Transcript for:
Degree of a Map and Homology

okay today I want to introduce the idea of a degree of a map so if we're given some function continuous of course from SN to SN then we know that this map induces a map on homology in particular I'll look at the enology group because that's the only interesting one they all the other ones die except for zero but in reduce mology all the other ones die here the interesting one is n equal Z and you know both of these are secretly a copy of Z know this is a copy of Z this is a copy of Z so you should think that this map is just sending Z to itself okay so what could it possibly be it's a homomorphism that sends Z to itself so what does it look like or negation those are two possibilities or times whatever just a homomorphism not an isomorphism right this map you can think of as just time D it's the times D map where the generator of this guy one you just see where he gets sent D and that tells you everything else gets sent as well right two sent to 2D 3 to 3D and so forth because this is a homomorphism so it's just times D we're going to call D the degree of the map so we call this D the degree of your map F okay there are lot of properties we can immediately identify about degree what is the degree I'll often denote this so I'll often denote the degree of f by just degree of f which is just equal to the whatever the number D is what is the degree first property of the identity map just one that's right because the induced map here is itself just going to be the identity map right the induced map just comes out to be the identity map on SN okay second property suppose f is homotopic to G then what can can you tell me about the degrees what does that tell you about decrees well we've already said if these guys are homotopic then their induced maps are the same so that the induced maps are equal to each other so then of course the degree of f will have to equal the degree of G okay that direction is straightforward it turns out that Converse is also true this this Converse is a theorem of half and it's proved in chapter four of Hat's book we probably we probably won't get there this semester um so this this shouldn't be obvious why it's true but but this is kind of telling you that degree gives you a complete classification of maps between spheres up to homotopy right so degree is really the thing capturing what the map is up to homotopy a few more properties probably fit them up here what is the degree of f * G well recall the induced map on f * G is just the induced map on F time the induced map on G such is a composition of whatever the induced map on G is with the um induced map on F and so if G is multiplying by some D1 and F is multiplying by some D2 then the composition is multiplying by the product of those so this will just be the degree of f time time the degree of G good okay um let's look at some particular functions what is the we're now space here so I come over here what is the degree of f where let's say F let's say f is a particular function let's say it's a reflection then you tell me what will be the degree of f so I'm thinking here I have something like maybe S1 or S2 um here's S1 and I'm just thinking that my f is a reflection you can choose what you're reflecting over let's say here I'm reflecting over this horizontal line so you can do either symmetry you want to but was reflect over the horizontal line so that points up top are being mapped to the reflection on bottom so this this is f what's going to be the degree of that guy well you might aely notice that F2 is just identity reflect twice that's identity so that means the degree of f times the degree of f should give you the degree of the density which is one so that gives you that the degree of f is plus or minus one so just by using this property we know the degree of reflection is plus or minus one but if you actually want to know what that degree is we got to think a little bit more about what actually is this map on homology so I can think of my S1 here I think of it as like being a CW complex then I have my top hemisphere I'll call that like a and my bottom hemisphere I probably shouldn't have used green b or more generally you know if you're thinking in high Dimension your S2 has some top hemisphere maybe I'll call that like big a that's this top bit and B that's the bottom bit and then there's your equator which you can think of as um however you want to with points and edges between them but if we down go down think on the level of homology we either one of you think about it I'll do the case of S2 it's very similar of S1 what is H2 of the sphere we've already said it's isomorphic to Z but like what is it generated by what is it generated by and here well you know depends we have put our orientations nice but if we orientate things that way we typically do I think we this example maybe like the first lecture or something this is just a minus B so then you think if f is the reflection map what does the induced map on reflection do sending you here to H2 of S2 and you're like well a minus B A minus B will get sent let's think what reflection does reflection just sends a to B and B to a so this would get sent to B minus a and so the generator here if you think of generator being like one will get sent to1 in the Z over here so my degree of this map is equal to1 so it could have been plus minus one but it comes out to be negative 1 you can think about this other settings as well like S1 or SN in general okay so that's the idea of the degree if you have some map from SN to SN you can look at the induced map on the nth homology group of the sphere we know that's just a copy of Z so it's it's a homomorphism from Z to Z so it's just multiplication