 So, module 7, so we have been studying various topologies like induced topology especially compactly generated topology and so on. And one of the important thing in studying function spaces is the exponential correspondence on the function spaces. Basically these things were done in part 1 very elaborately. Nevertheless, because of the importance of this part I will recall them as much as needed for our immediate purpose here. So first let us recall what is the meaning of this compact open topology on the function spaces. Given two topological spaces sectioned y, we denote y power x, the space of all maps from x to y, maps means continuous functions from x to y, this is given in the compact open topology which has a sub base corresponding to each compact set k and an open subset u, k is inside x and u is inside y, you define a subset k, u, put them in brackets and take all of them. So, what is this k comma u that is all continuous functions from x to y which takes f, this k inside u, f is from x to y, f of k must be going inside u, then you put it inside this k u, take all such k u, they are subsets of y power x, they will form a sub base, k u is all function f from x to y, this is f of k is contained inside u, this one sub base means what an open set, a set is open if and only if it contains, it is union of arbitrary union of arbitrary not finite intersections of k and u, so k1, u1, k2, u2, k and u and u, take the intersection, that will be an open set. So, arbitrary union of those things will be also open set. So, you should take only finite intersections like that will be form base, so that is the topology. Then that is an obvious map e from y power x cross x to y given by e of f comma x going to fx, if you fix x, f going to fx is nothing but coordinate projection, I would not like to call it that way because now we are not thinking of y power x as a product space, y taken x times, there is also that topology, whereas now we are concentrating on compact open topology. So, if you fix x, y x going to fx is obviously continuous even in the product topology, but what is important is as a joint function, this capital E from y power x cross x to y is a continuous function is what we want to have, that is the first part of the theorem here. This is precisely called evaluation map, what is the hypothesis on x, x is locally compact software, y and z could be an interpolation. Then the first part here is that E is continuous. The second part is really the name exponential correspondence, a function from z to y power x where z is an interpolation space, z to y power x is continuous if it only if the e composite g cross identity from z cross x to z cross y power x and then e. So, z cross x to z cross y that is continuous. Notice that g is continuous, the g cross identity will be continuous. If e is continuous, then the composite is continuous. The point here is on the converse here, namely if the composite continues, why g should be continuous? In general it may not be. So, the theorem says that it is continuous under this hypothesis, namely x is locally compact and also proofs are not very difficult. Nevertheless, let us go through that so that you will become familiar with the concepts of this compact open topology and the space y power x and so on. The first part is to show that e is continuous. Start with an open subset u inside y and you want to show that e inverse of u is opening in the product space y power x cross x. So, take a point f naught x naught in e inverse of u. What is the meaning of that? Then f naught is a continuous function from x to i and f naught of x naught is e of f naught x naught that is inside u. But now u is open and f naught is continuous means or some neighborhood of x naught is taken inside u by f naught. But now x is locally compact all the space. So, we can actually take an open subset v such that x is inside v contained inside v bar contained f naught of f naught inverse of u and v bar is compact. So, inside every open subset you can find a compact neighborhood is the local compact, local compactness part here. So, what is the meaning of this? This means that now f naught of v bar is concerned inside u which is same thing as saying f naught is in the bracket v bar u. v bar is compact u is open. So, this is actually one of the sub bases. So, it is an open set means v bar u can open set in y power x. We get an open set v bar u cross v in the product which will be a neighborhood of f naught x naught. Now, you look at e of v bar u cross v ok take a f here take a point here say any point then f of that x must be inside u why? Because if it is inside v it will be inside v bar and f of v bar goes inside u. So, therefore, e of f comma x will be inside u. So, this proves that a is continuous. So, that is part a alright. Let us go to part b. Part b there are two ways one way I have already told you if g is continuous g cross identity continuous the composite will be continuous because we have just proved that a is continuous. So, now, let us prove the converse ok. Now, I have assumed that this composite is continuous and I want to show that g is continuous. For this it is enough to show