. Take any w and any map g from z cross x to w, any function such that when you compose it with q cross identity, that is continuous. Then g is continuous. This is what we have. So, this is the diagram given. y cross x to z cross x, q cross identity, we want to prove this is a quotient map. So, take any w and any g here. This g will be continuous if you assume q cross identity and then composite with g, that is continuous. That is what we have. Now, how do we use the exponential correspondence? When you have these function containers, we know the same thing as saying that there is a map from y to set of all, space of all continuous functions from x to w. See what is that map? It is given by the exponential correspondence. This f is replaced by f hat here. Given by f hat of any point y, the same thing is f of y x. So, you get a continuous function. So, exponential correspondence says that f hat is continuous if phenomenon is continuous. So, this we are going to use now. So, once you have this, look at this also here, which is just y to z, the given map q. z to g, you have some function here, g hat, z cross x to w. You have some function here. I do not know whether it is continuous. But what I know is g hat composite q is f hat. This is just a set theoretic fact. g hat of z is defined as g of operating upon any x is g of zx. So, for that reason, this would be this g hat composite q for g hat is f hat. So, this is continuous. But I have started with q as a quotient map. Therefore, g hat is continuous. But once g hat is continuous, exponential correspondence says that this g is continuous. The proof is over. So, proof that product of any two quotient maps is a quotient map is still not yet over. That is not true. If one of them is locally compact, half-door or locally compact, regular, then it hurts. This is what happens. So, now we can use that to just give you satisfactory answer. This is not a field only kind of answer. But it is a useful thing. Suppose you have y1 cross y2 to z1 cross z2, you have q1 cross q2. So, two quotient maps and then you have taken the product. You can come to z1 cross z2 from here in two different ways. q1 cross identity of y2, then followed by identity of z1 cross q2 or first identity of y1 cross q2 and then followed by q1 cross identity of z2. So, the statement is if y1 and here z2 are locally compact, regular, then this composite is a quotient map because each of them is a quotient map or you may use the other hypothesis here namely z1, z1 say y2 is a quotient map, y2 is a regular, a local compact regular and here z1 is a locally compact regular. So, y1 z2 or z1 y2 or there is no necessary that both should be there, y1 and z2 or z1 and y2 that is the meaning of this. Locally compact order of basis then q1 cross q2 is a quotient. Now, we come to the main rarum in this section in this today's talk and that is due to my care. So, here is a partial converse. You may say start the regular space or maybe you can state the host of space, it does not matter, start the regular space, that is important. Then the following conditions are equivalent. So, it is only partial converse, regularity cannot be replaced, x is locally compact. Then every quotient map q from y to z, the product identity of x cross q is also a quotient map. One implies two is what we have to see. Converse, you have to two implies one. So, if this happens for every quotient map q from y to z, if identity cross q is a quotient map, x must be locally compact. This is much stronger than just saying that locally compactness under the assumption that you do not assume locally compact, the theorem is false. You can give that for that, you can just give a counter example. But here this is a theorem. How do you prove this? By assuming that x is not locally compact, we will extract a space y and a quotient map of y y to z q such that identity of x cross q is not a quotient map. So, this is the plan of the proof here. Here we are going to use the exercise 10.9 only. This is part of your preparatory assignments or practice assignments 10. This was not in the main part, but in the practice session. The main result is another not so familiar criterion for compactness. That is also part of this one. You have not done it or tried it. It is time that you should read the solution given. However, right now I cannot afford that one. So, what I will do? I will just recall this exercise here that I am going to use. Exercise 10.9. Look at this part 4 here. So, this is the main thing that I have to use. Let x be a topological space. Show that x is compact, spend only the following condition holds. For every family f of closed subsets, closed subsets of x with finite intersection property and linearly ordered by certain closure. We have intersection of all elements of this curly f is non-empty. Remember, if I remove this linearly ordered by certain closure and if I remove this condition and this one, this is a familiar condition for you. Namely, if every family of closed subsets of x with finite intersection property has a non-empty intersection. So, that is a larger condition. So, here we do not need all of them. Only those which are linearly ordered by certain closure. f1 containing f2, containing f3, you can take decreasing just like in cantorsets, cantors of the theorem and so on. You can take increasing but you can reverse the inclusion, you will get decreased. There