 Okay, we come back to the Biberback conjecture yesterday. I gave you one of the proofs of the first step of the entire story about this attempt of estimating the moduli of the coefficients of a univalent function unit that's normalized in all function. So, Biberback use another approach, but then I can show you that in fact, with some extra tools we have now, it is quite simple to have the Biberback theorem. But then Biberback conjecture is not easy to prove in general. When I was a student, so I was your age, it was debated if it could be possible to find them or not a general solution. So, there were some schools who considered the Biberback conjecture falls. So, in a sense, they were trying to find a counter example. Even though there were some evidence that for classes this is true. So, as I said, the Biberback conjecture was proved in the 80s. I recall here what I said yesterday that the main set is star like with respect to a point. If whenever you have another point, you have also the segment connecting these two points. It is convex if it is star like with respect to each of its points. Now, the functions are all of more so in this class of function, that is the function which are univalent or the morphic. So, the final unit this can also normalize. So, we start with this class. If the image, so the range of this is set with this geometric properties like convexity for the range implies that the function is convex. Similarly, it is star like if the image of the disk is star like with respect to the origin. All right. So, this is the standard notation. So, C, calligraphic C is convex, S star because of the star is star like, and S is simply the class of normalized univalent function in your disk. Well, this is of course true inclusion because for instance, the cubic function is an example of star like, but not the convex function. So, we have functions which are here but not here, right? We have functions which are not even star like. Then we, if I remember well, I introduce also the class P of functions, take values in the right half plane. So, such that the real part is positive, and it is normalized with this assumption that the P of 0 is 1. Then there is this estimates, this uniform estimates for the coefficient of a function in this class P. Class P is also known as class of caratadori function. When I say caratadori function, I mean a function univalent from the unit disk into the right half plane such that phi of 0 is 1. Normalize like this, and, sorry. Besides all these calculations, I want to point out that what we have as extremal function so that you cannot have better estimates for the coefficient. This is simply the function we use to prove that the disk is bilomorphic to the right half plane. So, in fact, the coefficients are all equal to 2, right? They are all real equal to 2, so the inequality cannot be improved. Okay. So now, some kind of theorization. This is interesting. So, you take a function, not even necessarily univalent, but only allomorphic fixing the origin and having derivative at 0 equal to 1. So, this function, of course, is not mapping the disk into itself. Otherwise, it would be obviously a rotation and not just one rotation, but the identity because of the Schwarz lemma. Right? We have that f of 0 is 0, f prime of 0 as module is 1, so it is a rotation, but that in particular prime of 0 is 1, so the coefficient is equal to 1, so f of z is z. So, in general, this function is supposed to map the unit disk into c. Otherwise, the class of the data, self-maps is very small, just one. So, we can prove that f is starlight, if and only if this function here, which is well-defined as we will see, is holomorphic and the real part of f is positive. Sorry, the real part of this capital F is positive. Again, another characterization for convex function is the following. You take this new function here, which is 1 plus zf's second derivative of f over f prime. So, it is some sense of high order as a condition, is with real part positive. This is an intermediate result due to Alexander, say, century ago, and it's in the same time by Biberbach. Bioneristic works on these functions. So, the function which is holomorphic and normalize as usual, then is convex, if and only if, zf prime of z is starlight. This is another case. So, the three classes are, there are relationships in the description of the classes, p, c, and s star. So, in short, I want to show you how these ideas can be obtained. So, we start from a star-like function, f. So, it means that whenever you have a point along the segment connecting f of z to 0, so t here is parameter from 0 into 1. So, it is, look at this left side of this quote. Okay, I take f of z. I know that f of 0 is mapping to 0, right? Then I take f of z, another point. In order to prove that the range, so the domain f of delta is star-like. It is to say that f is star-like as a function. I have to prove that whenever I take f of z, the entire segment is in f of delta, which means that when t varies between 0 and 1, there should exist a point here, right? Such that this is the case, right? Now, this function here is a function from the unit disk into itself. Clearly, since f of 0 is 0, I have that w of 0 is 0, because on the left-hand side, here I have 0 and here I have 0, and f of 0 is 0, right? This injective, remember that you are dealing with injective function. So from the Schwarz lemma, w of z as a module which is smaller equal to z, modules of z. What we do have is something not very much known in literature. So it is considered