 Last time we had started the study of product of CW complexes. We interact, we begin a little bit but today we will continue our study. So what we have done was just to study one single example of S1 cross I. S1 with a CW structure with one zero cell and one one cell and the interval with two zero cells and one one cell. Okay, so the product we have decomposed when we want to think of some structure on a given space all that you have to do is as I have told you earlier decompose the space into open cells. Of course in a systematic way start with zero cells one cells if possible two cells and so on. You have to start with zero cells no problem. So you take S1 cross I the cylinder and then look out for the structure of CW complex which is some way related to the CW complex of S1 and the CW complex structure on the interval I. Okay, so look at the two zero cells zero cell and one cell in S1 denote them by sigma naught and sigma 1. Similarly I could have denoted the cells by cells inside I by tau and tau and so on but instead of that I will use the natural notation zero and one as one zero simplex and the interval I as the one simplex. With that notation in the product structure S1 cross I we have sigma naught comma zero sigma naught comma one these two will be the zero simplex. Sigma naught is the zero simplex here and zero and one of zero simplex is in I. So these two will be zero simplex it's been the product of space of the zero skeleton of the two spaces. What should be the one skeleton? The one skeleton should be zero skeleton cross one one skeleton cross zero all possibilities. So that way what we get is the one cell and zero cell and one cell is therefore different whatever different possibilities we have to take. So sigma one cross zero sigma one cross one these two are the one cells because sigma one is a one cell in S1 and sigma zero cross I I is the one cell on the other side ok. So that is the kind of thing what we have done. This is there are two zero cells here sorry one zero cell and one one cell two zero cells here and one one cell here. So we start with two zero cells sigma naught cross zero sigma naught cross one ok. Then this itself will become a one cell this will be another one cell and that will be another one cell corresponding to sigma naught sigma one cross zero sigma one cross one and then other one is sigma one sigma naught cross I. So the remaining cell whatever must be one single cell which corresponds to what sigma one cross I that will be the two cell right. So the theme is that this example completely tells you the entire story no matter what X and Y are most general case. All that you have to do is X cross Y decompose it into zero cells one cells two cells and so on K cells where K cells of the product will come from pick up I and J such that I plus J equal to K take all possible I cells in the one and the J cells in the other one take their cartesian product. So those things will be K cells for the product ok all the write down all the open cells that will fill up because X itself is a disjoint union of its open cells Y is a disjoint union of open cells the product will be each member here will take each member here to take the product. So that is theoretically completely obvious thing what we have we have a slight problem here because we have modeled our cells on the round model disk. The disk is the unit disk set of all unit vector sorry vectors less than or equal to norm one inside Rn. So that is a round model but we are only interested in things up to homeomorphism anything which is homeomorphic to a disk will be called a cell. Therefore we are free to choose any model and here when you pass to product a disk cross disk does not look like a round figure right it is not. Therefore you have a problem there so you have to change the homeomorphism type anyway you have to fix up dn cross dm to dn plus m a homeomorphism ok. So that has to be done whether you do that every time or you choose a different model is left to you. So here is a way of handling it right in the beginning we can fix up a homeomorphism from interval no problem but as soon as dimension 2 you come square and the round disk ok cube and the round disk and so on. So what you do take j equal to minus 1 plus 1 the interval let us have this notation standard notation like I have put i for 0 1 take j for minus 1 plus 1 take j power m that will be homeomorphic to the disk dm but now beauty is j power m plus n will be j power m cross j power n very canonically ok you do not have to fix up another homeomorphism here this is just a way of writing first k m coordinates and then last n coordinates right this coordinate wise the same maps are the same maps so you can say this is identification with this model the product structure can be easily understood ok. So fix funds for all just to go to our old definition our old definition used the model dn remember attaching cells was done from disjoint union of the copies of dn when you say copies that is already this freedom that you can choose whichever model you want these copies of homeomorphism ok so that is what we are going to do here fix up hn from jn to dn whatever homeomorphism you have some natural homeomorphism so the characteristics maps of attaching maps and the attaching maps are respectively from jn and the boundary of jn ok naturally boundary of jn also will not look like a round sphere now but it is homeomorphic sphere ok and one more thing set theoretically you should know that boundary of jm plus n is boundary of jm cross jn union jm cross boundary of jn ok these are set theoretically totally obvious so let us make a definition here in the most general case given two CW complexes X and Y we define the product CW complex X cross I have put a W there Y so this is to denote the weak topology little what we are taking X cross Y it is only SS set but the topology will be the topology that you want namely weak topology so how we are going to do this underlying set is the product set for each cell sigma in X and a cell tau in Y with characteristic maps phi and psi respectively we take product cell sigma cross tau in X cross Y which is characteristic map the product of the characteristic maps in particular observe that the zero skeleton is nothing but the zero skeleton of X cross zero skeletons Y with the discrete topology as usual start with that the one cells as I have told you will be zero cells cross one cells as well as one cells cross zero cells all possibilities you have taken that is the definition after that what you do inductively the case skeleton will be what X cross Y is space obtained by attaching to K minus one all sigma cross tau where sigma and tau range over cells of X and Y respectively the condition is the dimension of sigma plus