is no problem. But the point is do not try f1, f2. That will give you that it is mislead you. It is going to, what is it, countable. There is no countability assumption here. This is just linearly ordered. That is the whole point here. Try to prove this one but later on anyway, we will give you a solution. The other thing is this problem 3 here, let x be, sorry, this one, namely problem 2 here, let l comma less than or equal to be a non-empty linearly ordered set with p belonging to p. Some point I have chosen, p belonging to l. Then there is a non-empty subset w of this l such that w is co-final in l and restricted to w, it is ordering is well ordered. Moreover, the point p that we have chosen is the restricted. So, I do not use this full force. What I need is every non-empty linearly ordered set has a non-empty well ordered subset which is co-filed. So, this is what I am going to use. To prove this one, you will need 1. To prove 4, you will need 3 also but that you read on your own. That is why they are the 1, 2, 3, 4. So, let us go back to Michael's theorem. So, as I told you to prove the unique equivalence of 1 and 2, we have only to prove 2 implies 1. 1 implies 2 has been taken care by the previous theorem. So, start with a topological space x which is not locally compact. We shall construct a quotient map q from y to z such that identity of x cross q or q cross identity of x does not matter. It is not a quotient map. Suppose x is not locally compact at some point x naught belonging to x, then it is not totally compact. At least one such point at which it is not totally compact. What does that mean? There is a family u a a belonging to a, I have just this a is nothing but just an index in set be a neighborhood system in x naught such that none of u a bar is compact. So, this is the meaning of that x naught is not locally compact. There will be neighborhood system such that u a bars are not compact. What does that mean? Each u a bar is not compact means what? Now, I am using this exercise 10.94 that I told you. For each a inside a, we have a family f a of non-empty closed subsets linearly ordered by an inclusion of sets such that the intersection of f, f, f belongs to f a is empty. Since it is linearly ordered, I do not have to take finite intersection property. It will automatically because I have said we have a family f a of non-empty closed subsets linearly ordered by an inclusion such that intersection is empty. So, this is stronger than saying that there is some family with finite intersection properties, the intersection is empty. So, this is where we have used x pi 10.94. For convenience, we should index this family. I could have just ran f a. Each f a has to be indexed. So, I will fix an a then I am indexing it by f t, t belonging to t a. So, t a is the indexing set that indexing set itself gets a linear order like as following. Namely, t is less than or equal to s if and only if f t contains f s reverse order. So, t and s belong to t a. So, in some sense I am taking this decreasing families. I cannot say decreasing sequence and so on. This is not a t s are not necessarily countable. They may be countable. They may be finite. I do not know, but they need not be countable. That is all. So, t is less than or equal to s if and only f t contains f s. See, I do not need to float a t a, but this is just a for the convenience and it will clarify the proof later on that is all. The f a itself is a family. So, if you take an element here, two elements here, one is contained in the other. So, that is all. But I need that instead of f a, using members of f a itself, I am using t. So, as soon as I have t belong to the other, there is a corresponding f t that is all. So, that is why I have to take this indexing separately. From 10.92 about partial order and so on, this linear order, we get a subset lambda a of t a, which is well ordered and cofinal subset of t a. Given any member of t a, say a t, there will be some s, which is bigger than t in this sense and belonging to lambda a. That is the meaning of cofinances. So, these two things right now, I will just from now onwards, we are just using just ordinary constructions here, quotient space and so on. So, we add one extra point to each lambda a. You can cause this as infinity a or 1, but I will call it as just lambda a. It is a very convenient notation. Take lambda a itself as your last element. This is a set. The set lambda a consists of elements of this one. See lambda a union lambda a is lambda a prime. Just one extra element, that extra element itself is lambda a. So, take this. Now, you extend the linear order on this well order on lambda a to the entire lambda a prime by just declaring this lambda a as the maximal element, the one maximum element actually. The lambda a is bigger than equal to all t belonging to lambda a. That is the meaning of it. Now, you take the order interpolation on lambda a prime. Because there is a maximal element also, this will be a compact host or space. It is always host or space. The compactness comes because of this extra point that we have taken. It is like one point compactification. Now, put y equal to disjoint union of all these lambda a's where a ranges over a starting with this neighborhood system which is indexed by a. That a comes here. Disjoint union of all these compactness. It is a huge set. Why itself is not a compact? It is disjoint