like a stupid lemma. Second, it's considered not a basic result. It is instead a very important fact. You take a circle inside the unit disk. So you take this R strictly smaller than 1, and this is mapped into a curve without self-intersections. This C R bounds a star-like domain, because we are taking, you see here this property. As z varies in the circle, this varies in a curve, but because of this property, this is true for any radius R. But what is the analytic interpretation of the star-likeness? The arc of f of z, that is the angle between f of z and the real axis, positive real axis, has to increase as z moves along this curve. So anticlockwise, it cannot decrease. Why? Because if you assume a contradiction that you have a function which has an angle for the image of, this is the real axis, and this is the arc of f of z, right? So maybe it moves like this, but the angle which measure the arc of f of z has to increase, because otherwise the contour would come back and forth in some sense. So that you would avoid a valley here. It cannot intersect itself. So the contour cannot intersect the contour itself. So if the arc first increases, then decreases and then increases again, then you would have some curve here, part of the curve and two points here cannot rejoin to the origin. So this is the interpretation. This is a good idea to have geometric characterization. In fact, the fact that arg of f of z increases along their circle in the positive direction. So for you here, okay? I move like this, the circle, and the image is something which is, but the arc increases as z moves. So it means, oops, sorry, sorry, so here we are. It means that analytically you have this function here is a real value function, right? So the derivative, it is increasing, right? So the derivative with respect to the angle as z varies between 0 and 2 pi in positive direction as to be non-negative. All right? This is the condition. Now, the function is restricted along the circle of radius r, where r is smaller than 1, and theta varies between 0 and 2 pi. Good. So we can actually write the arc in this way. You use the fact that, remember that the logarithm of something is the imaginary part which is exactly the arc, right? So because of this fact that this arc here can be written as image of logarithm of f of r, yeah, theta, you have this equality. Now you calculate this derivative, which is here, right? And while this derivative is nice to observe the world, this is precisely this null. First, let us consider the fact that we have the imaginary part of a function to be derivated. And this is exactly the derivative of the function, and then you take the imaginary part, right? This is the first step. Second step here, you have the logarithm of a function along a circle. Then the chain rule applies, right? So I have an imaginary part, and then this is what? This is the derivative of this function here. You see? Chain rule. So 1 over f, f prime, and then the derivative of this function here, which is i r e i theta. But the imaginary part of this, you see, in fact, this is z r e i theta, okay? It is i z f prime z over f of z. Well, this is i times something, right? Which has a real and imaginary part. Since there is this i in front, well, the imaginary part of this is the real part of this without i, right? So what does it mean? If you want to have this increasing property of the arc of the image, so this number is positive, then this function here, z f prime of f of z, as real part positive, or z f prime f of z, is, remember, this was what we have to prove, okay? If and only if this is the case. Well, similarly, you can repeat the argument for convexity. But this time, what is the geometrical condition you are considering? Okay. Well, you can... Convexity is a more complicated request compared to the likeness. So I have the circle I start from. I take the image of this circle. What I cannot have is that the image of this circle is a contour with some... If it is not increasing the arc, it is not even star-like. So we have to require at least this, because convexity is star-likeness with respect to any point. But what you can have is that, as I said yesterday, you can have a contour like this, a star, which is, in fact, star-likeness, which is star-like but not convex. So what I don't want to have is this kind of behavior along the contour. And I have to check that the slope of the tangent is not decreasing, because if it decreases, then increases again. Decreases you have this kind of contour, which is not good. So it is more complicated, since you are taking the derivative, so the slope of the tangent, you are taking a second-order condition, in some sense, as we'll see. Because we take the this number here, but of the derivative is positive. So it means that, as theta varies, the derivative as an arg, which is non-decrease, therefore, from here, you tend derivative. First derivative of the function will be here. And from this derivation here, you obtain something which requires a second derivation. So if you make all the calculation and complete the proof, or you look at a book, this is the condition which comes out from this. Make the calculation as before. Repeat the similar argument, obtain real part, imaginary part, then you obtain the imaginary part of the logarithm of f, of blah, blah, blah. And this is what you find out. Now, as an idea for the proof of Alexander, let me show you this. So you take this function here, g of z, z f prime of z. And consider this function here. Well, it is readily seen