 Okay. So, let's start. So, we define fundamental group. Now, we have another basic concept, coverings, and then we come back to fundamental groups. So, what is a covering? A covering is a continuous surjective map. So, let P from E to B be a continuous surjective map, like a projection, P-like projection, surjective map. An open set U and B, so that's the definition, an open set U and B, B as bases, P-like projection, that's the sum of the notation. An open set U and B is evenly covered. So, we want to define what means evenly covered by P by this projection evenly covered. If the inverse image P-1 of U is the disjoint union, is the disjoint union of open sets, again open sets, V alpha in E for some index set, alpha and J some index set. So, I write again P-1 of U, the pre-image of U of this open set, it's a disjoint union of this V alpha. So, this is a disjoint union, alpha and J. And also, these are open in E such that this happens and each restriction. So, we take P restricted to V alpha, so this goes from V alpha to B, but to U, if you want to U, it's a pre-image of U, and each restriction to each of these V alpha is a homeomorphic. So, this means each restriction for each alpha and J. This seems to be strange at first sight, we'll see examples, okay? So, this is evenly covered. An open set is evenly covered, if it's pre-image, is a disjoint union of open sets, which all look like U, okay? They are all homeomorphic to U. So, we give an example in a minute, but first I will recover. So, definition. Continuous, so objective map P from E to B as before is a covering. If each point B and B has a neighborhood U which is evenly covered. If each B and B basis, the basis, if each point in the basis B and B has a neighborhood U which is evenly covered. That's the definition of cover. So, we should discuss some example. So again, the covering is a continuous subjective map. Each point has a neighborhood U which is evenly covered and evenly covered means the pre-image looks like a disjoint union of copies of U, okay, which are called V alpha. So, the restriction of P to each of these is a homeomorphism, okay? It looks like U locally, okay? You know this point. So, example. The prototype example is the following. This is a prototype example. So, we take P from the reals to S1. So, what is S1? S1 is a unit circle, okay? Unit circle. So, S1 is in R2 in the unit circle. What is it? All points X in R2, the norm of X is one, no? Or if you want even better, all complex numbers, the absolute value is one. So, I really, sorry. I really want to be this in C, not in R2, okay? For the definition. R2 and C is the same, okay? So, it takes a unit circle, but complex, okay? In the complex in C. So, what is P? P of T, you have to define P from R to S1 is E 2 pi I T. So, this is a complex number, no? This is a cosine, what is the angle? 2 pi T. 2 pi T is the angle. Plus I sine 2 pi T is the angle. So, this is in C, in the complex numbers, okay? This is, so we have a complex word here. So, it's clear what it is, no? It's, well, I'll make a picture. So, this is a mapping from the reals to S1. It's continuous, okay? So, whatever this is. This is the definition, if you want, okay? Of A, E, 2 pi I T, you can define this, okay? But then, you have the standard rules for E, right? You have E, something times E, something else is E, this plus something else, no? E to the A times E to the B is E to the A plus B. Now, this is the most important rule for the complex, for the exponential function, okay? And this holds here, also, for this. Well, so, how does this map look like? So, we have the reals, it's 1 minus 1, minus 2. Now, we have this projection, and here's S1. Unit circle, in C, in the complex numbers, in R2. E to the 0 is 1, okay? So, this goes to 1. This is 1, no? I just write 1, no? In C, we have the complex numbers, no? It's real. And then, you go, how does this look like? You have this alpha, no? So, you go 2 pi T from 0 to 2 pi, okay? And you have the, this is the angle. So, some point here goes to, some point here, goes to some point, well, it looks like almost like this, right? And, well, I don't, still don't like it. Some point here. And here, this is the angle, no, 2 pi T, okay? 2 pi T, this is the angle, right? This is 0, okay? So, you have the T here, then this is E, 2 pi, I T, or cosine plus I sine, okay? The two coordinates. So, this is the mapping, and so, what happens? If you go from 0 to 1, the angle goes from 0 to 2 pi, so it makes one full turn, right? 0 to 2 pi. So, if you go from 0 to 1 here with T, then you go from 1, and you make one full turn here, the angle goes from 0 to 2 pi, and you return to 1. E 2 pi I is 1 again, okay? E 2, so if you have T equal 1, it's 1 again, okay? It's a famous Euler almost, no? E pi I is equal to minus 1, okay? That's the Euler relation, okay? E pi I, which is interesting because you have E, the exponential, the Euler number, I don't know, E. Then you have pi, okay? You have I, not 1 and 0, I, I, the imaginary I, okay? And you have minus 1, okay? So you have, that's the relation between 4E to the pi I is equal to minus 1, okay? That's the, so we have the four most important numbers in mathematics together in this relation. One or minus 1, okay? If you went up to 0, yes. So you go from 0 to 1, you go one full turn here, okay? Then you go from 1 to 2, and you get another full turn, okay? The angle, always. And the angle is not too pi, no? So if you have, here you're too pi, then you start again, then you go another full turn, then you are here, you go another full turn, okay? And here the same, these are all full turns, okay? The