 So, welcome to all of you. As promised today, we will take up one topic in point set topology and then we will take one more topic which we will discuss very thoroughly, point set topology because these are very much used in algebraic topology. The first one is our function spaces. Recall that when you have a homotopy from some x cross i to y, then you can think of just a family of functions from x to y indexed by the interval i. Each function is definitely continuous. Therefore, it belongs to the set of continuous functions from x to y. That can be thought of as a subspace of the product of y taken x times and then you have the product topology. But the product topology is not very good because it corresponds to point wise convergence. What we need is some kind of uniform convergence and the correct thing one has to cook up and that is why different topologies on these function spaces is needed depending upon your requirement. That is why the study of function spaces. The one which is most useful in algebraic topology is called compact open topology. So, today I would like to introduce you to that aspect. Later on, we will talk about quotient spaces also. Given two topological spaces, let us denote by y power x, the set of all continuous functions from x to y. In point set topology, this notation is used to set of all functions from x to y. Namely, the product of y taken x times, x many copies of y. Since we are not interested in all functions, we are only interested in continuous functions because x is not an arbitrary indexing set here, but a topological space itself and we are interested in only continuous functions. We will use this notation, the simplest notation for most use sort of concept. So, there should not be any confusion with this one. So, y power x will denote all continuous functions from x to y for us. So, now topology on this one is called compact open topology. So, the topology is described by describing a sub-base and sub-base consists of these like inner product notation, bracket KU. All those functions which are continuous from x to y such that they take the compact set KF of K inside U. So, this should remind you that we are more or less using some kind of continuity and epsilon delta condition. U playing the role of epsilon and K playing the role of this delta neighborhood or whatever, but there is no point reference. This should be for all. So, that is why this will become uniform continuity. This with respect to this kind of things, this condition automatically. Uniform continuity on compact substance. So, this is the catch here. F of K is contained inside U captures the uniform continuity aspect. So, these K comma U will form sub-base just means that I have to take finitely many such sets, intersect them and look at all such sets. They will form a base. Thus, the topology on this y power x is now such that any open set is arbitrary union of finite intersections of members K1, U1, K2, U2 and KN, UN. Ki's are compact and Ui's are open. So, this is the topology. So, just for a while, whenever it is unclear what topology we are using, we will use this notation compact open or compact open topology. Once you are familiar with this, we will not even mention that. Like we are all the time writing just R with its usual topology. So, let us look at the first position here. Its y power x with compact open topology is finer than the product topology. Therefore, it is Hausdorff respectively. It is regular. If y is Hausdorff, y power x is Hausdorff. If anything which is finer than the product topology, then that will be also Hausdorff. When I say just y power x, it is the product topology, right? y power x is finer, I am claiming. So, it will be Hausdorff. Similarly, product of regular space is regular. Then you take the finer topology, CO, it will be also regular. So, I have to see why this is finer than this one. To see that, you see the single point is always compact. Therefore, in this collection here, single point comma u, these are already there. And what are those? They are the sub-basic open sets for the product topology. Therefore, this collection has already more elements in it than the standard sub-based topology of space for the product topology. Oh, singleton x is compact. That is all I am using. So, the first position is very nice. So, you also know that there are more open sets in CO than in product topology. Therefore, convergence here will be stronger. Point wise convergence may not be a sequence which is point wise convergence function, may not be uniformly convergent. You know that. Whereas, if you have this uniformly convergent, then it will be point wise convergence also. So, if some sequence of functions is convergent in this topology, in the ordinary product topology, also it will be convergent because this is finer than the product topology. The central result that we want is summed up in one single theorem, three parts of that. The second part is the most important one for us which is of immediate use. So, let us be done with it and that is all about function spaces that we are going to discuss. So, this is called exponential correspondence. So, here is an important extra hypothesis here, restriction here namely x is locally compact hostile. y and z could be any topology spaces. For simplicity, you can assume y and z are also locally compact if you do not remember which one to be locally compact. But assuming y and z are locally compact does not help you at all, x must be locally compact and hostile. So, what is it involved here is evaluation map e y power x cross x to y, what is this evaluation? Evaluation of f at x, f is a function from x to y, the second slot is x, evaluate it at f of x. So, this is what