 Continuing with the construction of simply connected covering spaces, recall that we made a definition of semi locally path connected, semi locally simply connected spaces ok. Then we studied the path space over such a space and showed that the evaluation map is continuous there and it is an open mapping also. So this lemma today is a key to what we should expect, so where to look for the simply connected covering space for a given x, so this x is locally path connected and connected space. Suppose p from x bar to x is a covering projection with x bar path connected, take a point x naught in x, x naught bar in x bar such that p of x naught bar comes to x naught. Then the induced map p star from the path space of x bar to path space of x namely a path omega equal to p composite omega, so that is the map induced by p on the path spaces. This is a homeomorphism, now this lemma is not essential for the proof or for the construction of simply connected covering space, but it tells you where to look for the simply connected covering space. It tells you that it does not matter to take the path space where to take whichever covering space you do not have to construct this, namely all of them are homeomorphic to the path space of x itself. Therefore, the simply connected covering space is, covering space if it exists must be a quotient of the path space itself, path space of x itself. So this is the lemma which directs you to for that one. So the proof of this lemma that homeomorphism is essentially to proving that this p star is also an open mapping which is somewhat similar to that the evaluation map is open mapping that we have done. So I will presently skip the proof of this one and go towards the construction of a simply connected covering space, if time permits we can come back to the proof of this one at some other stage. So here is the proof, but I will skip this proof now, the proof has some nice diagram also here and so on. So as I told you we are now, we have already proved that the evaluation map from the path space to x is an open mapping, open, surjective because x is path connected. Therefore x itself can be thought of as a quotient space of p x x bar, p x x, ok. From the above lemma that just now which we skipped it follows that every covering space of x is also a quotient of p x x, ok. Why p x x is homeomorphic to p bar x x and x bar is quotient of p x bar. Therefore x bar is also quotient of p x. Therefore let us work out whatever we want to do with p x x itself, ok, that is the idea. So here is the final proof, so we start, ok, proof of theorem 7.7. Start with a semi locally simply connected space, I am just repeating these things. Take a point x naught belonging to x, ok, that is fixed point, ok, base point and this capital P, bold P, let it you know p x x naught again and again we do not have to write as a big symbol, that is all. This space is given the compact open topology, remember that also. We now define an equivalence relation in P by saying that omega is equivalent to gamma. Remember these are path is starting at x naught, ok, if we don't leave these two paths are path homotopic. In the other words omega 0 is already gamma 1, but end points must be also the same and it must be path homotopic, namely end points must be fixed during the homotopic, remember that, that is the equivalence relation. Let x bar be a quotient space of equivalence classes and let p from p to x bar be a quotient map, ok. So what the difference is, in the definition of fundamental group we took the loops at a single point. Now we are taking all paths, the end points could be, the other end point could be anything in the space of in set x, that is the difference, ok. So this is something much huger than just the set pi 1 of x x naught, it is not a group either. This is given a topology now, what is the topology, topology coming from the compact open topology of the path space as a quotient of that, this is x bar is the quotient space, ok. So by definition these two are path homotopic means the end points must be the same, therefore it follows that the evaluation map remember is just take the end point, ok. So that factors through the equivalence classes, on the entire equivalence class it takes the same value, therefore the e factors through a map, ok, like this, ok. This is x bar to p, then you get the map from, this map is what I am calling it as p now. Remember we have to construct not only the space, but also the projection map. So this p is nothing but e factor through the quotient, take any class here, it is represented by any path, the end point is independent and take the end point here, that is the p of that. So p of this is also just the end point, but it is independent of the path class, ok. Now one easy thing here is that p is an open mapping because we have shown that e is an open mapping, take an open set by the very definition of quotient space, inverse image is open, the image of that is same thing as image under p of the original open set here. So e is open, so p is also open, ok, we have to show that p is a covering projection, we have to show that x bar is simply connected, ok. So these two are the tasks now for us, ok. So let us start with v, any path