 Okay, so let me restate something that I mentioned last time because I forgot an important hypothesis. And just to review as well. So we had two coverings of a space x. And what I neglected to mention is that we want to, we need to assume that these spaces are connected. And so we fix base points y0, y0 prime, both going to x0 under the projections. These are coverings. And then the statement is that y is isomorphic to y prime if and only if the images of the corresponding fundamental groups are the same. So here isomorphism means that phi is a map that composed with p prime gives you p. And we want this to preserve base points. So y0 is taken to y0 prime. So I stated this last time and I didn't mention that they were connected. If you don't have that, of course, you can have trivial covers of x of all kinds of forms. You can take x itself or you can take many copies of x multiplied by arbitrary discrete set. And they would suddenly not be isomorphic. Okay, so this is going ways into this theorem that I've been hinting and which I'll finally state completely today. So we're going to associate to a covering y of x, a connected covering of y of x, the image or the fundamental group of y which we already saw is a subgroup by the covering map p lower star. And we're going to have a sort of a bijection between covers with a fixed base point and subgroups of the fundamental group of the base. Okay, the other issue is the existence of universal covers. Is there a question? Yeah? So an isomorphism, and he's asking what an isomorphism is, is a homeomorphism between the spaces that is compatible with the maps so that if you have, so this triangle, we have y, p, y prime, p, p prime to x so that these two things commute. So going this way, you save us that way. So the isomorphism backwards doing the same thing. Okay? So this, I wrote it in symbols, but if you want a picture, that would be it. And to make this, so I also want to include the statement about the base points to have the statement precisely as it is there. If you change base points, what happens to the fundamental group? We always say in, I didn't write the hypothesis, we want x to be path connected and locally path connected. So if you have such a space x, so if you have a connected, sorry, a path connected space, what happens to the fundamental group? How does this depend on the base point? It's conjugation, not just isomorphism, but conjugation, so that the groups are conjugate. Okay, so if I don't, if I'm not careful about the base points, if I ignore the base points, what happens is that these two groups, well, I can say is that they're conjugate to each other. But if we are careful about keeping track of base points, then the statement is that the groups are on the nose the same. Okay, so the existence of universal cover is something important. It's good to have a universal cover. Many things can be done with it and arguments simplify if you have it. So here I'm going to spend just a little bit on the technicalities of this question because it's going to be the case that in order to ensure that there exists an universal cover of a space, you not only need it to be path connected and locally path connected, but you need some other new technical thing that we haven't seen. Okay, so you'll need the three hypotheses, which are going to be satisfied by most spaces you ever want to consider, but they're technically completely logically independent of each other. Okay, so I'm just going to state it. I don't want to spend unnecessarily long time on it. Okay, so let's look at this question. So we have a covering. So recall that a universal cover means a cover of the space that is simply connected. We would like to be able to say that if X is sufficiently well-behaved space, it has a simply connected cover. And we've seen examples of this already. S1 has R. R has R. We looked at the figure 8-0 and we drew this picture. So we would like to know to what extent we can always be sure we have such a universal cover. Okay, so we have a covering map and let's assume that it is a universal cover. So P1 and Y with respect to some base point is trivial. Okay, so recall from the definition of covering that if you have a point X in the base, it has a neighborhood that covers, that is evenly covered. Okay, so if you have a path, sigma in U, we can lift it to Y to, well, let's say V and because Y is simply connected, this lift sigma total is homotopic to a point, to a constant map. So that means that the same happens downstairs because you can apply P to the homotopy and show that this path, which is sigma, is also homotopic to a point. Y, so that means in this, so what we conclude is that X, that every point in X has a neighborhood where all loops are homotopic to a point. Okay, so we're going to call this property semi-locally, semi-local path connected, simply connected. And again, I don't think we should lose our sleep on this technical issues, but this is the exact, so this is a necessary condition for the existence of a universal cover because the existence of the universal cover provides for you these open neighborhoods with this property and the point is that the converse is true. So a connected, locally path connected X has a universal cover, if and only if it is locally simply connected. So this is the technical precise statement and as I said, most of the spaces you want to consider are going to, that satisfy the first two conditions that are going to satisfy this third as well. But you can, of