that you take this ratio, which is this. And so since g is z f prime, g prime involves also the secondary of f. And from this obvious equality, this is an application of the definition. Well, you see that the two functions which are the two, sorry, the two ratios involved in the definition come out. So this is positive. The real part of this is positive if and only if. The real part of this is positive. So Alexander theorem does exactly this. f is convex if and only if. This, which was called g, is star like. G is star like if and only if. The real part of this is positive. But because of this equality, the real part of this is positive. This is z, the function f is convex, as we prove here. So convexity can be expressed in terms of second derivative of function, or in terms of star likeness of this auxiliary function. Now, we have some consequences in the investigation of the Beaver-Buck injector. Because of the estimates we have, if we start from a star-like function, then we have, well, this is, of course, a particular function which is in s. So this is the power expansion of 0, right? So z plus a2, z2, z2 squared, blah, blah, blah. I write this. I write the function z prime of z over f of z, which, as you can see, is well-defined. You might wonder whether this is a meromorphic function or a holomorphic. This is holomorphic, right? Because f of 0 is nothing but 0. OK, so for several reasons, this is OK. And furthermore, I know that this function here is with real part positive. Because I'm assuming that f is star like. This is to say, this function here is caracodorean. But for caracodorean function, I have also this estimates of the coefficients. So now the game is transfer the information on the estimates of this coefficients here and do the coefficients here. If I can do this, I have the estimates I need. What is z f prime of z in terms of the derivative? Well, f prime of z is 1 plus a 2 2 a 2 z plus 3 a 3 c square plus plus. Then I multiply everything times z, right? So I have z plus 2 a 2 z squared plus n an z to the power n. So I have this power expansion, right? So normally, I reduce by 1 the exponent, but I multiply by z here. So these are the coefficients explicitly written in terms of aj's here, right? Then I write, you see, this is a fundamental tricky mathematics. You divide and multiply by the same amount, which is in such a way that this function here turns out to be caracodorean. And this is the starting f. So we have this quotient here times this. And this is the power expansion of z f prime of z over foz. And this is the power expansion of f, right? When I multiply power expansion in a commutative setting, well, we have that this is a power series expansion as well, where the coefficients as n are obtained in this way. We obtain an plus the first of this time an minus 1 plus an minus 1, OK? So then we have c2 plus c2 times an minus 2 plus blah, blah, blah. And then a1 is, of course, 1. That's why here you don't have a1. But then you look, you have two power expansions. And here you have the qualities. That is to say, the coefficient of the same power of z has to be the same, right? In particular, Sn has to be NaN. Because two power expansions are the same if and only if they have the same coefficient. It's like the generalization of the identity principle for polynomials, if and only if they have the same, OK? But we have an explicit expression of the Sn coefficients here. And NaN is not known. But what I know is that, well, NaN is Sn. And from here, I obtained this. Look, I equate NaN. I move NaN on the left hand side. Are you with me? And then I obtain NaN minus 1 times NaN. And now forget for a while this modules, OK? So as I said, Sn has to be equal to NaN once more. So from this fact, only NaN on the right hand side is not obtained as a multiplication of some coefficient of Cj of this power expansion, right? So I have NaN minus NaN minus NaN. So this NaN minus 1 is this. C1 NaN minus 1 plus C2 NaN minus 2 plus, plus, plus, plus. When I consider the module I, the left hand side is like the right hand side, right? I have this as this equality. The modules of NaN minus 1 NaN, that is NaN minus 1 modules NaN, is the modules of this part here. Then I use the triangle inequality. And the fact that all the Cj's here in modules are smaller or equal to 2 because this function here is carotidory, right? So I collect 2 in front in this estimate and I have the sum of the Aj's with J smaller than N. And the last one is 1, right? Because it is A1, all right? Now, what can we do? For N equal to 2, in particular, we obtain 2 minus 1, 1. A2 is 2 times A1, module A1, which is 1, right? So we have this. So the Bielberg theorem is again obtained. But for this smaller class, what I want to obtain is that by induction on N, I can prove that each An has modules smaller or equal to N for function in the class S star. So for star-like functions, we can prove Bielberg conjecture. And let us check for 3. We have 3 minus 1, which is 2, right? A3 is smaller or equal to modules of A2 plus 1, right? But 2 is smaller or equal to 2. Sorry, A2 in modules is smaller or equal to 2. That is to say, this number is smaller or equal to 6. Or 2 modules of A3 is smaller or equal to 6. That is to say, A3 in modules is smaller or equal to 2. Now, assume that this is true for any k between 1 and M, as I said, by induction, right? Now, from this, I simply reported the same estimate we already used. We have M, M plus 1. Well, I consider M plus 1, right? It is by induction on M. So I assume that the statement