by some D we call that D the degree of the map the degree of identity map is one degree of reflection is negative one satisfies some properties like if two maps are homotopic they have the same degree and um the composition of two maps has degree equal to the product of the degree of those two maps we can do one more and then we'll use this to prove a really nice theorem well I don't know I yeah I think it's nice what if f is the anpal map so so often this is denoted minus identity the antipodal map so an typal just means some point in your sphere some point x is going to get sent to minus X well this is on the if this is on the front this is on the back where so if if you think of this is being embedded in in dimensional space in this case if it's embedded in R3 you would think of this is being some X1 X2 X3 well then this is going to get sent to just - X1 - X2 - X3 if you think here of your S2 as living inside of R3 so you can give these coordinates can you tell me what is the degree of the antipodal map Dimension right okay say more um well like with with like S1 that would be theal would just be kind of like a rotation by0 so that it would be hopics which is be one so so with S1 you want to think of the intipal map which sends like some X here to its pointx What If instead of thinking of a rotation is another way you can think of it I want to think of it reflection as a composition of Reflections that's right so there's one reflection that takes me down here and then there's another reflection that didn't draw it perfectly takes me here and if you just keep track of what's going on if you think of this X as being X1 X2 so here I'm thinking of my S1 as living inside of R2 the first reflection is just inverting I guess in this case it's the X2 so it's making you know one of these guys negative the second coordinate negative and then the second reflection is making the first coordinate negative so each reflection makes one the coordinates negative right okay so if I'm doing this map on SN you can think this is just equal to the [Applause] composition of how many Reflections yeah in S1 you need two Reflections in S2 you need three Reflections one for each coordinate in SN you would need n+ one [Applause] Reflections so this degree by this property right here would just come out to be -1 to the n+1 good okay so that's all we need I mean all we've done is we just used the fact that hn of SN is z but and then think a little bit about this idea of degree but that's the only tool we need to prove the hary ball theorem okay you've probably heard of this one before perhaps not um it simply states that if you're in a setting like S1 and you have little hairs on S1 what I mean by hairs I mean a vector field a continuous field of my tangent vectors then I can draw I can draw since a continuous Vector field so these are just all my tangent vectors nonzero vectors all the way around so I have a continuous field of nonzero vectors but in S2 [Applause] you could try to come up with some continuous field of nonzero vectors so maybe around the equator we just play the same game we did in S1 so you would have these little tangents and it's kind of hard to represent these on the back maybe we'll do something like that dot lines okay so you can do that and you can do that you know going up all the way around you can think there like circles around the sky circles going up Circles going down and on each one you can draw a similar continuous field but where this this go right as you approach the North Pole like this would tend towards what there wouldn't be like there would be no value you put a zero Vector but there's no nonzero Vector that you put there that wouldn't break continuity right so and same thing as South Pole are these poles it wouldn't be continuous and so one way you can kind think of this with hair is if you have hair growing off of a circle you can comb the hair around right in this nice continuous way but on a ball if you comb the hair around the ball you get CX at the top and the bottom right like on the back of your well mine doesn't typically your hair has a part repl where the hair parts right meaning that there's no nonzero Vector there okay now it's hard to imagine this for S3 and S4 and so forth but what turns out to happen is that SN has a continuous nonzero tangent Vector field this n here is positive if and only if and is odd so this result that holds true for S1 and S2 generalizes that you can find it for odd SNS but not for even SN I mean it's not even obvious that you can't find it for S2 right like maybe this you're like well maybe this is a really clever one you know and you can try to think about what that clever one might be but this say no matter how you try to comb S2 there will always be places where you break either continuity or you have to have a nonzero Vector right okay so so let's prove this let's suppose we have such a vector field where I'll call VX the tangent Vector associated with so I'll make XO Vector at the vector X which is some Vector in SN so so the picture I have in mind now I guess I could draw it here is if here is some Vector X which really means your vector is at the origin and it just points and it hits the sphere at this point so if this is X if that's what x from the origin comes up and hits on your sphere then you have this VX would be this little Vector pointed this way this would be your vector v of X right okay note whenever this is true whenever you have a vect a tangent Vector in a sphere we know that your vector X and V of X are orthogonal they're perpendicular to each other which is to say if you take this VX and draw it inside of the sphere you know okay kind of having a tough time now we can do this so here