course, it's an interesting challenge. You can see in Fulton's book, in fact, an example of spaces that don't quite satisfy all of the things at the same time. But the point I want to keep and given that we really don't have enough time to elaborate on all of this sufficiently well. So as you know, today is the last lecture that I give next time. Lothar Getcher will continue and he will start with homology. Shift to gears and it will be a slightly different topic. I mean, always within algebraic geometry but I don't think he will be mentioning much about the fundamental group. So it would require a lot more time to go through all of this more carefully. Okay, but the point that we should keep then from this is basically if we want to say it quickly if you have a well-behaved path connected space then it will have a universal cover. And because of the statement we at least mentioned before, if we apply this proposition to the case of a simply connected spaces, if both y and y prime are simply connected then pi one of them is trivial and therefore so is going to be the images and so any two simply connected covers that are connected are going to be isomorphic covers. So once you have the existence of a universal cover it's unique up to isomorphism. Okay, and again something I mentioned several times if you like or you familiar with Galois theory this is equivalent to the existence of an algebraic closure of a field. Yeah, that you construct it in somewhat abstract way and it's not, I mean there isn't a God-given way to express the algebraic closure, right? I mean you can create different algebraic closures but they're going to be isomorphic. Is there a question? Okay, so let me continue then with this story of the fundamental group I want to eventually end. So what I'd like to do today is end with the description of this equivalence between subgroups and covers that I mentioned and also talk about the van Kampen theorem which is possibly the most powerful tool that allows you to compute fundamental groups. Okay, so you could look as I said in Fulton's book for examples of these different concepts just one last phrase about this. Here we use the word semi-local simply connected and you could be wondering rightly what is local simply connected. And those presumably are different notions and they are. And here we're claiming in semi-local that there's a property about, so there exists a neighborhood with a certain property. In the case of locally simply connected you want this sort of standard idea of local which is that every neighborhood contains a neighborhood which is simply connected. So stronger than this one. Anyway, these sort of very technical matters I don't think we should be too much concerned with but in a spirit of being precise and giving correct statements that's what is required for the existence of universal cover. It's somewhat weaker condition than just having this every neighborhood having a simply connected sub neighborhood. Okay, so here now this is finally a bit more of meat to the story. So let's again consider the situation of a covering. X is path connected and locally path connected. So this issue was mentioned last time that I used the word connected plus locally path connected and somebody mentioned soon we consider it to be path connected. So should I take the word path out or leave it in? Well it doesn't really matter because the space is assumed to be locally path connected. So every path connected, every connected component of a locally path connected is path connected. So anyway, again this is getting too picky on the technicalities but this is a sufficient, it's a correct hypothesis for what I want to say. So what we've been saying is that we have a covering and I'm going to assume my top space is also path connected. So there's no funny issues with various slices or something. And attached to this covering we have an object which is the subgroup or the fundamental group. We'll fix some base point Y0. This is a subgroup or the fundamental group downstairs. So we have a Y0 mapping to an X0 here. Okay, so we're trying to understand what exactly is the connection between the subgroup H and the covering itself. And so here's what we have. So let's look at the group of automorphisms of the covering. That is to say, let's look at all the maps of homeomorphisms between Y, from Y to itself, that commute with P. So P of phi is P and preserving the base point. So taking Y0 to Y0. So this is part of the... No, sorry, I take that back. I don't want that. So this is true, but this is part of the definition of P. Don't need it in the set. Sorry, take it back. I just look at maps that commute with P in the sense. So this is called the group of automorphisms of the cover or also called deck transformations. So the claim is that the automorphisms of the cover, naturally isomorphic, to the normalizer H over H. So H is the subgroup of pi1 of X, the normalizer of H. This is in pi1 of X. It consists of all the elements G of pi1 such that G H, G inverse, gives you back or takes you back into H. So the thing to notice here that these are two groups that are pretty different objects. One has to do with homeomorphism of the space Y to itself that preserve the map P, the covering P, and the other one is something to do with the fundamental group of the base. So in particular, if Y is simply connected, what can we conclude? Well, let's see. If Y is simply connected, pi1 of Y is trivial. So our H is trivial. So this H is the trivial