angle, you can take model 2 pi, clearly, right? So what happens? So you see this mapping, okay? What is this mapping? This mapping is you start, 0 goes to 1, okay? And then you go here, you just go running around this, okay? That's the map, P. And if you go in the other direction, you run in the other direction if you want, okay? That's the map, geometrically. This is this map, okay? That should be clear, okay? This is this map. And this is, so this is the most important example of a covering, why is it a covering, by the way? When you take, what means covering? We have still the position. Any point of the basis, which is this, no? You should find, whatever. You should find, so here's a point, okay? Then you take a small neighborhood. So this is a subspace of C, okay? Clearly, okay? Subspace, subspace of R2, okay? Which is C, this is another version of C. Standard topology. The topology is the standard topology in R2, okay? This is the standard topology on the complex numbers, that's the same. So here you have a point, and now we have a small neighborhood, okay? Small neighborhood is an intersection with an open ball in R2, or in C, okay? So the intersection looks like this, right? That's this U, small enough, okay? And then the pre-image of this U, how does it look like? Well, the pre-image in this part looks like this, right? The angle between these two, okay? The pre-image in this part looks like this. The pre-image in this part looks like this. Here, this and this, and these are this V alpha, okay? So here you have V, what is a good notation, zero or one? Well, I don't know, V zero, here you have V one, this is V two, and this would be V minus one. Now here's the index, that is the integers, okay? In general, we don't know. And the restriction of P to each of these Vn, n, we don't have here, no, n. So this goes from these, Vn to U is a homomorphism, okay? That's homomorphism. Property, of course, of these functions, cosine and sine, okay? So it's a homomorphism. So it's a covering. So this is the prototype example of a covering, non-trivial covering. Of course, if you take the identity map from one space to the same space, that's a covering, okay? A homomorphism is a covering. But these are trivial examples, not interesting examples. A homomorphism is a covering, is a covering, okay? A homomorphism is a covering, but that's not interesting. This is not a, well, of course, you cannot take the whole S1, okay, as U, okay? That's not a homomorphism, okay? You have to take a small name. Well, you can take larger and larger neighbors, but you cannot take everything, okay? Then it doesn't work, clearly. This is the first example, and the second example is the finite version of this. This is our main example. This next one is, so we let n be in n, fixed, so fix the integer, n, so the natural numbers is one, two, three, not zero. So we take Pn from S1 to S1 now, and we are always in C here in the complex numbers. So I write Z for a complex number, okay? And this goes to Z to the n, n's power, n is fixed. So let's look at this geometrically. What does it do? So what is the n's power of this? Well, there's no n here, yeah. This n times the angle, okay? Of course, n times the angle, no? By this rule for the e function, okay? E, two pi i, e alpha, one, alpha i, one angle, times e beta i is e alpha pu beta times i, okay? The angle, yeah. So this is now two times S1. The basis is also S1, it's Pn, n is fixed. And now we have, here you have one, okay? One, n's power goes to one, of course. One goes to one, okay? And then now you start also here running in this direction, okay? And what happens if you have angle alpha here, you have angle n times alpha here, right? N's power, zeta to the n, no? If you have angle alpha here, then you have n times alpha here, right? So if this is alpha, then the image is n times alpha. This is the angle, no? So this point goes to this point, no? So you run also here, right? And then the first time, when are you back? At which point? So we start running here. And then we are back for the first time. I mean, we go here, no? In this direction. And I ask, when, here we have a full term. When or what's the length? Yeah, so the first one, which goes again to one, is e to pi i over n. So if you go from here to here, from this point to this point, then you go one time around, exactly. And then you go a second time. The next one is what, e two times. And so on, no? Two times two pi i over n. And then you have three times, okay? And when you have n times, and so, and so. So these are the n's root of unity, okay? n's root of unity, they go to one. Okay? The n's root, what do you finish, n's over n's root? So the n's roots of unity. What I say is, yeah, e two pi, e two pi i t, two pi t over n, that's the first one, okay? And then you multiply with m here, okay? So you get these, the first and second. You, how many do you have? N. Yes. One is one, okay, zero, and so on. Okay, so you see what happens with this map also? You go one time around here, okay? What happens down here? You go n times around here, okay? That's the mapping here. You go one time around here, you go n times around here, okay? With this mapping. e two pi i t goes to e n, n two pi i t, okay? n times, is the same, no? So you go one time around here, you go n times around here. So this is sort of the finite version of this, okay? Here you don't have S one, but you have the reals, okay? But here you never finish, so you go, okay? Here it's closed. This is the finite version of this, okay? This is the covering. This is also covering. And if you take a point here, then you have