was some Vector here is your X that takes you to some point here and you think at X you think that at X you have some little tangent Vector nonzero tangent vector v of X but if I scale this so it's unit length that's going to come out to be some looks like it's just going out like this some little Vector that hits over here somewhere so that these are orthogonal to each other so this this would be this scale to unit length so that's you have tangent Vector VX ID the magnitude of VX now it's length one so it lies on the sphere as well so now you have X and VX scale to unit length both being points line on the sphere okay what we're going to do is we're going to consider a function from SN to SN and what this function is going to do is it's just going to look at points that you can think of linear combination of these two so it's going to end up giving us something like a great a great circle of giving you points that depend on those two points so I'll Define it explicitly right now F of a vector X will be equal to oh well maybe I shouldn't use x i that that may not be good um I'm I'm fine using X cuz you have this these are defined for every point x as well at every point x um we're actually going to make F A family of functions over the parameter T so that this is defined to be at the order of these right yeah cosine of ttimes the the vector this is some scalar times the vector X Plus s of T time this unit Vector so it's the VX scale to be a unit Vector okay so what you should convince yourself of is that this is always of unit length so it's like what is the magnitude of this for any value T in any point x well these are both unit vectors and here you have the property that your coefficient here squ plus coefficient here squ sin square plus cosine square is one so this comes out to be unit length and so indeed this is always spitting out some value on SN this is just spitting out the values this is the parameterization of the circle so what you're doing is you're traveling along the circle well I guess you're traveling from we let's see exactly where we're traveling like let's pick some values of T like let's let t e equal to 0 then you tell me so to write that I can just call F of0 what is F0 of X well your cosine is one at zero and your sign is zero so at Time Zero you just start off at X now by the time you get to Pi / 2 this vanishes and your sign is one so you start off at X at time so you're here at time equals 0 and then at time Pi / 2 you're down here right so you can think it's moving you you're just moving along this Arc but then keep going don't stop at Pi / 2 let's go until we get to say time Pi at time pi where do you end up at yeah at time Pi cosin is 1 this is zero again so you're ATX so what we have here is a homotopy that starts off with the identity map and ends up with the antipodal map so what I've just shown is that identity is homotopic to the intipal map but if two maps are homotopic come back over here we said two maps are homotopic they induce the same map on homology and therefore they have the same degree so that means the degree so from this we get the degree of the idenity map must seally be the same thing as the degree of the intipal map but as we just said in the setup the degree of the identity map is one and the degree of the anpal map is -1 to the n + 1 and the only time minus1 to a power is one is when it's an even power so n plus one is even and if n plus one is even that means n is odd okay so that that completes the proof well okay that completes this direction of the proof and then you have to convince yourself um in the other direction works as well if n is odd well then you know these are equals so then there exists such a homotopia okay think through think through the other direction but we've certainly proven this direction okay good that's that's the har ball theorem that's our first application of degree um but it's not the main reason I'm showing you degree the main reason I've shown you degree is because I want to introduce a new homology Theory that's going to be defined using degree so up until this point in the of course we've seen two homology theories so far or two ways of defining homology the first one was simplicial homology so we had some H uh what was our notation how do we distinguish some do we have any notation to distinguish did we do like a little Delta complex here or something I think that's what we did so we had simplical homology and then we had singular homology singular homology and I'm not going to Hash through these definitions again you can go through the previous lectures but we Define these two homology theories and then was the last class we used the five LMA to sketch the proof that these are actually equivalent to each other so different ways to Define homology but gives you the same Theory today I'm going to give you a third homology Theory and um which is secretly equivalent as well but sometimes having several different ways of defining these homologies gives you different ways of calculating and sometimes it's easier to calculate homology into one such a count than another so what we're going to do today is define cellular homology and the way cellular homology is defined um is going to be defined for a CW complex so I I should take a few minutes to remind you what the C CW complex is you you go back I think it was the first lecture Le is zero we introduced this but maybe you forgot so a CW complex is a space a topological space that we're going to build up one dimension at a time and so you begin with the zero skeleton so so this is often denoted x0 it's the zero skeleton which is just a point or possibly several points it's your Zero Dimensional stuff okay I'll introduce the notation