group. What is the normalizer of the trivial group? Anything. The whole group. Anything will take, will, you know, G1, G inverse is always one. So any G will do. So this, the normalizer of the trivial group is the entire group you're doing the normalizer from. So in this case, N of H. So, well, H is one, the trivial group. N of H is the whole group. And what then can we conclude? This statement says that the automorphisms of the cover are isomorphic to the fundamental group. So this is an important fact. And in a sense is what we used to prove that the fundamental group of the circle was the integers. So you can think of this result as saying this is giving us a way to describe the fundamental group of a space by saying, well, you somehow find the universal cover and then you compute the automorphisms of the cover, which typically might be an easier thing to consider. Again, if you think in terms of, in the spirit of Galois theory, think of X as the base field, Y as the algebraic closure, and this will be the Galois group of the algebraic closure. So in this language, the fundamental group of a space is like the Galois group of the algebraic closure of the field over itself. So let's try to prove this. So let's try to get a handle on what these automorphisms are. So let's say we have two points Y0 and Y1 in the pre-image of our base point X0. Okay, so, well, first notice that if we have such a map phi, automorphism that preserves P, what should phi do to say Y0? Where should it go? Well, phi has to commute with P, right? So if you have a point Y0 that maps to X0, then here it has to go over there to some point that also maps to X0. So if you take a point in the pre-image in the fiber about X0, then this automorphism has to take you to another point also in the same fiber. So what it's going to do is going to move around the points on top of X0 or above X0 was arbitrary, about any point. Okay, but we want to connect this automorphism group with something to do with the fundamental group downstairs. So the space upstairs is path-connected. So pick gamma, a path in Y, going from Y0 to Y1. So what does this path do for us? We have on one hand the fundamental group based at Y0 and on the other the fundamental group based at Y1. The picture we have Y0 and it goes to Y1. So what do we discuss about these two groups when they're connected by a path gamma? They conjugate, right? So you basically follow the path to Y1, then if you have a loop at Y1 and you come back, that goes, allows you to give a path in Y1 to a path in Y0. So I think, let me see if I get this in the right direction. So if I have a path in Y1, I first do gamma, which is here. Let me call this path gamma twiddle. And then once I do the loop in gamma one, I take the inverse to come back. So now let's call gamma the composition. So pull it down. What kind of path is gamma? It's a loop, right? Because Y0 and Y1 are both in the fiber about X0. So once I project things down, we're going to have a path that starts at X0 and ends at X0. So gamma is a loop in X, excuse me, at X0. So if I take images by P of this equality here, I'm going to get P lower star of pi1 of Y, Y0 equals to the class gamma P lower star pi1, Y, Y1 inverse. So H is what we called this group. So now we could repeat this story for any gamma in the fundamental group of the base. It will lift the sun gamma twiddle. So the same identity will hold. So the point to make here is that gamma inverse H gamma is P lower star of pi1 of Y, Y1. So if we want to capture when such a gamma is in the normalizer of H, so this class is in the normalizer of H, well that means that this is equal to H. So this is equal to H if and only if that is equal to H because they're the same. So this is if and only if P lower star pi1, Y, Y0 is equal to P lower star Y pi1, Y, Y1. All right, so we want to characterize what elements in the full fundamental group normalize our subgroup H and we just saw that it translates into this statement. Now what does that statement say? It's an equivalent form of the statement that we discussed earlier. The first thing we did, or the second, now the first thing we did today. This statement is equivalent to what? That there is an isomorphism between the covers Y with base points Y0 and the cover Y with base point Y1. So there is a map phi that makes this diagram commute and phi of Y0 is equal to Y1. And it's equivalent to this. So we then can jump over and connect the first with the third. And so a gamma, a path gamma is in the normalizer, a path gamma that when you lift it goes between Y0 and Y1 if and only if there is a automorphism that takes Y0 to Y1. So we can look at this association a little more closely. And so what we just did is associate to somebody in the normalizer of H some automorphism of the cover. And it's not hard to see that this association is a homomorphism. And it's, why is it subjective? Yes? Suppose we have an automorphism. It takes Y0 to some point Y1, which are on the fiber of X0. Why does it come from some element in the normalizer the way we constructed it? Well, what would be the gamma you would pick is the path that goes from Y0 to Y1. And so if you follow this construction back, you'll get another possibly a different automorphism that takes Y0 to Y1. Why is it the same as the one you started with? Because Y is connected and basically by the uniqueness of all these liftings that we discussed. So once this map, any one of these isomorphisms is completely determined once you know where