neighbor u here, okay? Small enough, and then what happens? The pre-image, like this, this, this, and so on. And these are the v alpha, okay? You're finitely manning it. And the restriction to e, each of these looks exactly like this, okay? That's the homomorphism. From each of these small, so these are the v alpha, and this is the u, okay? Here we have only n, v alpha, okay? Here we have infinitely manning it. So the index set, we don't know. So these are the most important examples of a covering for us, for the moment, okay? If you see other covering, but not known. To get a feeling for coverings, I prove two small things. The first is, so you get a better feeling for this. Exercises, so observation, so remark. If p, so we have also the covering now. p from e to b is the covering, we consider. Then each, I call this fiber. Then each fiber, what is a fiber? That's just the pre-image of a point, okay? I call it fiber, which goes to the same point. So b and b. Each fiber, that's a very convenient way to say, instead of pre-image of a point, you say the fiber, okay? Of a point. Then each fiber, p minus one of b, b and b. So what do you see here? What topology for the fiber? Discrete, okay? That's a discrete topology. As a subspace of e, okay? That's a discrete topology. It's discrete, as a subspace of e, of course. No, it looks like isolated points, no? The fiber of one is one, this, this, isolated points, okay? And here's a fiber of, here, finally many, here, infinitely many, but isolated, okay? So this means discrete topology, okay? This is true for every covering. So the proof is almost trivial, but the definitions are proof. So we have a b in, so we have a point b and b, no? And we have the fiber, let b and b. So it has an evenly covered neighborhood, u, okay? Let u be an evenly covered neighborhood, neighborhood of b. What does this mean? That means that the pre-image of u. So this means, by definition, the pre-image of u is the destroyed union. This is for disjoint, yeah, it's not so standard. No disjoint, union of open sets, v alpha, alpha and j index, I don't know how many. And each p and p are restricted to v alpha. That's the definition again. From v alpha to u is a homeomorphism. In particular, bijective, okay? It's a homeomorphism. In particular, bijective. What does it mean? That means that each v alpha contains exactly one element of the fiber, okay? So this implies each v alpha, because it's a homeomorphism, okay? It's bijective, it's injective, okay? Each contains exactly one element, contains exactly one element. Each v alpha contains exactly one element of the fiber p, p minus one of b, p minus one of b. So this means that the intersection, you, so each v alpha, the intersection v alpha with the fiber p minus, in other words, the intersection of v alpha with the fiber is one point of the fiber, okay? It's one point of the fiber. But v alpha is open by definition, okay? v alpha is open by definition. v alpha is open by definition of a covering. That is part of the definition, by definition. That means the points of the fiber are open, okay? This is the subspace topology. Now this is open in E. This is the fiber intersecting, we get a point, and we get all points taking the right v alpha. So v alpha is open by definition of covering, and this implies that points of the fiber are open. In the fiber, okay? That's the subspace topology, no? So this implies points of the fiber. All points of the fiber are open in the fiber, not in the E, but in the fiber. So this is the discrete topology, no? Points are open, okay? So this is a typical feature of a covering. The fibers are discrete spaces, okay? There are, of course, other vibrations in mathematics, many, no, maybe you didn't see one, but you have, in general, the fiber might be a vector space or whatever, no? If you have a, you don't have differentiable manifolds, right? So far, or yes? Yes. Then you have the tangent bundle, okay? You have the tangent space, and you have the tangent bundle, and you have a projection to the manifold, and now the fiber is a tangent space, okay? Which is a vector space. This is not discrete, okay? This is a vector space, non-discrete. So that's the vibration. This can, you can interpret as a vibration, and the fibers are discrete, okay? This is, in some sense, a very simple case. So this is some, another thing. Okay. A second small property, which is an exercise. Maybe I give, I write this as exercise, exercise. Well, I will do it, exercise. So P from E to B is also covering now. So maybe you ask, what the hell is this now? We define fundamental group, and now we're doing something completely different, no? But the fundamental group of the S1 we want to compute, okay? We use this. And we talked about some rotation number, okay? This is something to do with rotation number, okay? If you have a path in S1, then you ask how many times does it go around S1, okay? And that you see in this covering, okay? Also, okay? The covering from R to S1, no? If you go from zero to the positive, then you just go around and go around. If you stop at n, you have something which goes n times around, okay? So there's some rotation number in this, okay? In this covering from, and that we will use for computation of fundamental group. So what is this exercise? P, E, A covering, B connected with B connected. Then each fiber, P minus one of B, B and B, each of these has the same number of elements. It might be infinite, no? So what does it mean? Number of elements. What do we say is the same? What is the word for this? Coronality. It might be finite or infinite, doesn't matter, okay? So