evaluation map, this will be automatically a continuous function. In fact, this is continuous even if you take even if you take here the product topology, this should be nothing but the projection map to the one of the factors. This x is an element of x, so it is x projection of this map. Since I have put a stronger topology here, automatically this should be also continuous. So, this is the way you have to see, but you can directly prove that also. So, evaluation map is continuous. Now, look at the function g from y to z to y power x, some function is continuous if and only if you take g cross identity from z cross x to y cross x y power x cross x. So, z cross x you land here with g cross identity and then follow it by identity, follow it by evaluation map. If that composite is continuous, g will be continuous. If g is continuous, the product is continuous, composite is continuous is easy. So, what is important here is if the composite is continuous, then g is continuous. So, this is the key thing which allowed the classical approach of saying parameterized family and so on. This was when formalized here. So, let us see what is the proof of this one. Before that, let us look at the third condition here. This is actually tells you the law of exponentials is valid inside these continuous functions just like the it is valid for sets. You must be knowing that cardinality of y power x power z is the same thing as cardinality of y power cardinality of z into cardinality of x product. This is same thing as a raise to b raise to c is a raise to b c for numbers. Same thing is true for cardinality of when you take these things are all functions, but this is also true now for continuous functions is the beauty here. Well, namely if z is also locally compact hostile, so x i position then the function psi from y power x raise to z. What is this one? This is set of all continuous functions from z to y power x which occurs in statement b here, b here. That corresponds to under b, it corresponds to a function from z cross x to y. If that is continuous, it will be continuous and vice versa under these exponential components psi of j is e composite, g composite, g cross identity is just like here. So, c is actually a consequence of b, but we will have to work out. What happens if this bijection is fine, but this is actually homeomorphism is what we have. So, let us go through this proof one by one. So, I am writing a full detail of why the function evaluation map is continuous. I have given you the idea why it is continuous. Let us write down some full detail here. Given an open set u inside y, we have to show that e inverse of u is open in y power x cross x. So, take a point f naught x naught belonging to e inverse of u. I must produce a basic open set which contains a point f naught x naught and contained inside e power u. Then I am done. What is the meaning of this belongs to e inverse of u? It implies that f naught is first of all continuous function because it is inside y power x. And f naught of x naught can evaluate it is inside u, e of that is inside u. Now, you see now because we have assumed that x is locally compact of x. There exists a compact neighborhood k of x naught such that f naught of k is contained inside u. This means that f naught is inside k u. Since k u is an open subset of y power x, we get a neighborhood k u cross k of the point f naught x naught and clearly e of k comma u cross k is contained inside u. So, k u cross k, k is a neighborhood compact neighborhood. e of I have formed a neighborhood here and a neighborhood here for x naught is contained inside u. So, you have to find a x naught f naught, you must find a neighborhood cross a neighborhood here. That is the product topology from y power x cross x. That is why. So, this is clear cut proof, no hand waving. Now, that is about to be. From a, we need to prove only one part because we have already proved e is continuous, e is continuous. If g is continuous, g cross identity will be continuous, composing with e will be continuous. So, only the converse part we have to prove. Suppose this is continuous, then we have to show that g is continuous. For this, it is enough to show that g inverse of a basic, sub basic open set is open whenever what is a basic open, k is compact and u is open. By sure inverse image of every sub basic open set is open then the function is continuous. So, how to show g inverse of k is continuous. So, take a point here, say z naught belonging to z. Let us say g of z naught is inside k u, that is the meaning of g inverse of k u. z naught is in g inverse of k u means g z naught is in k u. Then for every point inside k, what we get is e cross g cross, e composite g cross identity of z naught comma k by definition of what? g z naught comma k, e of that which is g z naught operating upon k. Remember, g z naught, g is a map from where x to y, right. So, g z naught, sorry, g is a map from z to y raise to x. So, g z naught itself is a map from x to y. So, g z naught of k belongs to this open set. So, there exist open sets vk, wk of z and x because this is a product, right. So, a neighborhood here and neighborhood here respectively such that z naught is in vk, k is in vk, wk and e of this product is contained inside u. If a point goes inside this one, the composite is continuous. So, entire function operating on a neighborhood contains this continuity of e composite g cross identity. Now, we use the fact that k is compact. We can pass on to a finite cover k contained is a i range of 1 to n vk i. What I have done? For each point k inside k, we have formed a wk. So, all these wks is a cover for k. Therefore, I say there is a finite cover wk i. k is contained inside wk i. So, let