is true for any k from 1 into M. So I prove this true for 1, 2, and 3, right? And now I want to obtain the statement for the coefficient M plus 1. So M plus 1 here means M, M plus 1. This is smaller or equal to 2. And then I have what? This is AM minus, sorry, this is AM, right? In here, in this game. Remember that N is now M plus 1. So I have modules of AM plus modules of AM minus 1 plus modules of AM minus 2 and so on and so forth. And by assumption, each of them, so each of the AMs has modules smaller or equal to M. So I have smaller or equal to 2M plus M minus 1 plus M minus 2 plus plus 2 plus 1. And this is the sum of the first M integers, right? So this is quite an easy exercise to have. This is 2 times M plus 1 plus times M over 2. So it is M over, sorry, M times M plus 1, right? But M is here, so we have this. Quite a simple proof. So for star-like functions, we have actually proved that it is as conjectured by Biberback. So the Mth coefficient, this is true for any N, OK? Because we use the induction. So the Mth coefficient of star-like function normalized is as modules smaller or equal to M. Here are some references. For those of you who can be interested in this stuff, in particular, the famous book by Durand was written before the branch proofs of the, so that you can see in some chapters some attempts, some other, but there are books so thick about this subject. I'm not saying that while we have the complete survey of the topic. However, it is nice to have a comparison. Well, this is also very old, conformal mapping. So this is before the branch there. So there are some more recent books. We have another approach, because most of the attempts were considered promising. Now they are considered not promising at all, OK? Because now they found the final result. OK, so first thank you for this. And now let me check if I can. So this is, as I said, this is nothing. But I wanted to give you the idea that the simple tools we have are extremely powerful, if used with some knowledge and accuracy. Now let me switch into another subject, which is more abstract. There will be no proofs about these facts. And I will, all right, let me check. For the next topic in complex analysis, I will need some topological backgrounds in covering. We have already encountered some topological instruments or characterization of domains or something else, which have influences in complex analysis. For instance, the existence of a logarithm is granted as soon as you have a known vanishing of a function. On a simply connected domain. The Riemann mapping theory, which is somehow the peak of one course in one complex variable, is valid only for simply connected domains. There are other restrictions and other dramatic conditions which can be better rephrased in terms of topology. So what I need now is the notion of a covering space. So we start from basic topology. And so the function involved in this part, in this topic, are essentially functions which are continuous. And only in special cases, we will see the function will become holomorphic. So for the time being, we restrict our point of view to the topological point of view. So let me give the very basic definition. You start from a topological space x. You consider another space, x tilde. And the continuous functions p, which turns out to be also subjective. So for instance, you can take x and x and take p, the identity. For sure, this function is a subjective continuous function. But furthermore, we consider p to have this property. You take any point in x, and for any point that exists a neighborhood, open neighborhood u, such that the inverse image with respect to p is a union of this joint open sets in x tilde. And each open set in x tilde, so when you restrict to one of these joint open sets, the function p turns out to be homomorphism. So p is not a global homomorphism, but locally, a tiny of such neighborhood it is. Now, for instance, the example of the identity works fine, so any space can be covered by itself using the identity. This doesn't give any extra information. Some terminology. The map p is called the covering map of projection. The space x is often referred as the base space, because this is the image. You have something here, and you can start something up the space, and then you project. And the total space, x tilde, is what will play the important role in the covering. Together with p, of course. So let me make you some example. Probably you already know this. Oh, first, yeah. In many, many textbooks, you find extra conditions on the space x. I only assume that x is a topological space, which is, of course, unavoidable if I want a continuous function. I have to know what continuous function means. However, most of the results are valid only if the base space is path connected, locally path connected, or something like this. So in some books, you find these assumptions for each proposition. In others, the assumptions are already condensed and used in the definition of covering. So since, essentially, no theorem can be proved without this weak connectivity conditions, some authors decided to add the condition directly in the definition. For most of the textbooks, the condition of locally path connectedness is assumed in the definition. This is just, OK. In case you find a different definition. Well, the very simple example, but the toy example, which always is useful to know, you start from the unit circle as 1 in R2. Well, you can cover S1 with S1 itself. We're using the identity, but this is the trigger covering. But you can