this is exercise, I will do this, proof. So it's a covering. So we fix the point. Let fix the point B, zero and B. And we consider the fiber. So P minus one of B, zero, well. So we have the fiber on this fiber of B, zero is a certain number of elements, okay? And now we consider all points in B, in the basis, where it has the same number of elements. So let X or whatever, X, B, all elements B in B, such that the fiber, the fiber P minus one of B, of this point, has the same. This is our reference point. It has the same number of elements as the fiber of this fixed point, okay? The fiber has the same number of elements, the same cardinality, the same cardinality. Number of elements seems finite, but okay, so I write cardinality, no. The same cardinality as the fiber fixed of our fixed point, okay? We prove that X is equal to B. And that proves the exercise, okay? This implies the exercise. All fibers have the same cardinality, okay? So what? So we have B is connected. So what do we prove? Yeah. So B is not empty. Since Y is not empty, B zero is there, no? No, X, sorry. X is not empty. Since B zero is an X. That's our reference point, okay? So it's not empty. So first is X is open in B. And then it's closed also, okay? And then it's everything, right? Because B is connected. The important thing is here that B is connected, okay? So the standard argument for connected spaces, X is open in B. No, let B be in B or in X, okay? B is our basis. Let B be in X. So it means the fiber of B has the same number of elements as the fiber of B zero, okay? Which is our reference point. But it's clear. And you, an evenly covered neighborhood of, and you, let you be an evenly covered neighborhood, neighborhood of B. So what does it mean? Again, the definition for the third time, no? That means a preimage, P minus one of U, is this joint union. This joint union of this V alpha, alpha in J. And P, the restriction is only morphism, okay? And the restriction to each V alpha, from V alpha to U, core restriction also, is only morphism. But this means, of course, that each of these, as before, no? Contains exactly one element of each fiber, of each fiber of an element of U, okay? Because this is bijective, no? So this implies each, P is a homeomorphism, each fiber, P minus one of B with B in U, no, in this U, has the same number of elements. Because what is the number? The cardinality of J, of the index, just no? With B in U, has the same number of elements. The same cardinality. Which is the cardinality of J, of the index? How many, do we have, okay? Which is the cardinality of J, of the index of J? Just the cardinality of J, okay? But for one of these B in X, this is in X, okay? So we have the right cardinality, okay? For one. So they have all the same, so this implies, they have all the same cardinality. And so this means X as U is contained in X. This means U is contained in X. This implies U is contained in X. For these, they have all the same, okay? If you have one point which has the right cardinality, then all has the right cardinality, okay? So U is contained in X. And this means X is open, no? Because we have B here, okay? And they both, so this implies, what means open, open means given a point. We have to find a neighborhood, open neighborhood, which contains X, so this implies X is open, okay? And now X is closed. Second, X is closed. Well, we do not prove this, we prove that X is, that X minus, no, what? B minus X is open. But the argument is now the same. That means we take a point here, okay? Then we don't have the right cardinality. But there's a neighborhood of this point where we have the same cardinality. So we don't have the right cardinality as before. Choose a point, B and B minus X. This does not have the good cardinality, okay? And U, and this now implies that U is contained in X minus B, all the same cardinality. And this means X minus B is open. And all together, I and two I is a typical argument, implies X is equal to B since B is connected. And that's the end, this is X, okay? So another kind of definition. So we have a covering. So now we have a covering and covering, I think from E to B, P, this is a base space, this is a space which covers the base space. So this is a covering as well. And now suppose we have X is something different. Now F continues and we want to lift this map. So a lifting of F, F tilde, a lifting, so this is given, P and F given. A lifting, lift or lifting, lifting of F is a continuous map, F tilde from X to E, such that this is commutative, okay? Such that this is commutative, this means first F tilde and then the projection is equal to F. So very easy, we have a continuous map into the base space and I want to lift it to a continuous map into the total space, into the space, okay? That is the lifting. Everything continues, of course, that's the lifting. And now we have to work a little bit, two facts. The first one is called the path lifting lemma. So this is the first we want to prove when there will be a second similar. So we have a covering, let, again, let P from E to B be a covering with P of, we fix points. P of E zero is B zero, if we fix some points. Then any path, F from zero one, a continuous mapping from zero one to B to the base space with initial point F of zero is our fixed point which is called B zero, okay? Starting in B zero, any path in B starting in B zero has a unique lifting. So this is lifting, unique lifting to a path F tilde from zero one to the E to the total space E such that uniqueness such that F tilde of zero. Well, we have a point which goes to B zero, it's E zero, no, no, initial point E