it takes 1.2. So this shows that this map is subjective and so now we like to show that the kernel is H. And so what would be the kernel? We associate automorphisms with paths upstairs and we just said that the automorphisms are completely determined by where they take 1 of the points on the fiber. And so in particular we're going to get the trivialized idomorphism if what? If Y0 goes to Y0 to itself. Once you know that Y0 goes to Y0 then it has to be the identity automorphism. And so what is the gamma path associated to this? It has to be a path that starts at Y0 and ends in Y0. So it has to be the class of a loop. And so that means that when you project it down that would be exactly somebody in the group H. So the kernel of this map is H because if so phi of Y0 equal to Y0 is the same as saying that phi is the identity of Y which is the same as saying that the path gamma belongs to H. Okay so maybe there's a little bit of thinking there to do but not hard. So let's review what is it that we proved because this is a very important fact. So we consider this group of automorphisms of the cover and what we argued is that it's isomorphic to the normalizer of this group H that we attached to the cover itself and possibly the most useful corollary or instance of this statement is the case where Y itself is simply connected in which case the normalizes everything and you identify the automorphism group of the cover with the fundamental group of the base. So let me finally state the statement that I hinted at several times and I think we've proved a significant amount of it. I think I'll just state it. So X is path connected, locally path connected, semi locally simply connected. So although this is a big mouthful and the statement is that if we look at covers of our space X together with a choice of base point which are connected or what's the same that they're path connected because in a covering you're going to be able, if it's connected you're going to be able to connect it by lifting paths downstairs up to isomorphism, isomorphisms that preserve base points, then there's a one-to-one correspondence with subgroups of the base, of the fundamental group of the base and the association is that you take a covering Y goes to X and you attach to it the subgroup consisting of the image by the covering P lower star of the fundamental group upstairs. And this is, in a sense, the exact analog of the fundamental theorem of Galois theory. So interpreted correctly, it tells you how to go between subgroups and coverings during Galois theory you go between subgroups and extensions. Of course if you do infinite Galois theory you have to be a bit more careful in the statement but roughly speaking this is the style of result. Now if you're not worried about base points then if you don't specify things about base points then you don't really have a subgroup of the fundamental group. What you have is a conjugacy class of groups because if you don't specify the base point there's no way to tell which exact group it is that you're talking about. You're going to get possibly different conjugates of it. So if you relax this statement and so if you ignore ignoring base points you're going to get a bijection between connected covers up to isomorphism and conjugacy classes or rather let me put it this way the same as before modular conjugation. And in this association so this would be sort of one form of it a little bit more precise and less precise form of this correspondence. In this correspondence the trivial subgroup corresponds to the universal cover. So you see that implicit in this statement is the existence of the universal cover. That's why you need all these three hypotheses on your base base X in order to guarantee that such a universal cover exists. So I think I'm going to leave this statement for that as like that for now because I only have a little bit more of time and I want to discuss the fine component theorem. So there's a bit more than one can elaborate on this theory that I'll invite you to read on your own or if you want to discuss it further you can always come talk to me later but I'm not going to discuss it in class because we're running out of time. Except that perhaps I typically like to think of examples with complicated concepts so why don't we give it a little bit of a few minutes of thought to try to understand this statement in the context of for example these coverings of the figure eight knot that we were discussing. Again I mentioned this several times I encourage you to actually do something like this take some examples like this coverings of a space and try to understand the various theorems that we proved, various facts in the light of the example so I don't make sure you understand the examples as deeply as possible as a way to internalize what the theorems say. But for that I probably should add one statement no maybe not okay let's see what we can do let's look at this cover number one can you see it? We'll write it down here we have the two points A, so these are examples A and then B which is the cover of the figure eight knot okay so what so this is a cover this is a Y, this is an X this group is the free group generated by A and B so this cover gives us a subgroup of this of what index? Two this is a degree two cover yeah what is the group of automorphisms of this cover automorphisms of the cover so this is our Y mapping to our X automorphisms Y, X I can't quite hear two elements he's saying that the two elements in the