also consider the function p from R to S1, which maps the point t, so the real number t, into the pair, cos t sin t, which, of course, lies on S1. And this gives you an infinite covering, as we say. If you move along the real axis, you go around S1 infinitely many times because of the periodicity of cos t. Another example is the following. Non-trivial example. So take the complex plane with the, well, it is like R2 minus 1 points of, for instance, the origin. The fact that it is the origin is not essential, but it makes all calculation easier. It takes as a function not the identity from C minus the origin into C minus the origin, but the nth power of Z. So now, each point has a neighborhood such that the inverse image is only n, right? This joint opens sets, where qn turns out to be a homomorphism. In the previous example, you take a pointless one and you obtain infinitely many in the neighborhood of it, and you obtain infinitely many intervals in R, so because of the periodicity. But here, you obtain exactly n inverse image of one point Z naught, right? You start for a point of the disk of the puncture plane, and there are only n inverse images, because you have a polynomial of degree n in C, and you have n roots, OK? Here are some of the properties of what are normally called the fibers. So p minus 1 of u is called a sheet. So the number of sheets is what characterizes the fibers. And this number is independent. It's a discrete, well, first of all, the fiber over a point is a discrete subset of it still. The fiber is the inverse image. It cannot be a continuous set. And the fibers are, well, as you see, every connected components, the fibers are homomorphic. If x is path-connected, so the base point is path-connected, this discrete space is the same for each x. And this is to say, the fiber is homomorphic one to the other. So they have the same number of sheets, and they are homomorphic. Well, the footprint image is then homomorphic to u times y. Well, if you don't consider the point y of the fact, but locally, you obtain this. So the cardinality of the fiber is equal to the cardinality of f, right? And this is called the degree of the cover and represents the number of sheets. So in the previous example, if you have the same base point, base space, sorry, and the same total space, for instance, the second example I gave you before, it depends on the function qn, the number of fibers, the number of the cardinality of the fiber, or the degree of the cover. It's not something which only depends on the space, but depends on the space and on the function. Yes, well, there are several ways to say the same fact. So the degree or the number of sheets or the unfolding covering number are introduced for several, say, you can see the covering in different ways. But you essentially have to know that what you are counting is the number of inverse image of one point, which with some path-connected hypothesis is the same. And in this case, in some books, you find that, well, this is the degree of the cover. In other books, you say, well, this means that the covering covers is an n-fold covering of the base point and so on and so forth. But in the sense, well, this is a triple cover. This is a double cover for what. So you find examples of double covering using, for instance, q2, as we did, of the plane minus the point, and n-fold covering in general. Now, what is also interesting to know is whether it is possible to get information from the base point so that we can lift information from the base point onto the total space using that, right? So since on the base point in algebraic approach, what you start doing is to study path and then loops in order to define fundamental groups, right? An amount of pins and so on and so forth. It is not to think of what is over the path to be another path in x tilde. Is there always the case that starting from a path in x, there exists a path in x tilde, such that the projection of this path in x tilde is the path in x. Well, the answer is, well, if you start from a path, this is a continuous line from an interval into x, there exists a unique path with the additional hypothesis, well, lifting path. That is to say, when it is projected as a path from x tilde, it becomes the path in x. But you have to add one initial condition because imagine in the general setting, you can have several levels of the inverse image, right? So there are several choices of the same possible lifting. But whenever you choose one, since these are disjoint, there is only one. And this is a proposition, which in some sense is the first step in the description of all lifting. These are called lifting properties. And the curve gamma is called the lift of the curve, so capital gamma of the curve gamma. As you can see, if x and y are two points in the base points, in the base space x, and then if you assume that x is path connected, you can join these two points by a curve, by a path, right? Then you go to fibers, and you obtain one lifting starting from one point of the first fiber into the other fiber, right? Because we have to consider this as a decondition for the lifting. Such a way to do a new project, the starting point of the lifting has to be, say, the point x, and the ending point of the lifted path has to be the endpoint of the curve. But there are several ways to do this, because each time I can choose a different point in the fiber of x, right? And this possibility of lifting with different initial points in different ways gives, in fact, the bijection between the points of