zero, okay? It's E zero, this makes it unique. We fix the initial point, okay? We can take any point as initial point here, okay? So then it's not unique, of course. Or if you fix the initial point, then it's unique. So this is a path lifting, you know? Passes can be lifted, and it's unique if you fix the initial point, okay? That's a special case of the situation, right? That's a path lifting lemma. So I will prove the path lifting lemma now. The main point is the compactness argument. I mean, we lift piece by piece, okay? A covering, it's a local homomorphism also, okay? It's a local homomorphism. Locally, it looks like a homomorphism, okay? You have U, and you have all these V alpha. And the restriction to each V alpha is homomorphism, okay? So locally, it looks like an homomorphism, okay? And then we use these U's to lift the paths piece by piece. In each U, we can lift, okay? Because it's homomorphism. That's no problem, okay? And then we have to pass together, finally, many of these U's to get the whole path, okay? To lift the whole path. That's a compactness argument, okay? It might be that they get smaller and smaller and smaller, but the interval is compact, zero, one. So by compactness, finally, many will solve. That's a Quebec number lemma again, if you want, okay? It's a compactness argument. So this is the whole idea. We lift piece by piece, the path, okay, in these evenly covered, that's no problem. And then we need only, finally, many to cover the whole path, okay? To lift the whole path. And that's a compactness argument, because otherwise it's a problem, maybe, okay? Because they might get smaller and smaller, okay? And then we finish and we, okay, so the proof. Choose a covering U of B, U's of the base base B, U written U, not printed U, okay? As usual, a covering by element, by open sets, which are evenly covered. Each point has a neighborhood, which is evenly covered, no? By open sets, which are evenly covered. By definition of covering, this is possible. Each point has a neighborhood, which is evenly covered. So choose a covering of B by open sets, which are evenly covered. What are we given? We are given this path, no, F. And then we want to lift to this F tilde, no? This is, by open sets, which are evenly covered. So the pre-image then, then the pre-image of this open covering. So F minus one of U, what is this? Well, of course, it's F minus one of U, U and U, okay? What is this? It's an open covering of the interval. It's an open covering, zero one. So what is zero one? Zero one is a compact metric space, clearly, not clearly. It's compact, and it's metric. Standard topology on the wheels. It's a compact metric space. So then we can apply the Lebesgue number lemma. The Lebesgue number lemma, what does it imply? It says there's a Lebesgue number associated to this open covering, such that each open set, which is diameter, smaller than this number, is contained in one of these. That means the image is contained in one of these, okay? And so we can find the subdivision of the interval in small sets, okay? So I write immediate as this. So there are points. There is a subdivision, a zero equal one, smaller as one, smaller, smaller as n minus one, smaller as n, which is the last point one, of the interval of zero one. There's a subdivision in smaller intervals, such that f of each of these small intervals, si, si plus one, is contained in at least one u and u, okay? So make the distance smaller than, make the length of these intervals smaller than the Lebesgue number, okay? Then it's contained here, and the image is contained in one of these. It's contained in at least in, so we're finitely many, now, this is the final point, okay? We're finitely many of these small intervals on, okay? Finally more. And now we can define our map, we're all lifting it, okay? Now we define f tilde. So that, of course, we have no choice for f tilde of zero equal to what was it, e zero. That we prescribe, the initial point, okay? That makes it unique, then, as you see. But it doesn't matter which point to take here, any point in the fiber, it's okay, you know? Let's say, suppose that f is defined, suppose that f tilde is defined already by induction, recursive definition, by induction. It's defined already on zero si, on the interval from zero to si, somewhere here, okay? And then we define it on the next one. And then we define it on the next one, okay? Then we define it on the next one. Then we define it, we define f tilde on si, si plus one, that's what we want, okay? And then we're finitely many, so we go on to the end, okay? And finitely many steps. Well, suppose, so this is all contained, let f tilde, f, sorry, of si, si plus one. So it's contained in one u in u, no, in u, I call it u. U element contained in u, element of this u. So for the first time, the preimage of u, these are evenly covered, okay? So the preimage is this usual disjoint union, disjoint union v, alpha, alpha and j. This point means disjoint union, sometimes. And again, p restriction to v alpha from v alpha to u is a homeomorphism. That's the definition of a cover. That's vf. But f tilde of si is already defined, okay? f tilde of si, let f tilde of si. This is already defined, no? Already defined, because suppose that f tilde is defined already on this interval, so this is the last point where it's already defined. So the projection of this is of course f of si, okay? The projection is f of si. So this is in one of these, so it's in the preimage of u, okay? It's in the preimage of u, f of si is in u, okay? f of si is in u, so this is in the preimage, is an element of pi minus one of