 Welcome to NPTEL NOC, an introductory course on points of topology part 2. We continue our study of compactification as a special topic today, we will study proper maps which are very closely hand in glove with compactification. Phenomenon that we witness in the above example is indeed typical. So, let us introduce a definition. A continuous function f from x to y is called proper if for each compact subset k of y we have f in our case compact ok. For example, quite often in a locally compacted R N and so on if you have a finite to one map not always ok, finite to one map it will be compacted, infinite to one map will not be compacted, will not be proper sorry I am talking about proper map. So, proper maps are a kind of tools to beat the non-compactness, study of functions continuous function from one non-compact space to another non-compact space often ok. So, I repeat the definition definition is just says that inverse image of a compact set is compact. Remember if you have a continuous function image of a compact set is compact comes freely now we want inverse image must be compact. So, you may say ok, you are talking about f being a homomorphism no, no, no f can be you know infinite to one map also no problem, but I want inverse image of a compact set be compact ok. I am not assuming that is one, one inverse function it is inverse function may not sense, but inverse of a set make sense that is all we are using ok. So, here is a theorem why we study the proper maps in the context of one point compactifications. So, immediately there is a result here let x and y be any two locally compact half-dobb spaces let f from x to y be a continuous function. Then the function f star from x star to y star defined by f star of infinity is infinity prime. So, I am using the two infinities here one is infinity another infinity prime ok and f of f star of x is f x for all x into x. So, f star is an extension of f just sends infinity to corresponding infinity that function is continuous if friend only f is a proper map ok. So, this is the motivation of defining the proper map here ok. Remember this what precisely what we did namely phi x which was defined from S n minus 1 minus the north pole into R n minus 1 S n minus 1 is a one point competition. So, on the other side also I take one point compactification then I extended it by taking infinity to infinity is north pole here on in the domain and the other one is infinity I have denoted by star. So, that was the model. So, that became continuous there we could verify it by just purely looking at the inverse image of what happens to the infinity there that is all ok. So, here we have to do this you know there is no geometry here by merely by definitions of this x star and y star this is also not difficult let us go through it. Assume that f star is continuous ok if f if k is a compact subset of y then k is closed inside y star also right because it is a compact subset of y star and y star is off the all. And then f star inverse of k will be a closed subset of x star, but closed subset of x star is again compact ok. So, but you have f inverse of k is nothing but f star inverse of k because k is a subset of y not y star ok. So, it does not have the infinity prime in this way therefore, f inverse of k same thing as f star inverse of k. So, we have proved that inverse image of every compact subset of y is compact inside x under now let us do the converse. If f is a proper map to show that f star is continuous at infinity this infinity goes to infinity prime. So, all that I have to do is take neighborhood of infinity prime show that f star inverse of such neighborhood are open inside x star ok that is what you are not sure ok. So, how are the neighborhood systems at infinity prime described they are nothing but y star minus k where k is a compact subset of y. Any take a compact subset take the take the complement that is an open neighborhood of infinity. So, I have to show that inverse image of this one is open in x star, but just now we have seen that this f star inverse of y star minus k is nothing but you know x minus sorry x star minus f inverse of k. So, it is just f inverse of k ok. So, this is same thing as saying that f inverse of k is closed and compact subset of x because what is this inverse image it is x minus f inverse of k which is a x star minus f inverse of k if you include star of because why star is there right. So, that is the neighborhood by definition inside x star of infinity here ok. Since we have assumed that f inverse of k is compact whenever k is compact this will be neighborhood. So, neighborhoods of of infinity prime inverse image are neighborhoods of of infinity inside x star ok because of the importance of proper maps in several areas of mathematics we shall study them a little more ok. We have introduced the properness via this wonderful property of Alexander's Compatification right, but the proper maps have their own proper life other than just Alexander's Compatification. Let us do a little more of this one indeed proper maps appropriately the concept of proper maps appropriately adopted in in algebraic geometry is very important because there you do not have house darkness at all. So, all these Alexander's Compatification etcetera do not make sense there, but proper maps do make sense. So, let us study them a little more start with any continuous function of topological spaces f from x to y is called universally closed if for every other topological space z no matter what it is f cross identity of z from x cross z to y cross z is a closed mapping see f cross identity is a continuous function it is a