fibers, which we were looking for. In this way, you can easily see that I don't want to prove anything as a sin. I promise you, I won't prove anything. But by using this lifting properties of the path, you can easily see that fibers are, in fact, have the same 39, so they are one-to-one correspondence. OK, I guess that all of you have already seen the notion of amotopy so that I simply collected to summarize the best effect I'm going to use in this proposition. So you start from an anthropological space. Nothing is related to coverings. But you take point x0 as a base point, and you consider the set of all loops starting, so closed path starting at x0 and then at x0, continuous path, right? And then you can form one into the other using the amotopy. And you said that two paths are equivalent if there exists an amotopy that is a function. Say yes. Continuous deformation of one into the other loop. So first loop to the other loop. So the set of all these classes is a set. But then on loops of bay points x0, you can also introduce a composition. You start from a path, then you have another path, and you say, OK, just the position of two path is one path. OK, so you move on one and the other, and it is a closed path again. Well, this definition is good in a sense that when you pass the classes, this is preserved. Classes of amotopy. And the set of classes together with induced operation of just the position on classes. So it depends only on the representative and the amotopy classes instead of a loop. It is a group. This was introduced by Poincare. I guess that this should be known. But however, this is called the fundamental group of x and gives you very many information about the topology. And so it provides you the tools to classify topological spaces. These are preserved by homomorphism and so on and so forth. And they are denoted by p1, x, x0. Because starting from p1, then you have also p2, pn, pn. So in general, even though you can define also the nth amotopy group, it's not obvious at all to calculate them. There are some open questions also for spheres. The pn of sk is not known in all possible choices of n and k. And what is surprising also is that you can have pn of sk. Sk is the kth sphere in R, k plus 1, the unit sphere. Well, not necessarily pn of sk is 0. So it is trivial, 0 in the sense of a group. It's a trivial group when n is greater than k. If some of you have some knowledge in homology, well, homology groups which are related to, in particular, the first homology group is very much related to the fundamental group. But for homology groups, you know that FPR sk is for some homology group of a set which lives in Rn. And you ask for hn plus 1 of something which lives in Rn. This is definitely trivial. So there are some differences between homology and homotopy groups. The relation, if you are interested in this, is that this group here is not necessarily a billion. It cannot be in general a billion. There are examples of non-a billion groups. On the other hand, the first homology group is always a billion and is, in fact, it can be obtained from this non-commutative group as the ablianization. You see this in the motion of the non-commutative fundamental group. So for instance, quite easily, take this simple s1. And you can see that any loop or starting point in s1 is somehow a multiple of a basic loop. You cannot have very many other loops. Just a multiple of one loop. So one sense, the inverse sense, the inverse sense, and so on. But the fundamental group is like z, isomorphic to z because this is the group associated to each loop, to each class of loops. You can associate a number of times. You are going around zero in positive directions or in negative directions. But if you take two circles, and you make this eighth figure, you can have one generator in one circle, one generator in another circle, and these are independent so that you can have loops which are like this. So you have two generators, and the fundamental group is what is known as free group z, so the free product of two groups. And the free product is not necessarily generally it's not commutative. Whereas if you calculate the homology group of the eighth figure, you obtain that this is the direct sum of z with z, which is commutative. This is if you want the relation. It is important to know, but not in our sense. So I wanted to give you one, well, D condition. There is only one condition in order to generalize the situation of lifting path. So in general, instead of starting from a function from an open interval or a closed interval of r into x into the base point, I might start from a function which is continuous from a topological space z into x. And I wonder whether you can always take, they can always lift f to a function from z to x tilde in such a way that this condition is true. So you see this generalizes the previous one. I'm looking for a continuous function from z to x tilde. x tilde is the covering of x. p is the projection or the covering map. And I want that the function which acts here from z into x is lifted to a function from the same z starting at the topological space into the covering in such a way that diagrams commute so that p composed to f tilde is f. And this is not always the case, unfortunately. So for path, it works fine. But in general, there is a very algebraic condition if you want. So you take the f star, which is the induced homomorphism on the level of fundamental group. So this is the fundamental group of z. And this is the fundamental group of x. And this is the fundamental