u, of course, okay? Suppose, no, no, not that. I wanted to write something else now. This implies this is in the preimage of u and I now let f tilde of si, so be in v alpha zero. It's one in one of these, okay? Alpha zero, it's unique also, okay? f tilde of si is in one of these, and this called v alpha zero. Okay, now we define f tilde, so for t in this new interval si, si plus one, define f tilde of t. So what do we do now? f tilde of si is in v alpha zero, okay? But on v alpha zero, we have a homomorphism, p restriction to v alpha zero. So I just take f of t and then I go back in the other direction with the homomorphism. So this is p restricted to v alpha zero, which is a homomorphism, so it's the inverse homomorphism, okay? We just lift by this inverse homomorphism, okay? By the, again, on the point si, we have the same value, we have the right value, okay? There's only, so then by the pasting lemma again, all the pasting lemma, f tilde is defined now on what? On zero, si plus one, no? You go one step further. It's defined on, and continuous on this, okay? And continuous. That's pasting lemma, I also continue. Continue. And also continuous. In finally many steps, f tilde is defined on zero. So in finally many steps, f tilde is defined zero one on the whole. And that's, this implies existence, okay? So this implies existence of f tilde. But we need also uniqueness, okay? So we want to prove. So uniqueness. It's important also. So this is the existence, and now also uniqueness of f tilde. So in zero, we have no choice, okay? f tilde of zero, we say we want e zero, that's by definition. So here's the, suppose it's unique. Again, the same argument, okay? Suppose, in the first point, it's unique. Suppose f tilde is unique, is unique on zero si, okay? We prove that it's unique also on si, si plus one. We prove it is unique. We have no choice. It's unique also on the next. Si plus one. And this of course implies an f tilde is unique on the whole. So this implies f tilde is unique. Zero one, what we want to prove. So it's unique on zero, we have no choice here, and then we have no choice for the next. I should see this because it's the same proof. We prove that it's also unique on si, si plus one. So as before, as in the proof which we did for the existence, f tilde of si, let f tilde of si, that's unique, okay? B, what was it called in the alpha zero, all right? The same notation as before. So we have P minus one of you, this is union of the alpha zero, right? Exactly the same notation as before, okay? So f tilde is in the alpha zero. F tilde is now defined on the next. It's defined also here, but we need uniqueness. So this f tilde, where's f tilde of si, si plus one? So f, sorry, first f, f, this is contained in you. Now this was a you, okay? f of this is in you and you. This we fixed in the first proof, okay? The same notation as before. So f of this is in you and you, and f tilde is in one of the pre-image of you, okay? This is exactly the same as in the existence proof. This means, of course, that f tilde of si, si plus one is contained where? It's contained in a pre-image of you, no? P minus one of you. What is P minus one of you? It's the disjoint union of this v alpha, alpha in J, okay? Okay? Important thing is these are open. All these are open. And now I say that this implies that f tilde of si, si plus one is contained where? It's contained in v alpha zero. This is one of these, okay? f tilde of si is contained here, okay? But then the whole has to be, but why is it? f tilde of si is here, right? That's our choice, okay? And then I say the whole interval must finish here. And not in this, we're also here, but in one of these only, okay? Why? That's the only interesting, what? I don't understand, what? And open, yes. It's in one of, in both of what? No, no, well, the point si is okay. Now si is in v alpha zero, okay? And then I say, then this must be all in v alpha zero. Why? That's one word. Yeah, connectedness, okay? It's connected. So this is connected. This is connected. If it's connected. And this is the disjoint union of open set. It must be one of these open sets, okay? Otherwise it would not be connected, okay? Since f tilde of si plus one is connected. But now here it's a homeomorphism, no? And this, we have no choice here, okay? So this means that f tilde restriction to si si plus one, what is it? It's a lifting of this, no? It's exactly as we defined, okay? So this f, it's f restriction to si si plus one. And then we go back with the inverse, homeomorphism. We have no choice, okay? P restriction to v alpha zero minus one, okay? That's how we construct it, no? And we have no choice. Because in each fiber, there's exactly one point. We have to take that point, okay? So this is the inverse method, okay? And that's it. So this means, that's the end of the proof, okay? So this implies what, it's unique also here on the next piece and then it's unique everywhere, okay? So this is uniqueness, very similar. The only interesting point is here disconnected. That's a crucial point in the second part of uniqueness, no? So this is the first lemma. Pathlifting lemma. And now we have to prove a second lemma, which is very similar also on si si plus one. This implies f tilde, so it's unique also on this one. Because here we have no choice. It's all in v alpha zero, okay? Not yet. Okay, so this is the first lemma and now the second lemma is called homotopy lifting lemma. So what? So again, p from e to b is a covering. Same hypothesis and we fix points. P e zero equal to b zero. Let f, so let f from no tactic, i times i. This is a unit interval, okay? I