that the g inverse of all these sub basic open sets is open where k is compact and u is open ok. Take only sub basic open sets inverse image of those things are open will mean g is continuous ok. So, let z naught belonging to z z was arbitrary space you remember the set g of z naught belongs to k u that is the meaning of z naught is in the g inverse of that. Then for every point k in k what have we have is e composite g cross identity of z naught k which goes to g of z naught comma k which goes to g of z naught operating upon k that will be inside u that is the meaning of this g of this is contained inside u ok. So, there exists open sets v k w k of z and x respectively such that this z naught is inside v k little k is inside w k e of all this w k v k cross w k is contained inside u. So, this is the continuity of this composite function e composite g cross identity ok. Now, k is compact. So, you can pass on to a finite cover k is contained i range 1 to n v k i each w k covers little k when you range k in k that will be an open cover for k k is compact. So, I extract a finite sub count and we call that as w and correspondingly I take v k i's here ok k 1 k 2 k i's are some points of capital K ok there are finite many of them. So, I take v equal to intersection of v k i's then you put intersection here the union here e of g cross w g cross identity of x of v cross w is contained inside ok that just means that g of v is contained inside k cross u ok k comma u the the this notation k inner product k u since v is a neighborhood of z naught we are proved. So, I have formed a neighborhood v here that goes inside that. So, k u g inverse of k u is open that proves this exponential correspondence theorem ok. Now, we come to the study of the compact generated topology ok. The basic problem here is that if x and y are compact generated in general the product topology need not be compactly generated. Therefore, we are looking at situations where this may be true that means you put extra conditions on x and y under which this may be true. However, there is no if and only if conditions ok. So, we will have to study whatever you know best we know about various conditions ok. So, first lemma is if y is locally compact and x is compactly generated ok whenever I say when it is compact all the time I assume host darkness of course and both of them are host of spaces then x cross y is compactly generated if it only if I want to come conclude here, but this is a lemma is a part of this one. So, if it only if every compact subset k of x k cross i is compact instead of saying the whole space is compactly generated first you assure that k cross y k cross y is compactly generated. If you do this one for every compact set then this will be fine this is the lemma ok. So, for this lemma also I am already taking assumption that second factor here is locally compact ok. So, already we are trying to see that if I just assume compact is generated compactly generated both of them I do not get x cross y compactly generated. So, I have to make more little more and more general situations here. So, this lemma says that y is locally compact ok that is the extra hypothesis remember locally compact or spaces are compactly generated. So, this is stronger condition than being just compactly generated ok. So, proof of this one you can try to do with point y is going here going there going there and so on, but exponential correspondence helps you in proving this one very elegantly ok. Consider this function x going to x cross y power y with the with topology whatever you know product with the weak topology ok remember that. So, take any space whatever topology you have then give the compactly generated topology that is the meaning of whenever you have topological space here writing a suffix w ok. The saturated function eta from x cross y to y does not need all these things it is just like the evaluation eta x y equal to x y ok. Remember eta is what eta is function from for each x it is a function from y to x cross y. So, I can I can evaluate it on y ok. So, what I have tell me what is the value value is x y. So, this is a definition of eta ok take this function ok give me. Now, look at the evaluation function x cross y power y cross y to x cross y this is the base space here this is just like instead of instead of x cross y y I have to do one more y here ok. So, this is the evaluation map E of take a function here take a point here ok evaluated at that point you get a point here because evaluation map we have seen that ok if y is locally compact of the space then this way evaluation map is continuous ok. Now, look at E composite take identity this eta cross identity of y operating by x y ok. This is eta of capital E of eta x comma y by definition this is identity cross identity eta cross identity means that, but eta is evaluation you have to evaluate E x on y so that is x y. We now use the theorem on exponential correspondence in both directions once you have established one ok. In between we also use the hypothesis x is compactly generated and y is locally compact. So, this much you have to do ok. So, we get x cross y is compactly generated this is