group of x tilde. And this is the same. So you take what is denoted here by f star of p1 of z. And want that this is a subgroup of p star of p1 of x tilde. So actually, this is an if and only if. So this geometric property of lifting a function if you want, which is geometry for the case of path, it is very geometric. So you want a curve to have another curve which can be projected over the curve we are starting from. And becomes, in general, an algebraic problem. So if this group inclusion is true, then the lifting, the lift is possible. Otherwise not. So there are examples where you cannot lift a function. For instance, assume that this p1 of the covering is trivial, which means that x tilde is simply connected. So this projection here, since p star is homomorphism, cannot be anything else but the trivial subgroup. And assume that this space here is not simply connected but can have very odd fundamental group. And if this is preserved and not mapped into the trivial subgroup, this condition is not fulfilled. So in general, what I'm saying is that if this z, of course it depends on its own f. But if this z has a complicated fundamental group and x tilde is very simple from the point of view of the motoping groups, then this condition can become difficult to find, to be fulfilled. Another, well, algebraic characterization of in terms of covering. So the number of elements in any fiber is the index of this subgroup in this group. Let me take this. As I said, p1 of x is not necessarily billion, so that this group is not necessarily normal. This is a subgroup of this, but this is not necessarily normal. A covering is called regular if the subgroup, the p star of the fundamental group of the covering, is normal in p1. In particular, if x, as I said before, if x tilde is simply connected, then this p1 is trivial. And then, well, the covering is, of course, regular. Because p star is a monomorphism of group and maps the trivial subgroup into the trivial subgroup. It cannot be different. And in this particular case, we have that, well, this covering is not only regular, but it's called universal. So the universal covering is a covering whose total space is simply connected. So once again, simply connected appears as a special case. In the previous example, and the only examples I gave you, we had a covering of S1 realized by the map t into cos t sin t. R is simply connected. This is an example of a universal covering. On the other hand, when I remove the origin from the complex plane, I obtain a non-simple connected domain. And the other example, so the plane minus 1 point covered by the plane minus 1 point and the function z goes into z to the power n, is not a universal covering. So in few slides, you will see why universal covering will play an important role, and also regular subgroups, regular covering, sorry, will be very, very important. When you have two coverings, say, you start from a covering of x, then you have a function from x into y, then you have a covering of y. So x tilde, x, y tilde, y. And then you have a diagram, two projections and a function. So you ask whether it is possible to consider a function between the total spaces continuing such a way the diagram commutes in. If this is the case, this is the associated covering function. And as we will see, this will be used several times and some consideration from Riemann surfaces. This is not assumed that the covering function exists given a function. It depends if the function maps fibers into fibers, or say points related to a chart. But in any case, it's important to know that there can be important. And a special case, if you want, we have the notion of that transformation, or covering transformation, or the morphism of a cover. So there are several ways to indicate the same stuff. So you start from covering x tilde p of x. And then you look for function from x tilde into itself, homeomorphism. So continuous with events continuous. And such that p composed of phi is p. So imagine that here you have a diagram with two projections, x tilde p, x tilde p. And here you have the identity if you want. So f tilde f is phi is the covering associated to the identity if you want. It can be seen like this in the previous slide. Or it can be also considered as the lifting of the projection of the covering function. Some basic facts. The set of all these transformations together with the composition. So we have invertible functions. So continuous with inverse continuous. Of course, the identity is always a possibility. Take the identity and you obtain a deck transformation. This deck transformations are important because they form a group under composition, standard composition of functions. And this group is called the deck transformation group. And they indicate it by the output. Normally out of standard notation. But it's important to now observe that when you have two functions with this property and you consider their composition, you can compose them because they are continuous from x tilde to x tilde. So you take another psi. Well, because both are deck transformation, also the composition is a deck transformation. It is easily same. What is the property in this geometric property related to this problem, to this condition here? Well, phi maps a point of the total space into another point. But the image of a point on a fiber has to be in the same fiber. Otherwise, this is not true, right? Take z. Phi of z is mapped, is projected into a point. But also P of z is projected to the same point because of this relation. So what