automorphism group is so therefore Z mod 2Z it can be in any no other group it has two elements why is that? and it has degree two what did we prove about the automorphism group of the cover? it's equal to then it's isomorphic to the normalizer of the group that it corresponds to the cover modulo the by the way notice that this is actually a quotient in the group's theory sense because a group eight, any group subgroup H is normal in its normalizer right you basically build this normalizer so that every element in the normalizer actually normalizes the subgroup so it's normal so it makes sense to talk about its quotient it's not as H is not necessarily normal in the whole group that would only be the case if the normalizer is everything so this automorphism group is isomorphic to the normalizer of H divided by H so what is the normalizer? well H is a subgroup of index two any group, any subgroup of index two is normal so the normalizer is the whole thing so this is isomorphic to Z mod 2Z so what could be this one non-identity homomorphism from the Y to itself we should be able to see it what do you think? what should it do to the points one and two to begin with? this is our phi, right? a phi goes from Y to itself and these two points are in the fiber of that point so what should this phi do? it has to permute these points if it fixes one of them is the identity so the only thing that is left it has to flip these two so if I take this cover and do this would that be a deck transformation? would that be a phi? an element of the automorphism group yeah, okay? it has to be okay let's look at the next one or number three in this list so what degree is this cover? three the index of H in the whole phi 1 is 3 okay? so what is the automorphism group of this cover? what are the possibilities? it has index 3 so the normalizer is something that contains it I mean the normalizer always is a group on top of the group you started with everybody in H normalizes H just because it's a subgroup so normalizer could be anything it has to be something possibly bigger it's either H or something all the way up to the full group so if it has index 3 what are the possibilities for the normalizer? it's either H or it's the whole group there's no other group in between so either it's the whole group in which case the subgroup H is normal or it isn't and the normalizer is H itself so the automorphism group has two possibilities so what could it be? if the normalizer is everything then the normalizer of H divided by H is isomorphic to what? it's a group of order 3 so it has to be cyclic of order 3 and if the normalizer of H is H the automorphism group is trivial so it's either trivial or it has a order 3 automorphism so stare at this picture and tell me what you think it is so for example this point will have to go over to some point that looks like it so this guy has a cycle of B attached to it so where could it go? it can't go anywhere there's no other point in here that looks like it this one doesn't have a loop B attached to it this one doesn't have a loop B attached to it so in fact it's the identity so this cover has no automorphisms and in fact if you switch the base point what you're going to see are three different subgroups so this means that the normalizer is itself which means that there are three different conjugates of this subgroup inside the whole pi 1 and that will correspond precisely as to picking the three different base points if you do that you're going to get a subgroup of the group downstairs that is going to be for each choice a different subgroup anyway I think I'll stop with this subject and move to the van Kampen theorem which is somewhat unfortunate because it's a big deal and we won't have time to dedicate it as it deserves but I think I feel pretty bad if we went through this course without even mentioning this statement so van Kampen theorem and I believe that Getsche is going to cover as I said homology and you're likely to see the mayor, Beatrice theorem which is a homology form of this theorem so the issue at hand is the following we have a space X which is constructed as a union of two open sets u1 and u2 and we want to relate the fundamental group of X to that of the three spaces u1, u2 and u1 intersect u2 in order to give the result in this theorem we need to discuss what it means to take the free product of groups so if we have two groups g1 and g2 we define g1 star g2 this is the symbol for free product of groups and is defined as follows the elements of g1 star g2 are words in g1 union g2 so think of words meaning strings of symbols each one of the elements of this word is taken either from g1 or from g2 so at the moment I just have long strings of symbols each one of them coming from either one of these two groups we can certainly define a multiplication of words by just concatenation if you have two words you just put them one after the other so that gives you a way to multiply words and that would not make you make this into a group so you also want to have some notion that a multiplication that came from the groups themselves so what we're going to do is we'll mode out by a relation which is that if you have an a1 and a b1 b i b i that belongs to g i and if you see a i b i next to each other in your word so this is our word w then this should be equivalent to the same thing but where now you replace the pair a i b i by the product of them where the product is done in the original group they belong so if I have an orange and