 Okay, welcome to all of you. So in module 10 as promised earlier, we will do what is called as Seifert Fan-Campan Theorem. The simplest form will be taken today. Later on we will see, you know, more advanced versions of this one. Seifert was a German mathematician. Fan-Campan was an American. Most of the time in America it is just called Venkampan's Theorem. They don't take the name of Seifert. But to be precise, they did it independently, not just a joint work and they did it in different assumptions, different contexts. So today we will try to cover up simpler versions, but which appeals to both the contexts. That is the whole idea. Okay, so this is the statement. Let X be written as union of two open sets, U and V. Both U and V are open and the intersection is path connected. Once since it is path connected, you pick up a point in the intersection, say X naught. And let us assume that the inclusion maps of U in X and V in X, eta and psi, let us call, they induce homomorphisms, eta check and p check. Okay, they induce homomorphism, eta check and p check, which are trivial homomorphisms. Everything goes to the identity element of pi 1 of X, X naught. They are trivial homomorphisms. They are inclusion maps on fundamental group level. They are trivial homomorphisms, eta check and p check. Okay, eta and V themselves are just inclusion maps. So this is the assumption that eta check and p check are trivial homomorphisms. X is the covered by these two open sets. Intersection is path connected. Under this, we can conclude that the fundamental group of X at X naught itself is trivial. You understand the whole thing? For example, suppose pi 1 of U X naught itself is trivial, then the homomorphism will be trivial. Similarly, pi 1 of V X naught itself is trivial. Then again the inclusion map will be trivial under inclusion induced map will be trivial when you pass to the fundamental group. Right? So if U and V are simply connected, the union is simply connected, provided the intersection is path connected. So this is one way of remembering it. The theorem is slightly generalization. U and V themselves may not be simply connected, but all those loops inside U as well as inside V separately thought of as loops inside X, they are not homotopic. So these two are different statements, right? U itself is simply connected, then they will learn a little bit inside U. So that is a stronger hypothesis. So under weaker hypothesis, namely inclusion induced maps are trivial. We will get that the fundamental group of X itself is trivial. In other words, if fundamental group of X was not trivial, some element of U, some element loop in U or something in V must have been non-trivially mapped into pi 1 of X. This is the meaning of all this. So let us prove it now by totally elementary methods, totally elementary methods. No covering space is nothing of that kind. So here is the picture of what is happening. So U and V are open subsets like this. Their intersection I am assuming is path connected. The entire space is your X. Now I have chosen the, in the intersection I have chosen a point X not. So the idea is if I take any loop inside V like this, when you use the entire space it is non-homotopic. Similarly any loop which completely lies inside U, the inclusion map will take it to X that it is there it is non-homotopic. That is the assumption. So suppose I start with a loop which is gamma not like this and I have divided this whole thing is gamma dot dot dot dot dot and it is coming here and then coming back here. Only the last two portions I have drawn with thick arrows, thick lines. So this is my gamma not. This is my gamma. So what I have done is I start tracing gamma not this gamma part. Now I see that it is going soon. It will be going into V part from U part. So here I stop and cut it. I mean I am making a subdivision. Remember we could make subdivisions. Now from this point onwards I am inside U. So somewhere here I have to stop because now I will be going inside V. So that will be say you can call it as gamma 1. Now I go here and call this part as gamma 2. Again I am entering into V. So I say gamma 3, gamma 4, blah, blah, finally gamma n coming back. What I have done? I have cut down the path, the loop into subdivisions such that each path is either inside U or inside V. Okay? Once I have that I use the path connectivity of U intersection V. Look at after going here I have stopped somewhere. Where do I stop? I have stopped inside U in seeing that I am now going inside V. So this point is already inside U intersection V. The first point on the path, I mean 0th point is this one, the first A1 is in the path that is already inside U intersection V. So this is only a path. Now I complete it to a loop by joining A1 to gamma naught. This entire path gamma naught composite lambda 1 is a loop inside V situated at x naught. Therefore by my hypothesis if you think of this as a loop inside x