wrote zero one, but i is not zero one as usual, okay? To b, be continuous. It looks like a path homotopy, but be continuous with f, f of, so zero times zero. This goes to the b zero, our fixed point. Our fixed point. Let f from e be continuous. Then there's a lifting. f tilde, then there's a lifting. Then there is a lifting which is continuous. f tilde from i cross i to a to e to the total space. Then there's a lifting f tilde from i times i to the total space such that f tilde of zero times zero is e zero. If in one point we have put that together. Then there's a lifting f such that if. It's also unique, but we are not interested in this, okay? In this point. It will be also unique, but here we don't care too much about uniqueness. However, what we say, if f is a path homotopy, okay? Then also f tilde is a path homotopy. That's what we're interested in. If f, what means path homotopy? If f is a path homotopy, what means path homotopy? Constant on the sides, okay? Constant on, what are these two sides? Zero times i and one times i. That means path homotopy, okay? It's constant on these two sides. Here the homotopy passes and constant on this. That's the definition of path homotopy, okay? If f is a path homotopy, which means constant, then also f tilde is a path homotopy. Finished. What is constant on? That's the same as before. Then also f tilde is a path homotopy, okay? So this is a homotopy. Path homotopy, not arbitrary homotopies, okay? It's a special case of i times i. So I indicate, the proof is very similar, so proof. So we apply again, as before, we apply the Lebesgue number lemma, okay? Lebesgue number lemma. What does it imply? So we take an open covering of B by evenly covered open sets, okay? Then we take the pre-image. Now the pre-image is in where? In i cross i, okay? i cross i, the pre-image. First was i, now it's i cross i. However, i cross i is compact. It's a metric, it's a subspace of R2. So it's a compact matrix space. So again, we have to do the back number lemma, okay? And what does it imply? There are subdivisions. Now we have two subdivisions, one of the first and one of the second, okay? You subdivide by these little rectangles, okay? You hold. There are subdivisions, whatever they're called. What are the subdivisions? S0 equals 0, smaller S1, smaller Sm, now, which is equal to one. And the second one, T0, which is 0 also, smaller T1, smaller, smaller T, well, it might be the same or not, it doesn't matter. Tn equal to one of i, okay? Such that each of these little squares, f of, and now I have to write this little square, si, rectangle, not square, si, si plus one times Tj, Tj plus one, f of this is contained in some, at least one, u and u is contained in at least one u and u. So u as before, okay? This is the first one. So you will open covering of, but as before. So now the picture is like this, now they have i cross i and we have here the S, S1, S2, and so on, Sm minus one, Sm, and here we have T, T1, T2, Tn minus one, Tn. And now we have these little, we have these rectangles. Looks like this, no? And this is one of these small stuff, okay? So now I prove existence of lift, okay? Existence of, we do the same as before. So we have, here in this point, we have no choice, f tilde of zero times zero is E0, no? And now suppose it's already defined where? So we start with this point, then we define here, then we define here, then we define here, then we define here, then we define here, then we define here. Now we have a choice, we may go here or go here, it doesn't matter, okay? Then we go here, then we go here, then we go here, then we go here, this is, then we jump again to this. Then the next one. Okay, we got to this point, now we have to do the same for the next one, okay? So suppose, the same proof more or less, suppose that f tilde is defined already on, it's defined already on, well, let's concentrate, no? This would be the next one, okay? So it's defined already on zero, now on I cross, so this is I times, this is Tj, no? Tj, let me write first, Tj minus one, this is, well, Tj minus one, zero Tj minus one, I times, this would be this, okay? Tj minus one union, and now we have this part, okay? So this is what? This is zero, S i minus one, this is this part, right? Times Tj, Tj plus one. That's exactly what I describe here, okay? I try, I hope it's okay, yes. Tj, Tj minus one, yeah, that's okay, then this is the next one, right? Is that true? S i, S i plus one, here we have S i minus one, okay? And then the next one is S i, no, S i minus one, S i. Sorry, it's not correct. I don't need i minus one, I need i and j, okay? Sorry, okay, that's better. So zero S i, and then the next one is S i, S i plus one. I don't need i minus one, okay? S i times Tj, Tj plus one, and here also Tj, okay? Then we want to define it on the next one, okay? Then we want to define it, F tilde on, and then it's exactly this. I want to get this as the next one, okay? On S i, S i plus one, times Tj, Tj plus one. And this is exactly this. And recall that this is contained in one u and u, okay? F of this is contained in one. Then we want to define F tilde on the next one. If you can do that, then we go on this, this, this, this, this, this, and we go finally minus one, no problem, as before, okay? Where is it already defined? It's already defined on, F tilde is already defined on, well, it's of course already defined on all this stuff, about here, on this part, no? It's already defined where, here, and here, right? So what is this? It's a concentration, how to write the right elixis here, the picture, no, it's more clear than that. So it's already