 Hello, welcome to module 52 of NPTEL NOC, an introductory course on point set of all V power 2. So we shall continue our study of function spaces, today the topic is exponential correspondence. So let me start with theorem because I have already introduced all the relevant notation, I hope you have done your homework for getting familiar with this notation. Start with a locally compact half star space, y and z are any two topological spaces. The evaluation map e from x cross c x y to y given by e of except the code f x is continuous, this is the first statement. A function g from any topological space that is some orbital x to z, z to c x y is continuous if and only if the composite e composite identity of x cross g from x cross z to y is continuous. So this is your criterion for determining functions into c x y when they are continuous. If z is also locally compact and half star, okay, locally compact half star is always a standing assumption on x. Now if z is also locally compact and half star, then the function c of z to c of x y to c of x cross that y, the exponential correspondence, the function that we defined last time, the exponential correspondence map psi, I am repeating it namely psi of j is defined as e composite identity of x cross j. This is a homeomorphism. Notice that by this statement b given any g here, psi g defined by this one is continuous, okay, if and only if you do not need, right. So the psi though it takes values actually in y power x cross z, it is actually inside this smaller sub pairs. So that is why I can write like this. The statement c is much more stronger, it just says that this is actually homeomorphism. So let us go through the proofs of these things carefully one by one. One point which you can remember is the following and how to remember. Whenever you take continuous functions from the domain must be locally compact half star or locally compact regular some locally compactness has to be there. In the right in the beginning we took functions from x to y, right. Therefore, x was locally compact all the there was no need to assume anything on the domain on the core domain. Similarly, here there is no assumption on that the function is taking values inside c x y both c before take c x y I would like to ensure that x is locally compact half star, okay. So that is the case when you come here because now you have to take c x y and then z to c x y. So I am assuming z is also locally compact half star. Now x and z are locally compact the product will be locally compact half star. So there is no extra assumption on this. So still y is a free topological space no condition on y, okay. So it is easy to remember where to put the hypothesis, okay. So first a I want to prove that exponential map is the evaluation map is continuous. Given an open set u in y you must show that e inverse u is open in x cross c x y. So take a point here x naught f naught belonging to inverse u. What the meaning of it belongs to inverse u? This implies that f naught from x to y continue the function and f naught of x naught is inside u, right. e of x naught f naught belongs to u the f naught x naught is inside u. Now use local compactness of x you can find a neighborhood compact neighborhood of k just means that an open subset that is closure is k and that k is compact such that this f naught of k is contained inside u, okay. Continuity will give you some neighborhood but inside that neighborhood we can take a compact neighborhood because locally compactness. So both of them are combined here. This means that once you have this condition this condition means that f naught is in this bracket k u. Since bracket k u is an open subset of c x y we get a neighborhood k cross k u this is the product of all of you by the way k cross k is a neighborhood of x naught k u is a neighborhood of f naught. So this product is a neighborhood of x naught f naught. Now clearly if you take e of this e of this by the very definition of k u, right take any x comma some f here f of x will be inside u. So e of that is inside you that is the mean, okay. So we have found out a neighborhood of the point which is completely taken inside you by that is the meaning of that e is continuous. Now let us prove b. So now we start with an arbitrary space z and a continuous function identity cross g is continuous e is continuous because we have proved it in a so composite is continuous, okay. So one way is obvious. Now what we want to do is assume that this composite is continuous then you have to prove that this g is continuous. See composite of two functions can be continuous without either of them being continuous you must know such examples. If both f and g are continuous then the composite is continuous that is the standard result that we have proved but it is possible to have neither of them are continuous but the composite is continuous, okay. So here we have to prove this non-trivial result. Assuming this composite is continuous you have to prove that g is continuous, okay. For this it is enough to show that g inverse of these basic open sets are open. So this is standard method we have been following. Instead of proving that every inverse image of every open set is open you can just take basic or sub basic open sets which are sub basic open sets here show that g inverse of that is continuous. So that will show that g is continuous. What are sub basic open sets here? k compact u open then you take bracket k u, okay. So start with a point z0 belonging to that such that g of z0 is in k u. So g of z0 is a function remember that okay because g is a function from z to continuous function from x to y. Then for every point k in k we have look at this composition e composite identity of cross g operating upon k, z0 is nothing but