the statement that we want to want to understand that it is equivalent to something else if and only if the identity map from x cross y to x cross y this is compact generated it is continuous ok. And remember start with a space like this when you make it compactly generated topology that will be always finer than this map. So, identity map this way is always continuous if x cross y is compactly generated already then you do not get anything more. So, x cross y to this one is continuous this is same thing as x cross y compactly generated ok. So, first part is just by definition itself follows. Second part is now eta from x cross y power y to w is continuous. So, this is the exponential correspondence now. Why what is the map here identity maps x cross comma y going to x comma y what is the corresponding map in the exponential it is a eta right. So, eta of eta of x y is x y. So, if you look at this eta is continuous remember then this is continuous. So, that is the correspondence eta is continuous ok this is the exponential correspondence if you consider. But now this is the same thing as saying for every compact subset k of x the restriction map eta on k restricted eta let us call it eta k. So, k to x cross y w is continuous ok this is a definition of the compact compactly generatedness of this part ok. So, I have from k or I from arbitrary x I have come to compact subsets now ok. So, that was my idea to come to compact subset. Now, you go back by exponential correspondence. So, this is true for every compact subset ok. So, but that is if and only if now I go back k cross y to x cross y w is continuous. So, this exponent y comes here k cross y is continuous ok by the exponential correspondence. But it this is if and only if is for every every k should be true if and only if k cross y remember y is locally compacted across space ok compact subset of k. So, this inclusion map eta this one is continuous ok k cross y eta k is continuous here which went on this k cross y to x cross y is continuous. But that is if and only if for every compact subset k the inclusion k cross y to k cross y w because this inclusion where does it go? I have taken the larger space, but it goes inside a k cross y to continue. If this continues from here to here that will be continuous because they have the same width apology here ok. So, for this is true for every compact subset which means that k cross y is compactly generated. That is why we started with x cross y to x cross w in the beginning. Now, we have come to the same conclusion with x replaced by k. So, this is the the lemma here ok. So, this lemma is proved now what will be lemma and this lemma. So, now, we keep using it very beneficially to prove all this A, B, C they are all different statements, but somewhat similar x and y are off-door spaces x cross y will be compactly generated under these conditions. Easy to remember first condition is both x and y are compact no problem. The product itself is compact right in a compact space if you give compact generated topology it will be the same topology. First part does not need anybody say any lemma or anything x is compactly generated and y is compact ok. This is the first thing that we would like to prove and then generalize this to x is compactly generated, but now instead of y compact, y is locally compact. So, first part is obvious, the second is also obvious, but that will be needed to prove the the part C and this is our aim. Remember in the lemma we had this statement if and only if something happens. So, we will prove that statement to prove part C here that is the whole idea ok. So, let us go through it again since x and y are compact so it is x cross y for any compact space compactly generated topology coincides with the given topology that is part A. B follows from lemma and A. See you want to show that x is compactly generated and y is compact ok. Then what you have to do in the lemma if and only if you take subspace is here k, k compact subset k cross y is compact, is compactly generated, but k is compact, y is compact by A it will be compactly generated. It is true for every compact subset of of of x then this will be compactly generated product will compactly generated A is the lemma. So, so lemma gives you immediately B ok. Now use this one and again the lemma now x and y role will be interchanged to show that x cross y is compactly generated ok. This time by taking subsets arbitrary subsets k of y ok that is what we will see. Every locally compact orders is compactly generated therefore C follows from B and the lemma. So, you interchange the role of where you take the compact subsets that is all. So, so what tricky but the the way the exponential correspondence helps is very elegant. So, we have come to a stage where it now we can study the topology of CW complexes namely when you take the product complexes x and y are CW complexes. There is a canonical way of getting a product on CW structure on x cross y. The only problem is the topology on x cross y namely the product topology will be in general different from this CW topology ok. So, that is the point we want to study we will do it next time. Thank you.