an apple I don't do anything with them I have an orange and an apple following but I have two oranges together I put them together and I make a double orange so you multiply elements of the group if they happen to be in the same group because you know how to do that and if they don't belong to the same group you just leave them as they are so again to do this formally it's a bit tedious but hopefully you can see how this works this will produce for us a group which is called the free product of the two groups and there is a categorical way to define this so if you have a g1 and a g2 then we have the free product of the groups which have maps because you can look at a word consisting of just an element of g1 or just an element of g2 and so there are homomorphisms like this and this construction is universal to the property that if you have any map from g1 and g2 two maps to a group g then there is a unique map that goes from the free product so it's sort of universal with respect to this property anyway but that's not quite too relevant for us at the moment just to get maybe a handle on this concept what happens in both groups g1 and g2 are the integers what do you think it's g1 free product g2 sorry let's go one at a time I heard one thing, what are you suggesting? Usain's z, how about back there somewhere free group generated by two elements something else well let's try to follow the construction and see what we would get so z is the same one so we want to skid these fingers then so we don't get confused so what would a generate of one over z is a and a generate of one of the other z is to be b so what would be the words well we just have a collection of a's to some power possibly negative power then some b's then some a's and then some b's and so on and if we see a whole bunch of a's we just twist them together and put a to the right power and if we see b's together we squash them together so at the end of the day I have now a word which consists of a to a power b to a power, a to a power, b to a power and so on and the only thing I can do is if I see an a next to each other I can add the exponents so if I see a to the 5 and a to the minus 5 well I can cancel those out so which we need to do to multiply things out and so what we get is what we were discussing before the free group in two elements so this is f2 free group in two generators and while we discussed and we did a little bit in towards proving this group is actually the fundamental group of what? of the figure 8 not and in fact now once I state the Fankampian theorem it will be a simple consequence of the theorem that that's the case we kind of argued bit loosely last time but this would be a different approach in a very unambiguous and rigorous way to do that so what's the theorem? so in the simplest form so simple form first you can probably guess it by now so I probably have to assume various things so u1, u2, u1 intersect u2 I want them to be path connected and so in the simple form I will also assume that u1 intersect u2 is simply connected and so x is the union and the theorem is that pi1 of x is isomorphic to the free product of pi1 let me put it x0 it's x0 pi1 x0 is somebody in the intersection so it makes sense to talk about these two groups so it's the free product of the fundamental groups of each one of the open sets okay so this is great and is a powerful tool to calculate and understand various fundamental groups of various complicated spaces so as I said unfortunately I'm not even going to try to give you a hint of how you prove it it's not particularly different because it's kind of long and tedious if you can look it up in Hatcher and he does it in his full generality full tone takes a slightly simpler approach if you already know that the spaces in question have universal covers then the proof is a little easier but regardless we're not going to do that here so I'll just go ahead and use it so let's take our space x to be our loved little 8 so I'm going to take say u1 to be this open set u2 to be this open set and then the intersection is this open set so what can we say about the fundamental group of the intersection is trivial because why? it's contractible you can contract every path so it fits precisely in the hypothesis of the theorem pi1 of u1 point and pi1 u2 point are both isomorphic to z because this little extra leg theory you can contract each open set looks like a circle so homotopy type of a circle so each one of them has a z and now what this theorem says is then the fundamental group of the figure 8 is the free product of z and we argued that's well that's exactly what we said before that that's a free group in two elements so if you like the definition of a free group of two elements you can take it to be precisely that is the free product of z so what would happen if you have a bouquet of three circles what is it sorry they why why this so I haven't written any conclusion so the theorem implies okay so please tell me again what is your question the theorem or so the theorem says now if you have two open sets that their union is our space x such that all three spaces the open sets and their intersections are path connected then and on top in this form of the theorem the intersection is simply connected right then that's an assumption although this up to here are assumptions this is the conclusions maybe then so if then okay so if the intersection is simply connected then we have the statement that the fundamental group of the union is the free