closed mapping what closed subsets of x cross z are taken to closed subset of y cross z. So, such a definition is universally closed if you do not take z at all x to y if a closed subset go to closed subset it is just a closed map ok. For all z and for products like this it is closed this universally closed that is the definition by taking z to be singleton space it follows that a universally closed map is a closed map there is no problem it is fairly obvious that the converse will not be true in general ok you look at examples on your own you easily can find however this will immediately follow from what we are going to prove shown therefore you do not have to worry the following lemma is a step in the right direction take any function continuous function from a to b ok where a is compact and b is house door then that function f is universally closed now it must have run a bell in inside you right from part one we have been studying this kind of situation we have proved that in the beginning from a continuous bijection from a compact space to a house door space is a homeomorphism then we also proved that any surjective map from a compact space to a house door space is a quotient map right we have been using this not only we have proved it several times now the secret is open this is precisely what it is it is not just a closed map but it is what it is universally closed from a compact space to a house door space a continuous function ok is universally closed ok so let us prove this what we have to do take any space c ok I am now change the situation instead of x y z and so on a b c deliberately so take any topological space c take a cross c to b cross c f cross identity you must show that this is a closed map ok so go to the graph of this function namely a going to a comma f a this is a graph of f inside where inside a cross b right so we know that this gives a homeomorphism of a with gamma f this map itself is a homeomorphism the image is the graph the domain is a ok moreover b is house door ok by criterion for house door gamma is a closed set of a cross b so this also we have seen part one not very difficult hence gamma f cross c is a closed space you see here there is no map now if this is a closed set crossing with a whole space here another closed set is a closed space of a cross b cross c ok so it follows that if you map a comma c into a comma f a comma c second factor is identity the first factor a comma this one is going to the graph right a comma c point a comma f a comma c this is a homeomorphism because the first factor is homeomorphism second factor is identity of a cross c onto the closed subsets gamma f cross c because a cross c is closed we have just seen that this gamma f is closed gamma f gamma f cross c is closed that is what we have seen in particular it is a closed mapping because a homeomorphism a homeomorphism is continuous it is open it is closed right so it is a closed mapping from where to where from a cross a to gamma f cross c on the other hand since a is compact so far we have not used anything other than house doorness of b ok now we come to use a is compact if a is compact the projection map a comma b comma c to b comma c that is from away from a that is a closed map so this also we have seen a projection map from any x cross y to x where y is compact it is always a closed map now f cross identity is nothing but p composite phi where phi is this map start with a c going to a comma f a comma c then you take the projection to second factor which f a comma c a a goes away a is compact a goes away ok so it is f cross identity identity of c so f cross identity is p composite phi phi is a homeomorphism and p is a closed map so closed composite closed map ok I will request you to pay attention to the technique which goes under this one ok this kind of technique is used in several topological proofs very simple idea but you know passing to products and so on and rise vice versa ok so pay attention to this yeah so now we have come to the properness let f from x to y be a continuous map of locally compact house door spaces both x and y are locally compact also then the following conditions are equivalent f is a proper map the map f extends continuously to the one point compactifications the map f is universally closed the map f is closed and for every point y inside y f inverse y is compact ok f inverse y is compact for each point see if this is a proper map singleton wise are always compact f inverse y will be compact but this is much weaker along with just closeness of the map then it is a proper map so this definition is used quite often ok rather than anything else for proper maps ok so when you don't want to mention compactness at all f inverse of y is put some restriction like you may ask for f inverse of y to the finite house so such functions are studied ok so this property pay attention to this is a wonderful property which you can modify ok so 1 2 3 4 3 different conditions are here they are all equivalent to proper again poofs are not difficult because we have already developed enough techniques here so 1 implies 2 we have already seen right that was our motivation to introduce something more properness was introduced like this so if it only extends continuously by taking infinity to infinity ok so we will have proved that 2 implies 3 let us prove 2 implies 3 from the lemma it follows that f star cross identity from x star cross z to y star cross z is closed ok which is same thing as that x star to y star x star is a compact space y star is a house door space any continuous function is universally closed for every z this is closed this is