union of all this. On each lambda a prime, I have the order of all this. Having taken this space y, now I construct the quotient space that by identifying all these extra points lambda a, lambda a will be equal to lambda b for a and b inside capital A. For all a and b, one single point. That I have denoted as z naught and the quotient space I have denoted by z. I repeat, how is z constructed? No other elements of lambda a prime are disjoint. Only the extra point that you have taken lambda a, all of them are identified together to single a point. That is z naught. So, this is a quotient map. So, this quotient map has a property that on z naught, you have all these lambda a's, the fiber. Everywhere else, it is a one on map. Inward image of every other point is just one single point. So, this is the quotient map. That means this is a quotient function. So, we use a quotient topology on z. That is all. We claim that q is as required is a quotient map such that identity of x cross q is not a quotient map. So, two things we have to prove. Actually, we have to only quotient map is fine. If it is not a quotient map, for that we have to prove something. So, here is a preparation for this. For each a inside a, what we have and each lambda inside lambda a prime. That means what? Lambda a prime is one extra element of a capital lambda a. So, you can take that for the lambda. Put E lambda equal to intersection of all f sigma. So, where sigma is less than lambda. So, if this lambda was capital lambda a, then what is this intersection? This intersection will be all these f t's, where t belongs to this lambda a, not lambda a prime. And what was the assumption? Assumption of that, remember this assumption intersection is empty. If I take anything smaller, then what happened? Because it is a decreasing sequence. In any case, first of all since the intersection of closed subsets, each E lambda is a closed subset of u a bar. E lambda, remember these are all subsets of u a, u a bar. E lambda is contained inside f lambda and f lambda is not empty if lambda is not the whole of lambda a. Each E lambda contains f lambda here already because E lambda is what? Intersection of all these things. It contains the next one because all of them contain f lambda. It is a decreasing thing. And f lambda is non-empty. Every member is non-empty. So, if this lambda is not the last element, then this is non-empty. If lambda is last element, then of course this is empty. So, that is all elementary observation. Now, we want to show that the map identity cross q is not a coefficient map. What we do? We take a closed subset of the total space which is the inverse image of subset that some subset below is not closed. So, that is what we want. So, put S a for each a, S a equal to E lambda cross lambda. This is X cross lambda a where this lambda a is inside only lambda a. I have not taken the last maximal element here. Put S a equal to E lambda cross lambda. So, each of them is a closed subset of X cross lambda a now because this lambda a has the order topology singleton sets are closed. There is a hostile space. We claim that S a is closed subset of X cross lambda a prime itself. The whole thing is a subset of lambda a X cross lambda a prime because lambda a prime is larger space than lambda a. In particular, it is a closed subset of X cross lambda a also. See, it is easy to see each of them is a closed subset. But why the union? This union is an arbitrary union over all the lambda. But we want to say that that is also a closed subset. So, how do you prove that? Take a point X lambda which is not in S a. Something is this want to show that this is a closed subset. The same thing as complement is open. So, start with the point X lambda which is not in S a. What is the meaning of this is not in any of this is a union of these things. So, this means that X is not in this first coordinate is not inside E lambda. If X is inside E lambda for any of this lambda then this would have been inside S a. This means X is not in E lambda. But E lambda is what? Intersection of all f sigma where sigma is less than lambda. Something is not in the intersection which implies that there exists some sigma less than lambda such that X is not in f sigma. Let R sigma with the right open ray inside lambda a prime. Remember these are all well ordered subsets. The right ray etcetera, left ray etcetera all make sense. So, once you have watched some sigma here R sigma means what? All those all those lambda is bigger than sigma. So, that is your lambda a prime that is your R sigma inside lambda a prime. Then f sigma complement f sigma is closed subset f sigma complement will be a open subset cross R sigma which is also an open subset f sigma complement contains X because X is not in f sigma and R sigma is the open subset which contains lambda. So, this is a neighborhood of X lambda in X cross lambda a prime. It is easily checked that this S a intersection this open subset is empty. What does that mean? That this is contained in the complement of S a. So, what I have shown here is that X comma lambda belonging to S a has an open subset which does not intersect S a at all open neighborhood it does not intersect S a at all. That means this complement of S a is open that is the whole thing. So, why this is empty for this you have to use the elementary fact here that f t is bigger than f s if t is less than s. So, take