 Welcome to module 35, Walman Compactification Continued, this time we will study universal property of Walman Compactification. So, here is a theorem, the statement is quite elaborate here, start with a compact half door space and a continuous function from x to y, then given an ultra closed filter f 1 x, the image filter f check of f is convergent to a unique point f hat of f in y, it is convergent in y and the point convergence point is unique, that unique point I am denoting it by f hat of f, that is the meaning of convergent to a new point here. After all it depends upon both the filter as well as the function f, association f going to this f hat of f which you have just defined because of the uniqueness of the convergence of this f check of f, this is a continuous function from the space of all ultra closed filters right, this is Walman Compactification to the given compact half door space. And it has the property that f hat composed phi is identity is identically f, the original function f. So, think of phi as an inclusion map of x inside w hat, this f hat is nothing but the extension of f, that is the way I would like to think of f. The third condition tells you the uniqueness of f hat itself, namely any continuous function g from w hat to y such that g composite phi is f is equal to f hat. So, this is the universal property of the compactification that we are discussing now. So, let us go through the proof of this one. So, this is the picture that I have in mind of what I am going to do. Starting with a continuous function of course, we have to assume y is compact half door. I want to get a function here f hat and I already described how this function is got. Take an element here that means an ultra closed filter, the image filter under f that is a filter on y, it may not be ultra closed or anything but it is what? It is convergent because y is compact and that convergence is unique because y is half door valve. So, that is the function f hat, we have to show that this f hat is continuous and this diagram is commutative. Such a f hat is unique is the last part, the uniqueness here indicated by a sheet. Now, suppose f check of f is not convergent in y. What does this mean? It means that for every y inside y, there exists a open able v y of y in y, it does not belong to f check of f. So, this is what we have seen earlier. Any filter does not converge means you have neighborhood like this. It does not converge for every point, you must have a neighborhood which is not in this. This is what we have used earlier also. But this is the meaning, what is the meaning of the f of, f inverse of u i, u i is a neighborhood of y inside capital Y. So, f inverse of u i that is in open subset of x now, f of that is contained inside u i. So, u i is also not inside f of f. If this u i, neighborhood u i does not belong to f check of f, means even this is not inside. Because if this set is there, then u i would have it, because it is a super set. But this means that f inverse of u i is not in f. If f inverse of u i is not in f, then image of f would have been in f check of f. If it is in f check of f, then u i would be in f check of f. So, that is the chain here. So, what we have got is that an open subset f inverse of u i of x is not in f. Since y is compared, we can choose finitely many y i's such that y is covered by this finitely many open subset u y i's. For each y you are getting this one now. But then what happens? If you take inverse images of this, they will cover x. So, x will be the union of iron 1, 2 and f inverse of u i's. And this x is inside f. Again we are using the last thing that we have proved last time and we used it. Now, we are using it again here. So, what does this mean? This means that one of them is inside f and that is a contradiction. So, we conclude that f check of f is convergent. The point is we are not proving f check is an ultra closed filter. If that is the case, then we do not need to repeat this proof. And it is not true also. It is not true that even it is a closed filter in general. Nothing is needed. We go back to x and do the work there and we conclude that the image check image filter must be convergent and the convergence is uniqueness follow because y is also that is an extra assumption. This proves one. So, the function is there now. Automatically what is this function? If you start with x here, then go to phi x. What is phi x? Phi x is the atomic filter f. What is f check of phi x? It is f of fx. It is atomic filter generated by the singleton fx. And that filter will not convert to any other point than fx. Therefore, f fat of f must be equal to fx. f fat of phi x must be equal to fx. So, if you come this way and go by fx, what you get is f is the conclusion. So, that is what I have done here now. I will repeat it. Note that if f is fx for some x which is the meaning of that this f is phi of x that is the definition. Since fx converges to x, it follows that f check of fx converges to fx. This just means that f fat of phi x is fx. I saw it in a different way because f check of fx is nothing but f of fx. The atomic filter generated by f. We need to show that f fat from W x to y is continuous with the topology here being the Walman compactification topology whatever you are defined. So, start with a u and neighborhood of y in f check of f. f hat of f. f hat of f is some point inside y for some f inside W f. So, I must produce a neighborhood such that that neighborhood goes inside u. Choose an open set v such that y belongs to v contained inside v bar contained inside u. I start with an open subset u and which is a neighborhood of y which is y I am assuming is f fat of f. So, how can you get this one? This is because y is compact, host of means it is also regular. Compact host of space is a t 4 space. So, in particular it is regular also. And hence what we get is a neighborhood v y belonging to v contained in v bar contained inside u. Now, we claim that this f is inside f inverse of v plus f inverse of v is open inside y f inverse of v plus with an open neighbor open subset in the Walman compactification. I want to say that this curly f that we started with this filter is here moreover under f hat this f inverse of v plus goes inside u. So, once I state it is very clear that this happened, but let us verify this one. Since y is f hat of f that means what it is the limit of f check of f and that limit is inside v. We have this neighborhood v is inside f check of v, n x the whole of n y is inside f check of f. So, in particular v is here. Hence there exists some a belonging to f such that this f a must be contained inside v, v is a superset of f a that is the way f check is defined. You take all image of members of f by little f and then take supersets that is the way f check is defined. So, there is some member here a inside curly f f of a contains is contained in v. So, that is the