then this is null homotopic. Because this path is null homotopic I can just ignore all this path and just go directly from here to here and look at the rest of them. This path is a null homotopic. You can add or subtract, add or you know delete this path there is no problem. So delete it. What has happened? The number of divisions in the gamma has reduced. So by induction the whole thing is null homotopic. What is the induction starting point that there is no division that means the entire thing is either inside U or inside V then it is null homotopic is hypothesis. So that is the starting point of the induction. So if I cut down this one then I have only from here to directly you know this gamma naught is not there. First thing is this whole thing is gamma 1. This entire thing is inside U now. Then gamma 2, gamma 3, gamma 4, gamma 1 up to n instead of 0 to n. So that is only n minus 1 path. Therefore by induction the proof is over. Okay. Now let me just go back here and see what details I have written and how to write down this. Okay start with a loop gamma in X at the point X naught. Okay. U and V cover X therefore gamma inverse of U and gamma inverse of V will cover the interval 0. Just like we did last time we can choose a you know Lebesgue number slightly and a number is slightly smaller than Lebesgue number for this covering. Then if I choose the length of these intervals I am cutting it into n parts that the 1 by n is less than this delta then what happens is the respective the intervals Ti plus 1 to Ti they will be all less than say delta by 2 modulus of this difference. Okay. That would mean that each closed interval is either inside U, gamma inverse U or gamma inverse V. It is the same thing as saying that gamma restricted to this interval is either inside U or inside V. The point starting point X naught is in both of them it is in the intersection. So the first part namely T0 to T1 this part is in one of them which one you just for that means it is a gamma inverse of V on the right side. Okay. By dropping some of the points this you might have cut it too much. Okay. So T1 to T2 may be also inside V then you better take both of them and don't cut together 0 to T2 all of it inside V. So like this you can combine all the things which are inside V till you come to a point next arc is going to be inside U. Okay. You can rename T1, T2, Tn in such a way that 0 to T1 is inside V, T1 to T2 I mean the part of the curve not the interval T1 to T2 inside the next one U and alternately U, V, U, V and so on. Okay. So that is just for saving some argument. So Ti to Ti plus 1 is inside V for all even numbers and for all odd it is inside U because I start with 0 to T1 inside V. The next one will be odd. So it will be inside U and so on. Okay. So that is the picture here. The next step is to show that gamma is homotopic to a constant loop, the constant path in C x, x naught. So we induct on the number of divisions required to express gamma in the above force. If n is 1 this already implies that gamma itself is contained inside U or V. Therefore, the hypothesis is P check is a homomorphism, a trivial homomorphism. So we are done. Okay. Maybe I should write here V. It should distribute V. Okay. P check is from pi 1 of V to this one. This must be V. Okay. It is a control. I assume now n is greater than or equal to 2. And that the claim holds whenever we can express a path in the above force with fewer than n divisions. So that is an induction hypothesis. Now put gamma equal to gamma restricted to Ti to Ti plus 1. So that what gamma looks like? It looks like gamma naught star gamma 1 star gamma n. Okay. Looks like means what? This is these two are not same paths. They are path homotopic. You remember that. If you divide then each of them you have to express all of them in terms of where path you have to be all the time parameter is from 0 1. So you have to take the composite of this. Then this is path homotopic. This is what we have seen. And gamma i are alternatively inside V and U. Okay. V, U. V, U and so on. That is the picture that we have seen. Alright. Now look at Ai which is equal to gamma Ti like A1. A1 was in the intersection of U and V. So choose a path lambda i from Ai to joining from X naught to Ai. X naught to Ai is lambda i. Then what happens is the loop lambda n minus 1 star gamma n. Okay. Bayes at X naught is completely contained inside U or V. So I check out the gamma 1 but you can check out the last one here. Deliberately I have done this one. I agree. So here also in the picture the lambda n minus 1 go like this. Come back gamma n that is null homotopic. Therefore if I by induction hypothesis this path all the way going up till here and coming back that will have only one less number of divisions. So this will be also null homotopic. So the composite of two null homotopic things is null homotopic. So you will get gamma itself is null homotopic. Okay. So this