connected open set x such that the inclusion induced map on the fundamental group from pi 1 of v to pi 1 of x is trivial, such a open subset around every point is guaranteed by the hypothesis of semi locally simply connected. Just for being a few, using this economically some words, let us call such a set as ambiently one connected, it is not simply connected by itself, it need not be. When you pass to the whole space, its fundamental group becomes trivial, that is the meaning of this, ambiently one connected, ok. So what we know is that x is covered by open subsets which satisfy this prep, they are ambiently one connected, ok. We claim that if v is ambiently one connected then v is evenly covered by p and that will finish the proof that p is a covering projection, ok. The idea of the claim is clear, alright. So we have to execute this one now. Given a path starting at x naught, there is an eminent of p such that the end point is inside v, v is chosen to be some open set which is ambiently one connected. We are going to define a set of, you know, subsets of x bar. So what are they, consider the set v bracket omega because this is going to depend upon v as well as the class of omega. Class of omega is an element of, what, is an element of x bar, remember that, ok. So all such classes, they are also elements of x bar which are omega star omega prime, path which look like omega star omega prime, wherein this omega prime is a small loop, you know, a smaller path completely contained inside v. Remember omega to begin with is a path from x naught, x naught to omega 1, ok. Omega 1 is what, omega 1 is some point of x, alright. Then I am taking this class, this class is an element of x bar. I am extending it by omega star omega prime by some path within v which starts at omega 1. Look at all those, that is going to be v omega, ok. The claim is, so I will give you a picture here, start with omega like this which ends inside this open subset v, then I can extend it by some omega 1 or omega 2 and so on. These extensions are completely inside v, ok. So this is my picture I can keep referring to again and again. So we have defined v omega like this, ok. Then p inverse of v, v is an open subset, p inverse of v some will open subset, right, of x bar. This is union of all v omegas, v omega defined that way, whereas omega 1 must be all those pathes I have taken, union over all those pathes, so omega 1 must be inside v, ok. Only then omega v omega makes sense by the way. So you look all the pathes which come inside v, at the end point must be inside v, then look at v omega that will be p inverse of v. Remember, p is nothing but the end point, right, the end point must be inside v. So if you have the point is already inside v and then you are connecting it with another path within the v, the end point of that will be also inside v, ok. So this p inverse of v is union of all these things obviously, alright. Second part is take a class tau which belongs to v omega, some member of x bar, this is an element of x bar, suppose it belongs to v omega, v omega which I am in that way. That would mean that v omega is equal to v tau. So this is like group theory wherein, you know, if x is in h, then its right coset hx will be hx itself, it is of that nature, ok. So let us see what happened. Suppose this thing here omega, this is tau, my tau, ok. If tau belongs to v omega, means this one, right, it is omega star omega 1 for some omega 1, right. Suppose this is tau. Now look at v tau, v tau will be what? All those paths coming up till here and then paths we go within v from this point. But I am taking the homotopy classes of paths. Or in particular, after going here, I can come back from this path up to this point. That will be homotopy to omega 1, therefore v tau is contained in omega 1 and v omega, v omega is contained inside v tau, so these two are equal, ok. So this is completely trivial but if you look at the picture it will be the same. So this will happen either some element is here, when both the classes are the same or it just means that v omega and v tau in general are disjoint. Either they are equal or they are disjoint, if they intersect they must be the same. This is like the cosets, cosets spaces inside, cosets at the inside of a group, ok. v omega intersection v tau is non-empty would imply v omega equal to v tau, therefore what we have proved is p inverse of v is a disjoint union of some v omega, ok. So you see, this is what we wanted after all. Then I want to say that each v omega comes to v which is obvious by a homeomorphism namely p restricted to v omega to v is a homeomorphism, this is what we have to verify, ok. v is path connected, it follows that p from v omega to v is surjective. Once a path has come inside, omega has come inside v, from there I can join it to every point inside v, so that gives you that p is surjective from v omega to v. Let us show that this is injective and this is where v, so far we never used this fact, namely v is ambiently unconnected, so far we not, this is where the injectivity we use that one, ok. Having said that I can leave it as an exercise but now let us verify this one. Suppose p tau 1 is equal to p tau 2, tau 1 and tau 2 are two elements in