defined on what is this piece here? That's S i times S i times, what is the interval? So this is S i times Tj, Tj plus one, Tj, Tj plus one. Union, this was this first, and now this one, what is this? No, S, no, no, first this and then times Tj, no? So S i, S i plus one, S i, S i plus one times, yeah, Tj, right? That's already defined, no? By induction, and now what is the main argument? What is this? Connected, this is connected, that's the main part. This is connected, okay? And then we have this FS is contained in this u and u, okay? I wrote, okay? It's in some u and u, I fixed this u as before, okay? And the P minus one of u is union v alpha, alpha and j, right, as before, it's the same as before. And this means this is connected, so this means, so of course f of, f, I don't rewrite, this is in u, okay? So f tilde of this is in P minus one of u, okay? So this implies f tilde of, no, I should give a name because I don't want to write this. It's already defined on, so what is a good name? X, it depends on i, j or whatever, okay? So why, let me just give you a name for this, okay? Without indices, without anything, so that's why. So this means f tilde of y is contained in a single one, it's connected, okay? So it's contained in, which one? In, where's my v alpha zero? It's contained in v alpha zero for some alpha zero in, this is a single one, because it's connected. It's contained in general in the union here, okay? But it cannot be, since it's connected, it must be in a single one, this is this joint union of open sets as before, okay? So it's in a single one, so we find it. And then we are done, no? No, define, so define f tilde on, what if another problem? Well, this is the next one, here's this one, right? I have to write it otherwise, si, si plus one, times tj, tj plus one, by f tilde, so what, maybe take f? So this is on u, no? F and then p restriction v alpha zero minus one, as before, okay? I don't give the arguments, okay? So this part you say, okay, it's the same as before. It's on infinitely many steps, f tilde is defined, okay? So now we are here, this and this and this and this and this and this. So this implies infinitely many steps, f tilde is defined on everything, on i cross i. And that's a uniqueness, so this is existence, sorry. Existence, not uniqueness. So we have existence of f tilde. The uniqueness, we are not interested. But we are interested that if f is a path homotopy, also if tilde is a path homotopy, okay? Suppose f is a path homotopy. That means f is here and here it's constant, constant, okay? Here it's constant and here it's constant. On this and on this, it's constant, that's path homotopy, okay? Which constant? This point anywhere goes to b zero, no? So it's b zero in this point, because this point goes to b zero, so it's f. Here I don't know, okay? So constant c b one doesn't matter, some other point. So okay, but we don't care. Constant c b one. And then we have f tilde. We already constructed f tilde from here to here. And what you have to prove is that it's constant here and here, no? That's what I have to prove, okay? If tilde restricted to this, let's consider this, okay? What is this? Restricted to zero times i. That's this side, no? Is a lifting of the path, well, of this path, right? It's a, this is a lifting of this. So it's a lifting of the constant path. Well, which one? C b zero in b zero, okay? C b zero in b zero in b. This is a lifting of the constant, because f tilde is a lifting of this, okay? If you project, it's a lifting of the constant path, C b zero. But the constant path, if you lift the constant path, then we remain in the fiber, okay? So the lifting, the lifting of the constant path, C b zero, a lifting, any lifting, a lifting of C b zero for a lifting of the constant path. The lifted path is in the fiber of b zero. Remains the lifted path. A lifting of C b remains in the fiber p minus one of b zero of this point, okay? A lifting of C b remains in the fiber p minus one, one of b zero. But this is a discrete space, no? This is a discrete space, which is discreet. There's a fiber in the discrete space, okay? Any path in a discrete space is constant. Any path in a discrete space is constant since i is collected, okay? The interval is collected. Again, since i is collected. So that means f tilde, so this implies then, f tilde, I respect it to, what was it, zero cross i is constant, okay? So it's constant here. What is this constant? Of course, this goes to, so it should be C zero, okay? In a similar way, I mean, it doesn't matter, no? So here we get C e one for some e one, no? In a similar way? Similarly, it's constant, so this is similarly, f tilde restricted to one times i is constant. Same, exactly the same argument. Okay, so this means f tilde is a path homotopy. So this implies, I don't know what to write. This implies f tilde is a path homotopy. If f is a path homotopy, f tilde is a path homotopy. That finishes the proof of the second level. And also the time. Well, it's perfect, almost. These are the work we have to do, okay? And now, next, tomorrow, we give, this immediately permits to compute the fundamental group of the circuit, okay? Which is that, by rotation number, we said, no? But we have to give a proof. And the proof follows, the work is these two lemma, okay? Lemma term. Now we have these two, and now it's easy, okay? So tomorrow, we prove that the fundamental group of the circle, unit circle, is z integers, okay? But given by rotation number. And then this has various applications, okay? This has, we will give at least two or three applications, okay? Okay, so, we'll see tomorrow.