us call this union as w, okay. On the other hand, let us take the neighborhood for this z naught to be intersection of vk i. Then e of g cross identity of this intersection here and union there that will be in consent inside it, okay. Because once it is intersection, this will be true for every element no matter where you take the second one to be in one of them. So, it is contained inside, okay. This implies g of v is contained inside k u because this w is inside k. The whole w goes inside u, okay. So, g of v is contained inside k u, okay. What do you mean by k u? Every element of g v takes k inside u. But they are taking whole w inside u. Therefore, they are taking k also inside. Since v is a neighborhood of z naught, we have done because we have formed a neighborhood here. g of that is contained inside k. g started with g of z naught contained inside k u and formed a neighborhood now. So, this shows that g is spontaneous. Is that clear? Yes, sir. The compactness, local compactness of x is used here. Now, you see in c, in statement c, there is local compactness assumption on z also. You know why? Because in this statement z is also going up exponential. It is also superscript. So, it is easy to remember whenever the space goes up here on the superscript, those things must be locally compact. The things at the bottom could be anything. This is y power x. So, if something should work properly, x must be locally compact. So, thing is raised to z. So, z must be locally compact. So, z cross x, both of them are locally compact. So, product is also locally compact. So, that is why z is also locally compact necessary. Why could still be anything? So, I am telling you a trick to remember. Things which go up must be locally compact. All right? So, let us prove c. The idea of giving all these proofs is so that you will get familiar with compact open topology. Otherwise, one can just take the statement and go away. Finally, we are going to use only the statements. From b, it follows that the map whatever you are writing here, psi, its map, how we define psi of g is e composite g cross identity. Whenever g is continuous, you have defined this continuous. So, you do not need the b. But even the inverse is continuous whenever this is continuous, this is continuous, that will need the part b. It is well defined and its inverse is also well defined. That is what it is. It is well defined and it is a bijection. So, for that bijective part, you need part b there. Applying b with y power x raised to z in place of z and z cross x in place of, this is in its not is, in place of x, continuity of psi is the same as continuity of, see I am applying b here, e composite psi cross instead of identity of x and identity z cross x. So, y raised to x cross z to cross, z cross x I take to y. If I show that this is continuous, then it will imply that psi is continuous. So, this is where I am using b very strongly. What is this map? You see some lambda here from z to y power x. Z is a point of z, x is a point of x. So, lambda z and x, you take lambda z, it is a function from x to y, you can evaluate it on x. This is lambda z x. By b, it will follow that, you know, if this whole thing is continuous, then this is continuous. But why this is continuous? Continuity of this one is obvious because it is again from b. So, this latter map is nothing but, you can think of this as y power x power z cross z bracket, I am putting the bracket other way around cross x. So, from here y, where are you coming, y power x cross x, this is continuous. From here to here again I will evaluate, it is continuous. So, this is given by lambda z x, lambda z you are evaluating cross x. So, it is like evaluation carried twice. That is why again from b, again from a it is continuous. From b this will be continuous. So, that proves that the evaluation map is, you know, evaluation map is continuous. That was psi is continuous. The proof of psi inverse is continuous exactly similar to actual tautology. Everything is if and only if we can go ahead. So, I will leave it to you as an exercise. You have to spend some time to get familiar with what is going on, playing with these exponentials and so on. So, you can check this that psi inverse is also continuous. What that in b? We do not need the local compactness effects to prove that the map g from z to n is written from continuity of the corresponding map. The local compactness effects is needed in the other way implications because it is needed in part a. One important special case of statement b is the homotopy version which we will be using all the time. This is the special case when x itself is i, z cross i to y, in cutoff z cross x to y. z cross i to y is a homotopy. This is continuous if and only if it corresponds to the function a little h. I have put little h here. Remember I have put little h t. Little h t is where all functions from z to y. So, now that family is whole family is a function from i to y. So, h itself is a function from z to y raised to y. h t of z is capital Z t. So, h is a function. h of z is a function from i to y operating upon t is h z. So, that is a meaning of it. This is a very special case and this goes in the in the suffix up here. So, this must be locally compact. But you have actually it is compact hostile. So, of course, it is locally compact hostile. Therefore, the joint continuity of this one ensures you that a family like this is a just one single function continuous function from z into the function space y raised to y. Set of all map all path is from path is in y with compact open topology. I think we will stop here because the next thing is a different topic namely quotient spaces which we will discuss in the next module.