is the way to write down a proof that is all. Any questions? Here I have written down full detail again. Start with gamma which is divided into n parts. These paths are not necessarily loops. Their end points could be different. Right? Gamma naught starts at X naught ends up somewhere in A1. So you have to convert them into loops. So first do not worry about this path. The last path you join it by lambda n minus 1. Insert lambda n minus 1 and lambda n minus 1 inverse and lambda n. This is the inverse of this so I can insert it because this is null homotopic. Okay. Then you use associativity and put these two things in a bracket. This becomes a loop in either U or V. So you can ignore this one. This is constant. I mean homotopic to constant. This path together this whole thing is will be inside the other one U or V. So it is one thing. Rest of them is one less so the entire thing is now n minus 1 only. Therefore by induction I suppose this path is gamma naught, gamma 1, gamma n minus 1 prime, n minus 1 thing. So this is also null homotopic. So together with this null homotopic the whole thing is null homotopic. So it is a way to write down. Okay. So proof is over. If two open sets are such that they are simply connected and then they intersect in a path connected subspace then the union is simply connected. This is a corollary to this one. Right. As I have indicated. Any questions? Let us now you know prove a very interesting result. We have computed pi 1 of S1. Okay. Now we are computing all the all the spheres. Pi 1 of SN for all n grad I am going to prove. In one go what is read all of them are simply connected. The fundamental group is trivial. Strictly speaking you should remember that I should put a base point here. If I am claiming that this is trivial for one base point it should be trivial for all base points. We have seen that because if we choose change of base point the two groups are isomorphic. That is what we have seen. So I have not written down what is the base point. That is all. It is not a mistake. This is deliberate just to cut down the notation. That is all. But in my mind it is there the base point. Whatever base point you take it is same. That is why I have not mentioned it. Pi 1 of SN is trivial. SN is simply connected. All right. For n greater than or equal to 2. For n equal to 1 it is infinite cyclic. For n equal to 0 what is it? For n equal to 0 it is not even connected. This S0. To take any connected component again the fundamental group is trivial. Okay. So in that sense only n equal to 1 is a distinct element. All other things are simply connected. A space which is not path connected is never referred to as simply connected. In the definition of simply connectivity you first assume it is path connected and then put the condition that the fundamental group is trivial. Therefore you cannot call S0 as simply connected. But fundamental group of S0 taking any point is trivial. That is true. Okay. Yeah. So what is the proof? Proof is very easy. All that you have to do is write the sphere as union of two open sets in a nice way. What do I do? I select the north pole 00001 subtract it, throw it away. Similarly take the south pole 0000 minus 1, throw it away. U and V are open subset because they are SN minus single point. Single points in SN are open. So these sorry close so complements are open. So once these things are open the union you have to check is the whole space. U misses one point that point is already inside V. So U union V is the whole of SN that is fine. Now comes how does SN minus 1 and SN minus minus 1, one single point looks like. So this is where you have to know elementary topology namely any point if you remove from a sphere the rest of them is homeomorphic to RN by stereographic projection. I hope you know these things if some of you do not know you can ask your tutors and you must know this before the end of this course. If you have learnt it in the beginning of the course if you remove one point from S1 what you get is a space which is homeomorphic to R same thing happens to remove one point from S2 you get a space homeomorphic to R2 and so on. In particular U and V are simply connected because they are homeomorphic to contractible spaces. So they are themselves contractible. So they are simply connected. Finally you have to look at what is how does U intersection V looks like. U intersection V is the sphere minus two points. Here you have to use n is greater than or equal to 2 not n equal to 1. If n equal to 1 and remove North Pole and South Pole what you get is two arcs disjoint arcs that will not be connected. If you remove one point from the sphere you get a RN n greater than or equal to 2. So if you remove one more point is still connected. Therefore what intersection of U and V is connected actually path connected. So all the hypothesis of the above theorem are satisfied actually stronger hypothesis. Namely pi 1 of