the minus one skeleton but that's just the empty set but we'll want a convention to say you have nothing so you begin with negative one you begin with nothing then you have you add Zero Dimensional things those are points then you have your one skeleton and the way you get your one skeleton is by attaching one cells to the zero skeleton so what is a one cell weall one cells are just Open onedimensional disc so these are open maybe I'll do like little open balls at the end these are open onedimensional dis which is open intervals and then what you do is you look at the boundary where the boundary points would be and you attach those you attach the ends here to the zero skeleton so you have some attaching map where this attaching map is here these two end points well the two end points you think of as s0 is just a map that's tell you how to send s0 onto your Zer skeleton onto x0 and so if I just say well attach both of these here and attach both here I just end up say with two loops with one loop with two loops on here that's your one skeleton and then you can build up to a two skeleton so how do you get a two skeleton well now I want some two cells so you want to attach a two cell well you can think of a two cell as just being an open two-dimensional disc open two dimensional ball okay but think of as being open and then it's attached around what the boundary of the closure would be so here this attaching map the boundary of this is S1 and that just gets attached onto your one skeleton so it's a map from S1 onto X1 and you know how might you glue this disc on here well you might have some instructions like if this was if this was called Like A and B What It's Like Glue along a and then glue along B and then glue along a backwards so these have orientations you can keep track of some orientation on these guys this was my a this was my B you have some orientation glue along a glue along B glue along a backwards glue along B backwards what does that give you the Taurus so then your two guy two skeleton now looks like a Taurus not not filled in not a solid Taurus just a empty Taurus where you can still think of okay I still have this structure from the previous points but it's been filled in now with a two cell and you keep building it up this way okay so so in general an N cell I'll denote as e to the N if there's several of them I might give them some subscripts you know I have subscripts for them so n cells I might give them like e the N1 e to the N2 and so forth and you have an attaching map each one of these are just n dimensional balls and you have an attaching map where this each e to the N um the boundary of it which is just an S to the nus1 gets attached onto your n minus1 skeleton and this is how you build your xn from the N minus1 skeleton by just attaching some n cells to to it does this sound familiar are we happy with this okay I'm reviewing this all because we're going to Def find our cellular homology which is just going to be defined in terms of some cellular structure so whenever you have a CW complex that means you have some way of building up your space by attaching cells will be able to from that CW complex be able to read off the homology so so this is quite nice it's it's kind of like simp homology but you don't need to have a a singular complex or a Delta complex you can just have your CW complex so in that sense more powerful but we're going to see it comes out to be the same okay so how do we Define this thing well with any homology Theory we always have some long exact sequence we have a long exact sequence and then we just Define the homology as you have these Maps between these a billion groups and each map you define as the Cardinal of the one over the image of the previous right so I need to give you a long exact sequence okay so here is the long exact sequence it looks like this this is the nth + one homology group relative homology of the N +1 skeleton Rel the N skeleton mapping to the nth homology of the N skeleton real the nus1 skeleton tapping to the nus1 homology of the nus1 skeleton well the N skeleton and so on you're like well how does this thing end well that's where um once you get to the zero skeleton it's the zero skeleton oh sorry this should be n minus 2 Rel the N2 once you get down to h0 for the zero skeleton it's ra the n - one which would be the minus1 skeleton and that's why here I just Define this to be the empty set so like if this thing keeps going it'll go to h0 of x0 x -1 but I just had to find that to just be the empty set which means it's just h0 of x0 because well nothing mapping them to zero so that's how the log actually equence ends right okay then we have boundary Maps DN + 1 DN and so forth where the composition of two boundary Maps where DN of DN +1 comes out to give you Z okay I am not going to prove that this is the long exact sequence I'm I'm I'm um this this will follow very quickly once I Define what this map is there so you know you can look at Hatcher nice proof it's actually very clever proof and okay I can I can give you like the very quick idea of it what you do is you say I already know that this shows up in a long exact sequence so like maybe what I'll do this is the crudest of sketches it's like I already know this shows up in a long exact sequence do you remember the long exact sequence that this guy shows up in it's the longa sequence we had for a pair a good pair this just comes out so just think like this is X relm a there was a longa sequence for that what would be next in the long sequence it would be hn of the a xn mapping to hn of xn + 1 and so forth so this itself is a long exact sequence we already know about so we we already have this long exact sequence and then you say ah these