x bar, ok. Actually I should assume that they are inside v omega, ok, where tau i's are inside v omega, that means what tau i is omega star omega i, i equal to 1 and 2, ok. But now I am assuming that p tau 1 and p tau 2, this means that their end points are the same. So we can go back to this picture, modulo that these two end points here of omega 1 and omega 2 are the same. What does that mean? Omega 1, omega 2 inverse is a loop where inside v 1, therefore inside x it is null homotopic. If this, this is null homotopic, then we know that omega, omega, this omega followed by omega 2, it is the same thing as omega followed by omega 2 all the way omega 1 and compare to omega 1. It just means that tau 1 is homotopic to tau 2, ok. This kind of thing we have seen several times, so I repeat this. So omega 1 equal to omega 2 means they are end same end points. I check from pi 1 of v to pi 1 of x is trivial, so we know that omega 1 composite omega 2 inverse is null homotopic, the class is 1. Therefore tau 1 which is omega star omega 1 is omega star omega 1, I can put omega 2 inverse star omega 2 because this is this is trivial, but now put the bracket this way omega 1 star omega 2 inverse is trivial, so this cancels out what is left out is omega star omega 2 which is tau 2, ok. So we have got p from v omega to v as a bijective mapping, ok. P is already continuous, what is p? We have shown that p is an open mapping, ok, p is also an open mapping, right. Only thing what remains is we have to show that each v omega is an open map, open set, ok. So that is that is that is that is remaining, then projection that p is a covering projection will be over because we have to show that p inverse of v is the disjoint union of open sets, each of them coming homomorphically onto v, ok. So everything else is shown except v omega must be an open set. So v omega is where v omega is in x bar, x bar is a quotient space of p. How to show something is open there? Show that v inverse of that set is open in p, ok. This is the definition of quotient ecology. So I have to show v inverse of v omega is open p. Take any lambda in v inverse of omega, sorry, v inverse of v omega. Around that I should produce an open set contained inside v inverse of omega, right. Yes or no? For every point inside this you should show a open set. What are open subsets? Compact open topology is what you have to show. You have to produce that. Some basic open set namely intersection of k, i, v, i, such a etcetera has to be found out now. So what is this? Look at this lambda which is a map from x naught to some point inside v, right. Because it belongs to p of that, phi of that belongs to lambda, that is the meaning. The class of lambda, phi is nothing but class of lambda belongs to v omega, ok. So this entire curve can be covered by finitely many, ambiently one connected open sets v1, v2, vn. The last v1 I can choose it as v itself. I can come back from that side. Start with v and then cover some other portions, some other portions. Actually you can take a infinite covering first and then take a finite covering because the whole thing is compared, lambda is compared, ok. And get a partition, 0 less than equal to t naught less than t n etcetera 1 such that each ti is in the, you know it is like a chain, is in the intersection. 0 to t1, t naught to t1 is inside v1, t1 to t2 is inside v2 and so on. So, this is the definition of ti, ti plus 1 is inside vi, ok. So, cut it, cut the entire lambda and cut it, cover it. So, then then cut it into partition, into these submitters. If w denote the set of all paths lambda prime at x naught, starting at x naught, at x naught such that lambda prime of ti, ti plus 1 is inside vi. That is the basic open set by definition for every i 1, 2, 3, up to n, ok. Then by definition of the compact open topology, this w is open. This w is nothing but ti, ti plus 1 bracket vi, right. That inner product bracket symbol is there and w is intersection of those things, ok. So, that w is an open subset of p and lambda, to start with lambda satisfies this property. So, lambda is in w. Claim is this w is inside p inverse of v lambda, ok. It is the same thing as taking that it is homotopic class lambda belongs to v omega, v v omega. The homotopic class of lambda belongs to v omega. Remember what is v omega? v omega consists of omega star sum extension omega prime within last v. v is one of them, v is vn. So, we have to show this homotopic property for this lambda, for all this such a lambda inside w, ok. So, here is the last picture. This was your lambda, ok, which you cover by these evenly covered open subsets v1, v2, v3, v4, v5 and this is vn, which is equal to v. This is lambda prime which also should satisfy that 0 to t1 or t1 to t whatever is concerned is a v1 and t1 to t2 is concerned is a v2 and so on, ok. Look at t1, image of t1 up to t1, 0 to t1, both the paths are inside v1. So, join them, the end points here, you get a loop that is the homotopic. Join t2 and t3 by a loop, then you get a loop inside v2, that is also an homotopic. Join t3 and t4, sorry, t3, lambda t3 and lambda omega, lambda prime t3, that will be also an homotopic. Anyway, all that you will get is this path is homotopic to this