precisely what I have done that is that was the lemma and now we have concluded this one now let f be a close subset of x cross z ok I have to do something from x cross z to y cross z not x star the f star is fine because we have got a compact so now I have to come back to x now right so let f be a close subset of x cross z if f bar is the closure of f that is the notation inside x star cross z so be sure where you are taking because inside x cross z closure in itself but f bar may not be close subset of x star z so you have to take the closure here right f may not be close so you take the closure of that then f cross identity of f bar is a closed subset of y star cross z so I should put star here f star but it is easily seen that f star cross identity of f bar intersecting with y cross z is nothing but f cross identity the infinity goes to infinity all the time in the first coordinate so if you remove that namely put only y cross z here only points which look like little y cross z not infinity cross singleton so then it is f cross identity of f ok use the factor f star infinity infinity f star of x is f x for x for this part and f is closed in x cross z ok so we have f star of identity f star cross identity of f bar intersection y cross z is this one this thing is closed in y cross z y star cross z so intersection with y cross z is closed inside y cross z because y cross z is a subspace of x star cross z ok so that proves 2 implies 3 namely we have proved that now f p c university clause ok now I have to prove 3 implies 4 of course university clause implies closed so only thing that is left out is inverse image of a single point must be compact this is what we have proved 3 implies 4 alright taking z as singleton z ok it follows that f is closed this is what we have seen earlier now given any y inside y singleton y is closed put k equal to f inverse of y that will be a closed subset of a locally compact half star space right hence it is a locally compact half star space this k itself is a locally we do not know I want to prove that it is compact now we have k cross z to y cross z this is a closed map because this is just like ignoring this singleton y it is just going to the whole k going to singleton y ok so k cross z to y cross z is closed this map can be identified with the projection map k cross z to z since this is true for all z ok from previous theorem it follows that k is compact see this is what we have to use now projection map from everything is compact then this is compact is what we have proved ok Michael's theorem 3 implies 4 is done now 4 implies 1 so what we have we have a functions closed and inverse image of a single point is compact from that we have to deduce that the function is actually proper namely inverse image of any compact set is compact so start with a compact subset of y put l equal to f inverse of k ok we have to show that l is compact so take fb the family of all closed subsets of l having the finite intersection property remember that if we show that intersection of fb is non-empty then it will follow that this l is compact that is that is the thing that we are going to use now we want to show that f has intersection property ip f ip is finite intersection property namely any finite collection of members of f when you take the intersection that is non-empty the entire intersection is non-empty is what we have to show by including intersections of any finite number of members of f to this family we may assume that f is closed under finite intersection ok to begin with any family you can take but then you can expand it to include take f1 f2 take the intersection put that also as a member of f it does not change our problem it does not simplify our problem as such but this assumption helps that is all ok all finite intersections of members of f are again inside f we can assume that ok closed under finite intersection means that now let curly g be all the images of f under little f f f belong to f so these are now inside y then g is a family of closed subsets of k y because f is a closed map ok they are all closed subset of y k because I start everything inside l here when you take f of that is they will go inside k ok subsets of k with finite intersection property because if you take f1 f2 fn here there is a non-empty intersection f of those f of f1 f2 fn they will sit inside ff1 ff2 ff3 so this is now again non-empty intersection so finite intersection property is true for curly g also therefore g has intersection property then intersection of all the members of g is non-empty that why belong into intersection of all ff where f is an f ok then put at z equal to f inverse of this singleton y that is a compact subset of k this is the assumption compact subset of x ok actually subset of k also subset of l because this y will be also inside k right in n say this is a compact subset now consider the family h which is f intersection z where f belongs to f so we have this family f already we are taking each member there and intersecting it with this z you see remember this y was chosen in the image of ff right for all f there is a common element y here ff so f inverse of y will be you know there may be there is there is nothing that is what we have to see now so I am looking at f intersection z where f inside f there is at least one one point here in f which belongs to f of that is equal to y right so look at this one this family is a non-empty closed subset of that each f intersection z will be non-empty f is closed that is closed is non-empty so they are non-empty closed family ok closed under finite intersections because we have assumed