guys together feature also in a long exact sequence we know about where here you would have had a h n of X so up here is like hn of xn minus1 and that Maps down here to hn of xn and then you get xn real xn minus one and then that Maps down to hn -1 of xn -1 okay I was trying to avoid sketching this proof um but then what you will do is do a little bit of diagram chasing to convince yourself if I Define this MTH just to be the composition of these two maps right so so if this is some you know I and J or whatever composition of those two maps so this is actually how you can Define this then what you can do is you can prove the exactness of this sequence and this crisscross pattern continues and so you can use the fact that we already know that these diagonal guys are long exact to prove that this is long this good activity to do so you can either like try to do this on your own or read through Hatcher's explanation of it but it's just just diagram chasing and thinking through that satisfies all the properties I don't necessarily want to take a long time on it is that believable is that good okay so this might look like a mess because it's a long exact sequence where your groups themselves are homology groups I haven't even defined the homology yet so this long exact sequence it's called the cellular chain complex let me put that up here this is called your cellular chain complex before we had chain complexes this is my cellular chain complex cellular chain complex but if I want to define cellular homology my definition will be hn of cellular homology this is my notation for cellular homology so this is a cellular horology will be defined to be the kernel of DN over the image of dn+ one so your groups themselves are homology groups well then you take you find your cellular homology on this so so these are relative homology groups and on that you define homology so this might seem like excessively messy complicated ugly but these relative homology groups are very nice what is hn of xn R xn minus one so you should think this is just like collapsing down everything in the N minus one skeleton only leaving the stuff you added in the N skeleton so what are you left with like think back here you know if I Collapse down everything just leaving this like two cell but I Collapse all of this what am I left with well the boundary of this have been glued to this so it's just like the two cell with the boundary collaps down to a point so what are you given spere it's a sphere so each two cell will contribute a sphere and so this just comes out to be a wedge of spheres so this is just hn of a wedge of spheres where you have one Sphere for each two cell so so here I made index I but I'm trying to say it's this is indexed over your two cells you have one Sphere for each two cell or in this case n cell in cell okay so what do it come out with what's the homology of that have we talked about this o did we talk about homology of a wedge product I'm not sure if we did ah that would have been a good exercise okay it's the obvious thing what is it it's just the direct product of Z's so the homology of w sum is just the direct sum of the homology of each of the Summit so it's just the direct sum of Z over I so you just get one copy of Z for each two cell so if you prefer it's like it just comes out to be Z it's just Z not two n cell to the number of n cells you know this is the number of n cells that's that's what we end up with okay okay I still need to show you talk about what this map DN comes out to be and we'll do that in a second hint it has to do degree that's where we spend so much time talking about degree but I think as I talk about that it would be good to set up a concrete example so let's try to find the hology of a Taurus the cellular homology of a Taurus now I'm not going to have time today to even begin to sketch this so I'm just going to do maybe two or three examples to round out our time today but spoiler just like before your cellular homology will come out to be the same as your singular and simpal homology so I'm not going to have we're not going to we're not going to prove this that's fine what I want to do is actually show you you know you refer to Hatcher you can read the proof it's very nice um but I want to just show you how to actually calculate using this homology Theory so for a Taurus so so what these Maps become then are just copies of Z for your zero cells one Cells Two cells and so for so your exact sequence you just think you you have zero cells you're going to have one for one cells you're going to have one for two cells and so forth well your Taurus we just drew it over there it doesn't have doesn't have anything higher so everything higher is zero how many two cells do you have just the one right it's just a copy of Z how many one cells were there two so that's a z s and then zero here is there was a single point Z so remind you of the construction you had a single zero cell you could call that X then we add it in our one cells one one cell going this way and call that a one one cell this way you can call that b and then we GL Inu in the two cell creating a Taurus right so that's my attempt at drawing a Taurus okay so that comes from gluing in your your two cell okay what I need to tell you is what this map D is so we can actually calculate homology so here's the formal definition of DM DN is a map from hn of xn to xn -1 to hn -1 of xn -1 to xn - 2 but we are just thinking of both of these as being a wedge of spheres one Sphere for each n cell so this you're thinking of secretly as being a wedge of n cells of spheres that become SNS and this you're thinking of as a wedge of nus1 spheres so if I take a generator over here a generator one of these generators