path, composite this path going back, composite going back, composite going back and so on, exactly similar to the Venkampan's theorem that we have proved right in the beginning, the proof of Venkampan's theorem if you remember, introducing in between paths here, ok. So, it follows that this lambda prime is homotopic to this lambda, lambda and particular lambda prime, lambda arbitrary sets, you know, it applies to omega also. So, they are all homotopic to omega star something, ok. So, this completes the, completes the proof of that p is a covering projection, ok. What remains the proof of simply connectivity of x bar? So, let us complete that one. Connectivity follows because p is connected and is a portion of map, ok. There is no problem, it is path connected because p is path connected and it is a portion of map. To show that x bar is simply connected, let lambda, capital lambda from i to x bar be a loop at the constant path at x naught, ok. You have to choose some point, base point which comes to x naught, right. So, I choose a constant path at x naught where cx naught denotes the homotopic class of the constant loop at x naught. In order to show that this loop is null homotopic in x bar. So, lambda is a loop, loop of loops. Remember that it is not a ordinary loop now. Loop taking values inside the classes of loops, classes of path history, ok. So, in order to show that this is null homotopic in x bar, by the injectivity of p check. p check is already we have proved that it is what? It is a covering projection. Therefore, p check is covering projection, is a injective map. So, you go down to x and show that, you know, p composite lambda is null homotopic. The p check, in injectivity of p check suppose show that let us write it as omega which is p composite lambda. This is null homotopic in x, ok. So, we are using the partly proved statements to prove further things here. This is what it is, ok. So, now what is this lambda? Ambrose of i proved x bar, we have come here, ok. This is a loop in Cx naught has to be used, alright. So, let capital omega from i to p be defined by omega t s equal to omega t s, which we which we have considered for contractibility of p, ok. So, define, now the path defined by this one is such that to take the end point of, end point of this one, this will be omega. Here, what I have taken is the same omega I have taken, ok. The end point of this for each you take, it will be s is equal to 1, omega t will come. So, the whole path will come, s equal to 1 I have to put, this is omega t. So, it is a path now, the end point path is a path. This was a loop of parties, ok. Therefore, p composite, p composite omega is nothing but omega. This is, this p composite phi remember, e is by definition, you know, p is by definition such that p composite phi is e. So, p of phi composite omega is omega. Thus, we have two lifts. One is capital lambda, another is phi composite omega of this omega in x bar. There are two lifts. Moreover, phi omega of 0 is the constant loop Cx naught, which is lambda of 0. Therefore, by unique path lifting property of the covering position p, it follows that this lambda is nothing but phi composite omega, this path we have. So, we took lambda, came to the, came to x under p, then we identify what is this original lambda, it is phi of omega. This omega is now all the way inside p, ok. So, you see, we have used the property that arbitrary path is inside x, can be lifted to the p very easily, ok. And when we quotient it out, the lifting property is still retained by covering projections. This is what is happening. So, here we have used the lift. So, this omega has been lifted to this capital omega, and that under quotient map it becomes this lambda. Therefore, phi of omega 1, this lambda 1, this Cx naught, which means that omega equal to omega of 1 is no longer ok. Look at omega 1, what is going to be equal to 1, that path, ok. So, that omega, that is omega 1, right, when we put s equal to 1. So, this is null omega 1 is what it means, right. Phi, phi is what? Phi is the homotopy class of this one, omega 1 is a loop, right. This Cx naught, right. So, this class is same thing as Cx naught means omega 1 is null omega 1. But omega 1 is nothing but omega, ok. So, most of it is like a tautology. Actually, all several constructions in mathematics, when you have in a blank where to go for is tautological. Right in the beginning of, you know, construction of real numbers out of Cauchy sequences of rational numbers. You want every Cauchy sequence to be convergent, what did you do? You took equivalence classes of Cauchy sequences and declared them as real numbers, over. Now, Cauchy sequence of Cauchy sequences convergent is what you have to show. So, this proof is similar to that. You want all kinds of lists of pathways inside a simply connected covering. So, you declare the covering itself to be set of all pathways, but that is too much. So, you have to do equivalence classes. So, it is in simplistic language, this is, I could have taken the example of, you know, metric completion of a matrix base or construction of real numbers. This is similar to that. Okay? So, let us stop here. Let us come back to other things in the next session. Thank you.