by definition gz0 the first coordinate remains k and then evaluation. Evaluation of gz0 on k will be this far, okay. So gz0 belongs to k u, k is inside k so that is inside u here. So by continuity of e composite identity of x cross g this is continuous is the hypothesis. There exist neighbourhoods wk cross vk inside x cross z of this pair a k, z0 such that identity composite e composite identity of x cross g operating on this product neighbourhood is goes inside u that is what we want that is inside u. Since k is compact we can pass on to a finite cover k contained inside i into 1 to n wk i I will call it as w what we have done for each k we have got such a thing right there is a neighbourhood wk. So these wk's cover the whole k and k is compact therefore I have got a finite cover here which I am denoting by wk1, wk2, wkn. On the other hand I take v to be intersection of this vk1 vk2 vkn. Then all that I need is w cross v for this union and here intersection that will be contained inside wk i cross vk i taken union i in 1 to 1 to n. Any point here will be in one of them k is the same indexing on the other side. So second coordinate is in all of them so it is inside that one so this is contained in this union. Therefore if you apply e composite identity of x cross g on this one that will be inside u now because g identity of each of these is contained inside u. So the union will be also contained which implies that g of v belongs to w, u this bracket here that is the meaning of this one. But w bracket u is contained inside k bracket u because k is contained inside u w is larger okay. Anything which brings the whole entire of every inside u must be also bringing k inside u. Since v is a neighborhood of z0 we are done okay we have found out a neighborhood v of z0 so the g of v is inside this basic sub basic open set that is what we have started with okay. So we have to go through these steps to prove the converse part here namely just assuming that this composite is identity so sorry this composite is continuous we proved that g is continuous so that is part b. Now we will do the last part see namely exponential psi is a homeomorphism as pointed out before first of all because of b psi is well defined I repeat this is what I told you right in the statement here because of b this makes sense that this is a function from here to here otherwise I have no I am not have liberty at defining psi, psi has to be defined like this because g is continuous this right hand side is continuous we are here so that is the first remark again this is repeated here that is all okay. So psi is well defined let now I will have two more exponential maps here so for evaluation maps here remember this capital E of from x cross c x y to y similarly I have this z cross c z c x y to c x y and e 2 is from x cross z cross c of x cross z y to y so domains are repeated here okay this is domain that is the core domain the last thing will be core domain that is the pattern for evaluation maps everywhere so there are two such evaluation maps okay because we are assuming that z is also compact locally compact and half dot okay the statement a is valid for both of them that means that these evaluation maps are continuous okay both of them are continuous. Now again from b so that is from a b with c of z c x y in place of z the b gives you criterion for continuity into a function space so now you apply to instead of z you take this one okay in place of z and x cross z in place of x you look at continuity of psi okay what is the domain of psi it is c of z comma c of x y what is the core domain it is c of x cross z comma y right so that is why I am taking this in x cross z in place of z continuity of psi is the same thing as continuity of this e 2 composite identity curve psi identity where identity of f cross z now if I prove this composite is continuous psi will be continuous so I find is directly proving psi is also possible okay that will also have cumbersome notation so I find this way it is easier to state it once you have done the groundwork here namely b you can keep using b again and again so I will show e 2 composite this identity of x cross z psi this is continuous the domain is x cross z cross c of z comma x cross z to y okay from x cross z identity map x cross z psi is from c of z to c of sorry psi is a map from where psi is a map from c of c of c of z to c x y to c of x cross z to y so that is how I have taken this okay this is c of z cross x cross z to y so given by here is x here is z here is lambda okay so that will give you lambda z of x so this later map is the composite of the two maps identity cross even from x cross z cross c of z comma c x y to c of x cross y for c this x remains as it is this this is one evaluation here z part disappear okay x part has remained so c x y we have come then you apply again evaluation map here this ordinary evaluation map namely e x cross c x y to y that is what we have started okay so that is the same thing as this evaluation map this composite of these two each of them is composite is what we have just seen this is even and that is e so these two are continuous therefore this is continuous so that establishes that psi is continuous exactly similarly you can show that its inverse namely phi is continuous in fact we want to show that p is a homomorphism same thing as psi is a homomorphism note that as a set function we have seen that this psi is a bijection and its inverse is phi therefore we have no other choice but to show that once a function is bijection it has a unique inverse no matter which smaller subset you take on each smaller subset the inverse will be corresponding restrictions that is all so I have to take the same phi show that this is continuous now then it will prove