is the action of the transformation, deck transformation? They permute the elements of each fiber. And this defines a group action, which is related to the automotive group. And what is also interesting to observe is that if you interpret this condition as lifting property of the projection, then none of the deck transformation but the identity can have a fixed point. Why? Because as I said, lifting property, well, it's clearly fulfilled here. No problem because, well, the P star of x tilde is containing P star of x tilde. So it's the same. Good. So you can lift P star, P into a function at phi, right? In particular, the lifted function is not uniquely determined unless you fix one value, right? As I said, it depends on the starting point, say, of this function. So if you assume by contradiction that there is a phi in the deck transformation group fixing that point, then necessarily it is the identity because the identity is also a lifting of P. And it is the one which fixes the point. Since there cannot be two lifting with the same starting point and the identity is for sure one of this, the other has to be the identity. And this is probably what summarizes the importance of deck transformations and the other stuff. So this group here is isomorphic to the question group of fundamental group of x over the normalized of P star of the subgroup P star P1 of x tilde x0 tilde. This is, in general, a subgroup which is not normal and cannot consider the quotient of a group with subgroup. So this explains why regular coverings are very much appreciated because in this case, you start from a regular covering, the P star of P1 x tilde x0 tilde is normal. So you don't have to consider a normalizer. You can consider this quotient. And this is a subgroup which is isomorphic to out P. In many cases, well, it seems strange, but it can be a way to calculate this. So in some case, for instance, for the calculus of the one possibility is to calculate the fundamental group of the projective space knowing that the only automorphies are the identity and the antipodal maps, which are acting on the double covering of the projective space. So you obtain information of the fundamental group of this base point, base space by knowing, by describing the de-transformation group. But here is what is the best you can have. Assume that you are considering a universal covering, which is in particular regular. So this is definitely trivial, not even regular, even more trivial because this P1, so the fundamental group of x tilde of the total space is trivial, which means that any loop can be deformed to the constant loop. This is the definition of simple connectors. So the classes reduce to points. So when I take P star of trivial group, and then this quotient is necessarily P1 of x, x0. So we have, for a universal covering of x, the de-transformation group is isomorphic to the fundamental group of the base point, base space. And this concludes the slides for today. Now, let us start from possible application to our setting. Assume that you have domains, something which can be covered by objects which lie in the Riemann sphere. So your freedom is to choose something which lives as a total space in the Riemann sphere as a topological space. And then the final function to cover domains or other objects in the complex sense. For instance, we have seen that the puncture plane can be covered by itself using several functions. And puncture plane is a non-simple connected subdomain of the Riemann sphere. However, what is important to know is that any good object, topologically good object, has a universal covering. And so in our setting, this means that we can refer to the possible choices of simply connected domains as covering of these objects, good topologically objects. What are the classes of simply connected domains which can play the role of total space in this game? Well, bounded simply connected domains are biomorphic to the disk. So I can consider simply the disk. Either the disk or the plane or the Riemann sphere itself. All right? That's why we have started the automorphism then of these domains. We have started the automorphism of plane, of the Riemann sphere, and of the unit disk. What are the characteristics of these three domains? They are all simply connected. They cannot be biomorphically identified. There is no homomorphic function which is non-constant from the plane into the unit disk. And similarly, from the Riemann sphere into the unit disk and so on and so forth. But the Riemann sphere is also compact. So we have also another character. So we have a simply connected compact total space. And this is the only possibility, Riemann sphere, or the plane, unbounded, or the disk. All right? So the next task is, next and final task is, can we describe the objects we can cover by this universal coverings a little more by studying the automorphism or by the action, or precisely, of these automorphies on these total spaces? This is what we'll do. And we will describe all possible objects which can be covered by these three spaces. In terms of quotients, of actions, of groups, of goods, of groups of automorphism, I will describe everything. And we will reobtain the results like Picard theorem and so on and so forth using this covering backgrounds. Well, I think that I can stop here today. And on the Monday, I will complete with some more specific details about the quality of the request, topological request of the objects, of the base points to be covered by the disk, the plane, or the Riemann sphere. OK. Thank you.