product of the fundamental groups of each one of the open sets that is that clear okay so if we had a bouquet of circles so the figure not eight anymore but that was the fundamental group of that yes we can do it as we did before and what would you get sorry let me finish we would get the free product of the four circles that we see indeed so we would have for each one of the petals of this flower we would get a copy of z and if you iterate the argument that we did just now you would get z free product z free product z four times so the fundamental group here will be esomorphic to the free group in four um generates okay which is the same as z for z for z for z for z good so we would take a lot more work to try to figure that out without this term so this is a pretty um powerful theme alright now how about what's wrong with this so take x to be s1 and take u1 to be this open set and u2 say with a one base point there and u2 to be this each one of them is simply connected right okay so does this contradict the term why not right if you take the intersection this consists of two pieces which is not path connected this is very crucial then for the hypothesis of the term that you are in the situation where all three pieces of this are path connected um let's do another example um let's take x to be the two sphere okay we already proved that the pi1 of the two sphere is trivial we did it for any sphere of any dimension bigger than one but let's do it again using van kampen theorem we will see that it's a lot easier than all the maneuverings we need to do before so how would you do this we can take the sort of top half and the bottom half so going down a little bit of space below so they have a non-trivial intersection so we have this is u1 and u2 except that I now realize that I haven't we can't do it yet what's the intersection it's a little ribbon right so it's not path connected but it's not simply connected so I cannot apply the theorem as we have it so I'll have to postpone it because I need now to give you the form of the theorem that is the full form without the assumption of simply connectedness of the intersection okay so let me give you the general form okay so hold on a minute for this example so the general form of the theorem so the hypotheses are the same except we no longer assume that the intersection is simply connected and so the conclusion in this case is that pi1 of xx0 let me write it out and then we'll discuss what it means so there's a bit of a mouthful of symbols there so to say this in words you would say this is the free product with amalgam or just an amalgam okay and to describe this let's go back to our groups g1 and g2 they have maps to this free product but there is also possibly a subgroup h that may be maps to both so if we follow h down the top arrows we'll end up with a subgroup of the of the free product if we if we follow it from below we get a different subgroup and so while we like this force is the fact that we want the two subgroups to be the same to have the same image so an h viewed through this path on the top or the maps on the bottom should end up giving us the same element in the image and so what we're going to do is just impose those relations I'm trying to find the formal definition sorry so I suppose you have the g1 and g2 free product with amagen so what we're going to define is g1 free product g2 with an h below okay and in h is going to be another third group that will have a map to both g1 and g2 okay and we want those images to coincide in the in the group that we're constructing so what we're going to do is take the free product of g1 and g2 and divide by the subgroup normal subgroup generated by those relations so let me I want to give you the precise form written it down but I don't have it so we want if you want maybe I'll put the h here what we want of this group is is the universal property that if you have any map from g1, g2 to another group g so we have a group g and a map two maps like this that agree so we should give these names this is i1, i2 this is h1 and h2 this is j1 and j2 so if h1 composed with y1 i1 is equal to h2 composed with i2 so going this way is the same then we want the existence of a map here then there exists the map so it's universal with respect to the property not only that you have maps from g1 and g2 but those maps agree once you restrict them to the image of the subgroup h so one way to think of this is what I was saying earlier is that you basically take the words in g1 and g2 to form the free product and then you divide by the smallest normal subgroup so that you can make a quotient of groups such that the images of both of h through both of these maps into g1 and g2 agree and of course if the h happens to be the trivial group the trivial subgroup then the images are the trivial subgroup and then there's no condition you know the trivial group will always be the trivial group no matter where it comes from and so the so if h is the trivial group then g1 g2 is just g1, g2 the free product so the statement that I wrote for the Fundcampian theorem is as the other case a special case of this one in the case when this is simply connected this is 1 and then I don't need to worry about this amalgam business okay alright so let's see if we can get a sense of this theorem in the remaining time so let's go back to the case that I had to stop because I realized I don't have the yet so here's our sphere yes what is the relationship for the maps that I wrote look