automatically that both p and psi are homomorphism okay so let us prove that p is also continuous so proof is even simpler here but anyway you have to go through these steps namely given any continuous function f from x cross z to y we know that for each z in z the partial function f z namely you are fixing one coordinate z and taking x go into f of x z those things are continuous that you know already joint continuity implies partial continuity therefore we get a function this function f hat for each z you get f z so I am denoting by f hat so given by that going to f z this function itself is continuous y by b now we apply b okay so you have to see z to c x y continuous you have to take x cross z to you know that evaluation map you composite with identity cross this one and we see okay so if you apply that it will follow that f f hat is continuous because finally what you get in f of x z okay when you evaluate f f hat okay this means that first of all that under this phi c of x cross z y goes inside c of z comma c x y because it goes to f hat and f hat is a continuous function from z to c x y okay so this is just the justification that phi has a correct domain and codomain now we have to show that phi is continuous that part is still there okay from b the continuity of e the same thing as continuity of even composite i z cross phi okay so what is that function that is z comma f going to f z again by b this latter function is continuous because when you evaluate it x further x f comma z comma f which is just f of x z and this is continuous okay because f of x z is continuous this is an evaluation map from x cross z to whatever c of x cross z y okay so this is continuous so this continuity implies this is continuous this continuity implies phi is continuous alright so we have established the card you know cardinality in a very strong sense homeomorphism homeomorphism types of you can just write instead of continuous function y power x power z is the same thing as y power x cross z okay you can easy to remember this one but everything continues and the whatever goes in the exponents those spaces must be locally compact so remember this much okay here is a remark note that in b we do not need the local compactness of x to prove the continuity of g to g from z to c x y okay this is needed okay in the in the proof of a okay the local compactness of x is needed in the other way around implication only because it is needed in a if you want to apply other way around then you have to use a also the proof of a is the same if we assume x is locally compact and regular which is slightly more general than assuming locally compact host or because we have seen that locally compact and host of implies regularity but the other way may not be so we can also do with locally compact regularity all that I have used here is that points of x have arbitrary small neighborhoods which are what compact neighborhoods this the set of compact neighborhoods of a point forming a fundamental system at each point so now we shall do one more one more justification for introducing this compact open topology all right let yd be a metric space we say a sequence fn from x to y is compactly convergent or uniformly convergent on each compact subsets of x so this is a longer wording so people call compact convergent that is no other no other justification that is all okay this is a neat wording this is very long so compactly convergent to a function f what happens I mean when you said happens namely if for every epsilon positive and a compact subset k there exist n0 such that distance between fnk and fk is less than epsilon for every k inside k and n bigger than n0 if this happens for each k and then this n depends upon k that will be just point wise convergence so the same n works for all the points of a k that is uniformly on k a k is compact that is why it is compactly convergent is the name okay so this is not my definition this is a standard definition I am just recalling it now we come to something let y be a metric space x be locally compact or regular locally compact hostile or regular whatever one compact hostile or regular one of them a sequence fn from x to y of continuous function is compactly convergent to a continuous function f from x to y if and only if as a sequence in c x y c o that with the compact of puttapology it converges to f here the hypothesis is sequence is continuous sequence of continuous function and the limit is also continuous okay then compactly convergence implies the usual convergence in this topology that is what final conclusion here we are not proving starting with a sequence of function we are not proving that f is continuous if indeed that is true we know already in the case of metric spaces okay if it is compactly convergent sequence of continuous function then f is continuous but we are not proving that statement for compact open deposit there we are assuming this one the statement is that namely fn converges to f fn is continuous f is continuous the convergence can be in the general sense of in metric space here you can term it in terms of compact open topology that is the whole idea okay statement is clear I hope now let us work it out it takes a little bit of time but this is routine there is absolutely no new ideas here all these things are standard methods in analysis suppose fn is compactly convergent to f let f belong to w where w is an open subset in the compact open topology we have to show that there is some n0 such that fn belongs to w for all n bigger than n0 so this will prove that fn converges to f in the compact open topology okay so one way implication will be done so let us do this one let ki ui i rank 1 to k be such that ki's are compact and ui's are open and fp is in the intersection