will correspond exactly to one of the end cells so one of these generators would be like n of you know whichever one they're indexed so there's one corresponding to each one so this is one of my n cells that gets mapped onto my n minus1 skeleton the boundary of one of my n cells where that n cell corresponds to the generator here the boundary of that guy is just a copy of SN minus1 and you have an attaching map that tells you how to attach that Snus one to your n minus1 skeleton so that's your attaching map now once you have xn minus1 what we can do is we can apply a quotient map where we collapse everything down to a single point except for a single n minus one cell and it's like you can pick whichever n minus one cell you want um but you know it's going to end up with a single n minus one cell with everything else collapsed and so the boundary of nus one cell gets collapsed as well so you end up with a copy of SN minus1 and let's say this corresponds to index beta for whichever n one cell that was what you have then is you have a map from this copy of nus one of SN minus one which was the boundary of your n cell here down to an um another Snus one which corresponds to a collapsed n an N minus1 cell with its complement collapsed so you have a map from sn-1 to sn-1 and if you just think about that map from sn-1 to sn-1 the composition of these maps that map will have some degree so you know this map is f you know if you call this F there will be some degree of F and that's what our DN is now this might seem like a very weird abstract definition but let me show you how it plays out in practice in in practice let's think about this guy here your Z2 is has two generators an A and A B that correspond to this these two one cells that are mapped in with then their end points um you know identified with X to give you two circles A and B we can think about where A and B get sent so so like this a you can think this a corresponds to you can think this a corresponds to like some copy of S1 right you can think this B corresponds to some copy of S1 if you want but I'm just thinking about what's going to go happen with this a and this B so so this a is a copy of S1 so it has some boundary the boundary of a is just an x minus an X so when you map where that boundary of a goes to it just goes to x - x and so this a just gets mapped to x - x which is zero which is telling you here that this a has degree zero oh okay I made a small mistake here DN um the the way we Define DN um is like this is correct but you have to do it for every possible um cell so so the proper definition here is DN of an N cell um uh an N cell with index Alpha will come out to be just some correspond to some n minus one cell with index beta with in front of it some D of a alpha beta this this D of Alpha Beta is just your degree of f sorry but then Alpha um this this n cell you could collapse different quotient Maps you have different quoti Ms to all the other nus1 cells so that's why you have this index beta CU it's all the different nus one cells so you just send it to the sum of all the different nus one cells so this is indexed over the sum of all n minus one cells beta okay but it's not going to matter here because the same thing that happens to a just like how a gets sent to zero B also just gets sent to x - x and so that's zero as well and so your generators both go to zero because you just look at this attaching map it's just your attaching map here you're looking at your attaching map and it's just okay respect to orientation it's just x - x so they both get sent to zero and if both generators get sent to zero then the whole map is zero likewise we can come over here and we can ask well how about D2 let's think about this two cell so there was some two cell we attached here some two cell we glued in which we had thought of as the boundary we attached it along the boundary of some um two-dimensional guy here so let's give that two cell a name I'll just call it um uh Big C uh I guess we can call it E2 for my two cell well how did we attach this what's the attaching map the attaching map is literally you glue along a and then along B and then along a inverse and then along B inverse so if you think about where this guy gets sent if you quotient so you just end up with a copy of a then he'll get sent to just gluing along a a and then gluing along a inverse well that's trivial so he'll get sent to zero copies of a but if you instead quotient along b along B he would also once you quoti out everything else he would also get sent to glue along B and then glue along B inverse so it's zero copies of b as well so this is also the trivial map so both are what that means now is your homology is straightforward the cellular homology is just the kernel which is everything over the image which is nothing so for zero it comes out to be a vurus S1 cross S1 for zero it comes out to be Z for Nal 0 for Nal one it just comes out to be z^2 because it's the kernel whichs everything over the image which is nothing and for Nal 2 it's also just Z because the kernel is nothing over everything so for 0 and 2 it's Z for um Nal 1 it's z^2 and then is just zero trivial otherwise okay so that's how you calculate this stuff so so I almost think the example of a Taurus like is too simple to actually have you understand this map so like maybe this is a little bit mysterious still because everything just gets s to zero when you attach it so let me end with one more example that's very similar to the Taurus but the attaching maps are slightly different which will actually create something interesting in our homology Theory so let's end with one last example which is the Klein bottle so K is my cim bottle so give me the skeleton for