meaning of that v belongs to f check of f, but this means a is contained in f inverse of v and this a is inside f that means f inverse of v must be inside f because f is a filter. That means this filter is in f inverse of v plus one part is over. Now, I have to show that this f inverse of v plus is mapped inside u by f hat. What does that mean? Take any ultra closed filter belonging to this set f inverse of v plus. Look at its unique point to which it converges that convergent point must be inside u that is the meaning of f hat. So, start with f prime inside f inverse of v plus which is just the meaning f inverse of v belongs to f prime. What is this mean? f of this one will be inside f of f prime. f of f prime is contained in the f check of f prime. This is a set here, but that is contained in each member here. So, this is set of set of set. Each member here is contained inside this one belongs to f check of f check that is another family. So, in particular this f of f inverse of v is a member of f check of f prime. Since f of f inverse of v is contained in v contained inside v bar it is a closed set. So, v bar must be inside this one. From the remark 7.43 which we have used several times it follows that the limit f hat of f prime the limit of this one f check of f prime that must be inside v bar. And that v bar is inside u. In fact, it is inside v also. For every v if this one is there the limit point must be inside that one. So, v bar inside u this completes the proof of the continuity of f hat and hence that of the proof that of statement 2. statement 2 is wanted to show that the association defines a continuous function and the commutativity is already seen very easily. Now, let us prove 3. The 3 is a consequence of fact that phi x is dense in w x and y is half-dolph. Remember that if you have half-dolph space suppose you have two continuous functions from one space to a half-dolph space y. Then the set of points wherein the two functions agree that is a closed subset. Now, apply that to this situation. You have two functions x to y. What are they? You have actually two functions w x to y. What are they? They are both extensions of the same f or you may say that phi composite f hat composite phi is f is also equal to some g composite f. What does that mean here? If you think of phi as phi x as a subset of w x both of them agree on phi x but phi x is dense. Therefore, phi x bar is the whole space. If they agree on phi x they must be agreeing on phi x bar because phi x bar is the closure of phi x. So, a closed set containing phi x must be containing phi x bar. That phi x bar is the whole space. Those are the two functions agree on the whole space. So, this theorem is what we mean by universal property of w. It would have been fantastic if we could prove such a thing without y being also just y t1 space. Unfortunately, we are not able to do that. That is one of the drawbacks of Wallman's Compatification I will say. Returning to our remark 7.72, let us suppose that the Wallman Compatification w x of the space is Hausdorff. So, I just remade this remark. Take a special case namely w x is Hausdorff. Since a compact Hausdorff space is a Ticknow of space and Ticknow of properties hereditary, it follows that x is a Ticknow of space. So, I said it would have been wonderful and all that, but the very fact that the Wallman's Compatification is meant to get a compactification of a larger class of spaces. So, if you want it to be Hausdorff, you are forced to do it only for Ticknow spaces. So, this is inherent if you want to think of this as a weakness, then it is inherent in what you are trying to do. There is no other way. You cannot expect w x to be compact, w x to be Hausdorff all the time. That is not possible. That was not the intention of Wallman at all. So, that is the point of making this remark. So, we have to accept it. Having done that, let us see what we can do when it is Hausdorff. So, x is already in our Ticknow of space because w x is Hausdorff. The moment it is a Ticknow of space, we can also get the ToneCheck Compactification e, e x bar. So, this makes sense. So, what is the difference between these two compactifications is the natural question that we want to investigate now. So, this is answered by the two universal properties of both these compactifications. So, take e, e x bar denote the ToneCheck Compactification. By the universal property of e, e x bar, we have a unique continuous function phi hat from e x bar to w x such that phi hat composite a is equal to phi. This is similar to what we have done for Wallman Compactification just now. Moreover, since e x bar is a compact Hausdorff space, we can apply the previous theorem just now that we have proved. We get a function e hat from w x to e x bar such that e hat composite phi is equal to e. So, we did it for x to y, a function f and then we call it f x hat. Now, this is f phi e hat. So, I have what e hat here. So, what we have got is here x is your space which is a Ticca space. It is sitting inside ToneCheck Compactification here and Wallman Compactification there. So, this is the ToneCheck embedding here and this phi is the Wallman embedding x going to the atomic filter effect. So, what happened? This function gets extended to e hat from here e is like this. So, it gets extended to e hat like this and this function gets extended to phi hat here. Now, what happens if phi hat composite e hat that will be an extension of e inside e x itself. But the universal property of e x bar says that there cannot be two different extensions. Identity function from e x bar to e x bar is already an extension of e. Obviously, e composite identity is e. So, there will be two of them namely phi hat composite come back by e hat and the identity. So, there cannot be two that means that this composite this is identity of e x bar exactly same way e hat composite phi hat will be identity of w x. What is the meaning of this that these are homeomorphisms not only that they are homeomorphisms they are commuting with these embeddings. The embedded object x goes to the embedded object. The homeomorphism if you think of this as a subset of both of them the homeomorphism can be thought of as identity on the subset. So, it is in this strong sense that we say that the Stone-check compactification and the Walman compactification are the same for a Tick-nob space such that its Walman compactification is Hausdorff. We have to assume that that is the conclusion. The general question is when is Walman compactification is Hausdorff. Suppose you assume it is Tick-nob space suppose you grant that will you immediately say that x is w x is Hausdorff that is not that is not a answer. You may need some more conditions ok one does not know that. So, we will stop here this is a good point to stop this study here we cannot go on doing that. So, next time we will start a new topic. Thank you.