of ki ui contained inside w this is because these things make a sub base so whenever you have an open subset and a point you have a neighborhood coming from a you know base this is intersection of finitely many elements of sub base okay so there is such a set ki ui this is bracket ki ui intersection now for any a any subset of this metric space y and r positive we have the standard notation long back we have used this one b r of a is set of all y inside y such that the distance between y and a is less than r it is like if this a is singleton then this is open ball otherwise this is in a r neighborhood of this a distance between y and r a is r it is less than r now if fp is continuous function and ki's are compact f of ki's are compact fp is here so f of ki's are inside ui right where each ui is open therefore you can find an epsilon i positive such that this open neighborhood of epsilon neighborhood of f ki itself is contained inside all that you have to do is take the distance from distance between f ki which is compact and complement of ui which is close that's a positive number to take any epsilon i smaller than that this will do so if you have done it for all the i i read 1 to k you can take this epsilon to the minimum of them once you have an epsilon you get an integer n i such that for all n bigger than n i we have distance between fn x and fx is less than epsilon for every x inside ki this is the compact convergence of f the sequence fn to f okay for each ki which is compact I will get an n i here but what is the meaning of this one this means that you see fx x is inside ki so fx will be inside f of ki and epsilon is same epsilon so if you take the epsilon neighborhood of this f of ki this fn of the entire ki all the points must be inside here okay fn of x must be inside here that is true for all ki so fn of ki must be here okay this happens for all the i i equal to 1 2 3 up to k now if you take n0 maxima so I have I have this this n is bigger than n i corresponding here but now if I take n0 to a maximum of n1 into nk and n is bigger than n0 then all of them will be simultaneously true right therefore fn of ki will be contained inside ui for all of them which just means that fn is intersection fn is in the intersection of this bracket ki ui but that is inside w okay so one way we have proved we started with f such a name all the fn's n greater than some n0 they are also inside that is what we have proved okay the converse converse is more or similar but slightly different now you will see suppose fn converges to f in cxy given k contained inside x compact and epsilon positive by continuity of f and local compactness of x I have to use this one somehow for each x belonging to k choose neighborhood vx of x the state vx bar is compact and f of vx bar is contained inside some neighborhood of fx I will use epsilon by 3 so that is epsilon by 3 is just a afterthought you know this kind of thing I could have chosen some epsilon finally I will get 3 epsilon otherwise I that is why I just start in the beginning epsilon by 3 that is all okay so again local compactness and continuity of f is used so far fn's sequence hasn't entered here choose finitely many points x1, x2, xr belong to k such that k is contained is a i range 1 to r this vx size for each x inside k I have got vx and vx bar has this property they cover the compact set k so I get a finite cover okay this is again routine thing now fn converges to f in cxy implies that for each of these i there will be ni such that fn will be inside this open subsets this brackets vxi, b epsilon f of xi okay vxi bar comma b epsilon 3 fxi bracket this is an open subset so fn should be inside n bigger than n9 okay because f is there so after certain stage fn must be there so this happens for n bigger than n9 again take n0 to be maximum of n1, nr then if n is bigger than n0 and x is inside k I am going to verify this uniform convergence part dfn of x and fff of x is less than or equal to fn of x and go to fn of xi keep n the same change x to xi plus now fn of xi to fxi plus fxi to fxi so there are three quantities here if we choose i such that see x is somewhere okay x is inside k so it is inside one of these vxi choose particular vxi say x belong to vxi then each of these quantities on the right hand side will be less than fxi by 3 because I have chosen i here okay if they happens to be j you have to put everywhere j that's all so they are epsilon by 3 so hence we are through some totally in this distance is less than epsilon so I made a remark saying that if you have a sequence which is convergent you know in cx a sequence of this one which is suppose compactly convergent okay so that will be function some function would that be already continuous see I said I am not proving or I am not addressing this problem in this theorem but that problem binds me but we can you know it can bother me right as a function it is some function from x to y so that is the same thing as saying that under the standard topology is this one a closed subset of y power x okay this is the question so y power x also I am giving you compact open topology only not the product topology would it be closed that is a question compact open topology on this one compact open topology but is it a closed subset this is a question okay a completely satisfactory answer will take us to yet another concept which we have not introduced here that concept has a name uniformity okay so which we have not introduced so far so we shall skip it there is no time for doing uniformity if you are interested then you can look into many books special book for me and mine book is Kelly's book okay thank you so next time we shall use this compact open topology to do some interesting application okay thank you