U and pi 1 of V are themselves trivial. What is the conclusion? Pi 1 of SN is trivial. Is it okay? Yes. Now here are a number of exercises. Each of you should try to solve them by yourself. Then only you will know that whether you are understanding this course or not. So you have to rely more on your self-assessments here. Whether you take an exam or not. So the tutors will help you in understanding things by if at all you communicate with them by telling whether your answers are correct or not. So let me just go through these exercises. These exercises will be separately sent to you in a PDF format. Right now you do not have to write down these things and so on. So the first one is suppose X is path connected and pi 1 of X is abelian commutative group. Okay for some A inside A, A is a some point in X. Then for any B inside X any two paths is tau 1 and tau 2. Joining A to B, you have h tau 1 and h tau 2. Right? These two homomorphisms are the same. This is what you have to do. Okay, h tau 1 of some omega is equal to h 2 of some omega where omega is a class inside pi 1 of X A. Second exercise is take any homomorphism from Z to Z integer to integer group homomorphism. Okay? You can think of this as induced by a map from S1 to S1. Given the continuous function from S1 to S1, you can fix up some point say 1. You can also assume 1 goes to 1. Okay? Then pass to the fundamental group pi 1 of S1 to pi 1 of S1. But they are infinite cyclic groups. Right? Z to Z. So that homomorphism will be alpha. Every homomorphism is given by some map. So this is what you have to do. Okay? Moreover, up to homotopy, there is a map is unique. Suppose you choose two such maps F and G. Such that their F check and G check are the same on the fundamental group. Then I want to show, I want to claim that the maps themselves F and G are homotopy. Okay? Then there is a remark that this result is true for all spheres but not for arbitrary spaces. So for spheres, this goes under half degree theorem which we may not be able to do in this course. Okay? Take a subspace A contained inside x and a point A inside x. Inclusion induced homomorphism A to x is subjective. If we know if every loop in x based at A can be homotopped to a loop in A. Suppose omega is a loop in x based at A, then there will be a loop to tau inside A and these two will be homotopy. That is the meaning of this one. Further, if A is path connected, then show that this statement that pi 1 is subjective is equal to saying that every path in x with end points in A can be homotopped to a path inside A. Okay? So this is, this also, these are straightforward exercise you have to work out. The next exercise is about the product. The product of two spaces pi 1 of x cross y to x naught y naught is pi 1 of x x naught cross pi 1 of pi 1. Product of two spaces, the fundamental group is also a product of the corresponding space. So product of the fundamental group of the corresponding space. In particular, you have to write down what is pi 1 of s 1 cross s 1. Now that is, that is very obvious once you know the new state. This notation is isomorphism of groups here. Next, once you have computed this, the next exercise depends upon, that is why they are branched together here, depends upon this exercise. Look at a map f from s 1 cross s 1 to s 1 cross s 1 given by z 1 z 2. These are unit complex numbers. Okay? So you can multiply them. Z 1 z 2 comma z 2. Z 2 remains as it is, z 1 gets multiplied by z 2. Look at this function. What happens to this function when you pass to the fundamental group level? What is f chain? You have to compute. All right? This you can do once you have done this one correctly. This exercise, you do not have to do worry about right now. Only after several, some more lectures have been done, namely after we have done a live session, you can discuss mobius band. Okay? So I have told you some, some of them, but we can, you can take away all of them also. So for example, this one says if two functions from x 2 any sphere. Okay? This is sphere s 1, s 2, s 3 whatever. Okay? These two are maps such that a point of f, f of x will never be the negative of g x. Okay? No for no point. f x and g x are not antipodal. Then f is omotopropagic. Okay? So there are enough exercise here for you to work out. The next module will be some set theory, sorry, some set topology, point set topology. These point set topology in an elementary course in point set topology might not have been covered because they are somewhat advanced topic. So I will cover it to the extent we will need it in this course. Though the title of the function space and so on, we are not going to do the entire function space theory as done in point set topology. That will divert the course. So we will do only things which we need. Okay? Rest of them if you want more, you will have to pick it up from some point set topology book. All right? That is for the next module. Thank you.