a clim bottle you start out with a a what single a single point yeah okay so maybe well you know so maybe you just think about the class representation of a bottle like this and this will help you so it's like how do you do a c bottle well it's like a Taurus you put a Twist on so if these maybe there's a Twist on the A's you leave the B's on Twisted but notice this point is going to be identified with this point which gets identified with this point identified with this point so it's really just a single point these are all the same point x so you can think it was just a single point x that you added two one cells to and then you glue in a two cell where your attaching map here is going to be like a b a b inverse so the attaching map for your two cell is a b a b inverse following the orientation here and what should I call my two cell I'll call it E2 okay so so let's draw out the homology of this guy so again if I'm just thinking of my number of zero cells my number of one cells and my number of two cells cells this is simply going to be a copy of Z CU there's just one two cell two copies of Z for the one cells CU there two one cells and a copy of Z for the zero cell and zeros at the end but now these Maps this map D2 and this map D1 get a little bit more interesting cuz let's think like how we attach these one cells there were two one cells here your one cells are a and and B but like how did we how did we attach them well I don't think this is too terribly interesting like how how did you attach a to to your point x well it was just awesome both ends went to X right so you can think if you started off with just a single point x the attaching map for a also just attached a like this a was just attached with both end points CU both end points are on X and B also is attached to both end points are on X so the one skeleton for the clim bottle and the Taurus are the same so the these both again map to zero these both map to zero just like before and so you end up with degree zero and you end up with the this map is zero but where this gets interesting is the way you attach the two the two cell so let's say this is my two cell E2 okay so I have like some some two-dimensional ball E2 and I have an attaching map along the boundary of E2 so so this is some map from the boundary of V2 which is S1 onto this one skeleton how do you attach this onto here well just read off the word right the attaching map is given by the instructions a b a b inverse so when I Collapse along a where does this get sent to how many copies of a when I Collapse along a oh I mean when I Collapse so a is the only thing that remains I Collapse the complement of a so I Collapse everything so it's just a copy of a remaining what do I end up with I would have two copies of a so this would get ma to 2 a whereas where does this get mapped to when I Collapse so it's just B that remains okay so C I Collapse everything that's not B so B is now just a copy of S1 because everything else got claps down to a point and what am I left with well I glued along B then Glu backwards along B so I would end up with zero copies of B remember degrees is equivalent up to up to homotopy so attaching along B and then backwards is the same as as just going to zero degree zero okay so so this is kind of helping you understand what's going on here how how these Maps work and then you know you can extend this map over any combination of you know A and B's and so forth but but like let's calculate the homology now what is the cellular homology for zero of the kleim bottle each zero of the kleim bottle well that's just your chinal of your image that's just Z nothing interesting there cuz your image here is zero how about the cellular one homology of the clim bottle well that is this kernel which is all of Z2 over this image up now this is getting more interesting it's the aelan group generated by A and B because that's the kernel of this map but the image of D2 is two copies of a so you add the relation 2 * a is 1 so what what group is this the B there's no relation on B so B is just giving you a copy of Z but the a has has the identity of 2 * a is 1 so the a corresponds to a Z2 summon that's z m 2 Z and then when you calculate C2 the cellular homology the kernel of this map what's the kernel of this map well the kernel of this map is just this copy of E2 cuz he gets sent to z b a is z there right right the the this kernel here is Trivial and the image is Trivial so the H2 is just zero very good okay making sure that that was right yeah um E2 gets sent is the only generator here and he gets sent to you can think E2 is being sent to like 2 A Plus 0 B if that helps um so this um has has no non-trivial kernel so the cronal is just zero so your second homology group is zero okay and there you get the homology of the client bottle I think you calculated the singular homology of this at some point Didn't you or the simpl homology did some of you do this homework exercise and you got the same thing right and you know so you had another way of seeing it but but this is showing you hopefully the power of the power of cellular homology so next time I'll either give a few more examples of this or I'll move on but I encourage you you know lots of lots of spaces to calculate the cellular homology get used to get used to these calculations this definition seems quite abstract with degree but really what it's saying is just keep track of how you're attaching maps are attaching things that's that's what's really that's what really this definition is saying just keep track of how you attach map is gluing things and and then you can quickly calculate homology so this becomes a very fast way to calculate the homology of a CW complex we'll stop there