the i's and the j's and so on what is this map this is just a way to state what the group g1 free group g2 over h is you can define it in this categorical way that gives you a clean way to identify it but it doesn't tell you quite what it is so the universal property that I want is that this group is a group such that if you have two maps h1 and h2 into any other group g such that the compositions all the way to h agree then there is a map from this one to the g so it's the universal one in the sense it's the smallest group that has this property any other group you can achieve this way you can reach this way you can do it through this one so it's a typical sort of universal type construction unfortunately it doesn't necessarily give you a big insight as to what it is I tend to like to think is is the idea is that you have these versions of elements of g1 and g2 where whenever you see an image of h that came from one group it has to equal the image of h that came from the other group so let's try this if we go back to our sphere we had these two semispheres so what is pi1 of each one of these two open sets is one right this is a contract is a disc is the same as a disc is a contractable space so this is one what is pi1 of this z but in fact it doesn't matter because what is this theorem telling us the van Kampen theorem says that pi1 of the sphere is isomorphic to the free product of the trivial group with itself over the image of z so the problem is that the notation is kind of bad because it's not z but sort of the image of z by these maps and the image of z by these maps have to be trivial because both are the trivial group so basically we're saying we take the free product of the trivial group with itself which is the trivial group and we identify in it the images of z which has to be the trivial group so we don't do any identification you take the quotient of the trivial group you're never going to get anything smaller than the trivial group so the fact that this was z or whatever is for this example not relevant so fortunately there's a bit of a long story that I had to cut to a little bit of time but I just felt that this was ludicrous not to cover this theorem in this class so I'll finish with one example and we'll stop there let's compute let's try to compute the fundamental group of the torus by using this theorem we know it already we've seen it in a different way in a simpler way but let's try this so I think of the torus of my square with identifications so let's take u1 to be this open set that contains a point that is going to be our base point and u2 that is a little bit smaller whole that is the complement so the union of these two open sets is the whole torus so we could use front campus theorem path connected and everything else what's the intersection a little little annulus because we take this a little bigger than the other one so when you cut this out of that one we're going to see this so what are the fundamental groups here let's be careful what is the fundamental group of u1 I didn't quite understand what you said just zed the integers everybody happy with that it's trivial I'm less happy this is the same as the fundamental group of the torus minus a point which in fact is something worth thinking about well what happens is that we cut out and make it bigger expand it all the way to the end so what do you get at the end you get a square with those identifications nothing inside what is that are you sure well we have an a and an a and a b and a b never mix so if you look at this thing we're going to have the a that goes to a point all the four points in the corner are the same point in the equation and the b is another one so what you get is actually the figure a I see your problem and I take it back I'm thinking of this example sorry I just got confused sorry let me I apologize this is the trivial group we're talking about the shaded stuff so I was thinking of this sorry I missed misspoke are we okay now okay so how about this this is an annulus so this is z so this is the free group in two elements this is trivial and this is z now that in itself doesn't tell us what the group is because the maps are the thing that matter you need to know how this so this z maps to there and maps to there but you need to know how it maps there before you can think of what the result is just the fact that a z and the free group doesn't tell you much what it tells you is that the group in question is well what does van Kempe theorem say it says we take the trivial group f2 with the amalgamation of this copy of z in it this is the same as f2 modulo the normal subgroup generated by the image of the circle of z so what is the image of this circle that we see here and see what it looks like in the fundamental group of this right so this loop there inside the torus what is it this is our base point well is the same as going around around the outside you can open this up and make it look like this so this path here is ab a inverse b inverse so the conclusion is that the fundamental group of the torus is isomorphic to the a group generated by a and b that's the free group modulo the relation that ab a inverse b inverse so what this is is z cross not star z not just the normal partition product so ab a inverse b inverse equal to 1 means that a and b commute in the quotient and so you have a free group with two commuting generators and that gives you z cross z alright so we'll stop here and you'll continue with homology next time thank you