that g has that property ok so close sorry f itself has that property right so therefore this is closed in the finite intersection hence it has ip because they are subsets of this compact subset z ok but then the problem is over because this non-empty subset is a subset of all the intersections without taking that into consideration this is a larger intersection so this must be non-empty ok so I told you that you know any arbitrary closed map need not be a university closed set and so on right or you can you can use the fact that a non if you have a non-compact set then projection map will not be closed and so on right so here is a very simple example by the way a few maybe two years back I taught this course to a couple of students and one of them came up with this idea his name is Vidit so I put his name here ok so here is a very simple example take any space x such that the projection map is not closed ok for example you can take x equal to r any non-compact space this will do ok x equal to r if you are not sure of that you know that the projection map from x cross r r cross r to r is not closed ok now take x inside x not inside x one any point the constant map x going to x not is clearly a closed map this we have used earlier right constant map is always a closed map ok now consider f equal to identity cross the constant map from x cross x to x cross x not this is my z whatever you want to say x not so I have taken x cross x to x cross z not say this is y x to y whatever so constant map is there and identity cross constant map I have taken if g from x not cross x not to x is the homeomorphism which ignores this x not ok x comma x not going to x then g composite f which is projection pi one from x to x ok just look at g composite f what happens f is x comma x not x comma y goes to x comma x not right that is f you know then x comma x not goes to x so whole thing x goes to x comma x x comma y goes to x so this is a projection map right the first projection map this is the first projection and that is not closed map ok that is how we started with that one it follows that identity cross c is not a closed map because if f where closed g is closed the composite will have to have been closed the composite is not closed one of them is closed the other one cannot be closed that is all ok so identity cross c is a not closed map where identity of what all that I have taken is that x is a space which is possibly a non-compact space for example x equal to r such that the projection map is not closed that is all ok the one point compactification of locally compact half star space is smaller than all other half star compactifications remember we have put a partial order on compactifications right so some of them may not be comparable at all but here is a one conclusion namely the one point compactification is smaller than all of them in particular each one point compactification is smaller than all other one point compactification but we have we have verified arbitrary one point compactification one is smaller than the other one is the other smaller than the need not imply they are equally that is it is not always possible right so be careful about that but smaller makes sense so so what is the meaning of smaller to see that let eta one x one be any half star compactification e and eta extra the one point compactification consider the homomorphism eta composite eta inverse eta one inverse from eta one x you come to x then for a eta so you have a data x extend this to a mafie from x one to x star both of them are compactifications right so by sending all the points not in eta one x so single point infinity so this is a nice way of extending it right now don't disturb the x part at all so all the points which are outside with x you collapse it to one single point the only thing you have to check is that why this is continuous okay we need to check only the continuity of it at points other than eta one x in the compactification right once you do that you have got an equality this just means that x one is bigger than equal to extra that's all okay so let us see why this map is continuous if you use an open neighborhood of infinity next hour then by the definition f which is extra minus you must be closed and compact subset of eta x that is the definition of the extra one the Alexander of Compactification hence eta one composite eta inverse of f is a compact subset of eta one of x okay you take this this is a homeomorphism okay so f is compact so this is a compact subset of eta one x since x one is half star this will be a close in x one also okay any compact subset of a half star is closed opposite that without half star of x one we would not be able to conclude this we have compact but why should be close so half star we have used that it is closed now hence the f inverse of u f is remember it extends the identity map on the entire of x star by pushing all the extra elements to a single point infinity the f inverse of u is nothing but x one minus eta one composite eta inverse of f that is an open subset because this is a close subset so we have proved that the the map phi is continuous at all the points away from x star they are all going to the point at infinity in the second in the in x star so this is this is this completes a proof okay yeah so so one of the fallouts of this study is the universally closed concept of universally closed functions and proper maps okay there are many other applications of one point compatibility themselves but let us go to other things now so today we will stop here next time we will study another compatibility stone check compactification thank you