a point here it cannot be Z S a for any of them is all that you have to see. So, we have got a set here which is a closed subset of X cross lambda a prime. So, in particular it is a closed subset of X cross lambda a also because it is contained in X cross lambda. Now, take S to be H of S a and take union of over all a inside a. So, each S a is a closed subset of X cross lambda a disjoint union of those things with the disjoint interpolation is Y. Therefore, disjoint union of all these S a's that is a closed subset in Y cross lambda sorry Y itself Y itself is the disjoint union there is no Y cross. So, that is a subspace Y. So, Y has this disjoint union of closed subset that is a closed subset in the disjoint in topology there is no problem. But S is its image H of S a as a rings over a is the same thing as H of all these disjoint unions that is subset of X cross. Because first coordinate I have taken as X here. So, we claim that H inverse of S is a closed subset of X cross Y, which is nothing but I just told you disjoint union of all these sorry here all these S a's it is a subset of X cross is Y here not Y X cross Y. Yeah. So, what we have what is that H inverse of S is closed is why we have already seen, but S is not closed in X cross that which will show that H is not a closed map H being identity of X cross cube. So, again I am repeating here H inverse of S intersection is X cross lambda a prime is these S a's therefore, they are closed. So, H inverse of S is closed. This I have repeated I have already told you all that I need to show is that its image inside X cross is not closed. For this we want to check that X naught comma Z naught is not in S because Q inverse of Z naught is all singleton says a inside a. However, we claim that X naught Z naught is in the closure it is closure is taken in S cross Z something is in the closure and not inside S just one just this is not a closed set that is all. X naught comma Z naught would have been inside a you see what is S by definition is H of S is X cross that it will never hit the point Z naught at all. So, for this start with X naught Z naught inside U cross V where U is open in X and V is open in Z. I want to show that this is in the closure we want to take any neighborhood I could take standard neighborhood show that it intersects S that is all. So, take a basic neighborhood U cross V of X naught Z naught in the product apologize. Let a belong to a be chosen such that U a bar is contained inside U. So, what I am using here that this family of U a bar is a neighborhood system at a. So, if U is a neighborhood of neighborhood system at X naught sorry if U is a neighborhood of X naught there will be some a such that U a bar is contained inside U. It is a closed neighborhood none of them is compact that is how we have chosen. Note that this lambda a is in Q inverse of V intersection with lambda a prime the top thing will be there because Z naught is there inside V. So, Q inverse of V will have this lambda a prime lambda a itself it is in lambda a prime. And hence we will have some lambda belong to lambda a that means we something less than that such that Q lambda will be inside V because it is a neighborhood inside inside this lambda a prime Q inverse of V is open. Q is a quotient map it is an open subset contains this point this is the maximal element. So, there must be some right ray which contains that starting from lambda may be something smaller than. So, you can take some lambda belong to lambda is that this Q lambda is inside it or you can say that something less than lambda say lambda prime open to capital lambda a closed that entire thing is contained inside Q inverse of V same thing as Q of that is inside. It follows that now H of E lambda cross lambda to come down under the quotient map is contained inside U cross V. See this H it just identifies all these lambdas when lambda is smaller to the same elements here there is a identity map. So, that will be inside U cross V intersection S which completes the proof. You just recall that S is also H of S. So, this E lambda H of E lambda cross lambda is inside U cross V that is a point of reason. So, I will just make this remark we kept the statement of the theorem as simple as possible unlike the original statement given in Michael's paper. Indeed the quotient map that we have constructed satisfies a number of properties. Let us say it belongs to a special class capital P of quotient map is class capital P is class of all quotient maps from X to Y where X is a disjoint union compact of stop spaces and fibers of Q are all singletons except one fiber which is discrete closed set. So, this is actually this X is our Y and Y is Z I have deliberately changed them because you should be able to do that on your own also. What we have what is the quotient maps are very peculiar only one point inverse image may be having too many points rest of the points inverse image has consist only one point such maps you take and such maps if for all such maps identity cross is a quotient map if you take the then X will be locally compact that is the statement that we have actually proved but we do not want to insist on that one. So, we can add these three also in the theorem 1, 2, 3 are all equal. This will be much weaker statement than 2 I mean looking you know because it is a subclass okay that is all thank you next time we will do something here.