dimension of tau should add up to K so that will be the case skeleton finally we give the weak topology on the union of all these case skeletons what is the weak topology something is clothed if it is clothed if it is clothed if it is clothed if it is intersection with the case skeleton is closed in the case skeleton inductively the attaching maps we do not need any new topology new definition because if Y is known and attaching maps are known then X is known so each case skeleton gives its topology already inductively but on the union I have to define it as the weak topology so this completes definition of what is the meaning of X cross Y as a CW complex the CW product so X cross Y is a CW product is taken this way once again if you are dealing with finite CW complexes or even finite dimensional CW complexes then you do not need this extra definition because this is finite dimension this is finite dimension say this is m dimension n dimension you will stop at m plus n dimension m plus n dimension skeleton is obtained by m plus n minus 1 dimension skeleton therefore there is no need to redefine the topology by using this okay so this is an easy observation here so the new this redefining the topology is a necessity only when infinite dimensional spaces are enough okay and quite often that is the case therefore we have to study this infinite dimensional case properly and that will be the gist of whatever comes next alright so before crossing further let me examine one more interesting example instead of S1 cross I now I will consider S1 cross S1 the torus once again take the simplest CW structure on S1 on both sides namely 1 0 cell and 1 1 cell on both sides then as usual of course you can think of S1 as a subset of C okay unit vectors or unit complex number and S1 cross S1 is subset of C2 just for fixing notations okay so what is the 0 cell you can declare what you want here I would like to take the minus 1 as a complex number sitting inside S1 as the 0 cell and the characteristic map from J because I am using minus 1 to plus 1 as parameter for the 1 cell so phi from J to C take phi t equal to e power pi IIT okay so we now look at the product silver structure on the S1 cross S1 okay so there is one just 1 0 cell namely the point minus 1 plus 1 you see in the case of S1 cross I you had 2 0 cells so here is only 1 on both sides there is 1 here 1 here there is only 1 possibility and that will be minus 1 minus 1 whereas there are 2 1 cells mainly 1 cell here and 0 cell there and interchange 0 cell here and 1 cell there okay so and what are the characteristic maps for that phi 1 of t is first one phi 1 t comma the constant map minus 1 phi 2 t is minus 1 constant and phi t in the second coordinate so this phi 1 and phi 2 t finally there is just 1 2 cell namely phi cross V J cross J which is just e power pi IIT 1 comma e power pi IIT 2 so that I have written down but that is obvious okay once you have that see look at this one this is S1 cross S1 suppose to be whatever so this is e0 and this is e11 okay and then there is another one e12 so this will be e0 cross e0 okay this will be e1 cross e0 and that will be e2 cross e0 or e0 cross e2 then finally e2 is which is e11 cross e12 so this is 2 cell which will be sitting there okay quite weird way to be sitting so this inner circle here I mean this is nothing inner circle this is a round thing but this is not a part of the skeleton this is not any cell there is only 1 0 cell 1 and 2 these 2 are 1 cells and then 1 2 cell so 0 and 2 and 1 so there are only 4 cells you know okay so this is the way you have to just decompose the given space okay automatically you would know when you write down parameterizations the attaching maps so here let us do that and get a nice output okay so it is interesting note that attaching map of this 2 cell trace is the first circle S1 cross i S1 cross 1 then the second circle 1 cross S1 then again the first circle in the opposite direction and again the second circle in the opposite direction so if you denote these elements say S1 cross 1 the class of that and 1 cross S1 in the fundamental group of this S1 union S1 you see this circle union that circle with one single point common so there is a wedge of 2 circles there is a subspace of the torus right on the subspace of the look at the fundamental group of that then what is it happening is the boundary first goes around this one then goes around that this time it comes around the other way around and goes around that way what is just vacation you can just write down what happens to the p1 t comma p2 t right p1 t by i t by i t but on this one how it is sitting in S1 cross S1 you can work out t power i mean c cross c or there is another method I will tell you so first think of this as the loops in their class of the loops okay if you S1 cross 1 and 1 cross S1 they are X and Y are the same then the loop that is attaching attaching loop is nothing but XY X inverse Y inverse in the fundamental group of S1 edges S1 okay so this is the picture you all know that the torus can be got by attach by identifying this edge with this this edge with this edge and this edge with this edge of course when you identify the arrow is on the other side here here the arrow is on the other side upward okay when you do that this becomes your S1 cross 1 along with that that is also S1 cross 1 that will become 1 cross S1 but now look at this one the boundary of this square you can trace it from here to here to here to here to here this is going to be the boundary of the two cell that I am going to attach attaching map what is the attaching map first it is this one let us call this X that is what I have done then it is Y but now X is this way so when you are tracing this way it is X inverse similarly Y is this way when you are tracing this way it is Y inverse so when you draw the picture on the mother's face of which these are quotients everything will be very clear so this is the lesson you have to learn so this is XY X inverse so when you have attaching XY X inverse we have done in the first part that the fundamental group gets quotiented by this element XY X inverse which means that you have to take the normal subgroup generated by this element go modulo that that will be the fundamental group of the mapping cone the mapping cone is one cell is being attached to the one skeleton S1 cross S1 which has free free group on two generators that is the fundamental group now go modulo XY X inverse Y inverse which is the same thing as declare X commutes with Y what you get is the free abelian group so pi 1 of S1 cross S1 will be the free abelian group over two generators of course this we know already by another method namely we know that pi 1 of X cross Y is pi 1 of X cross pi 1 of Y therefore pi 1 of S1 cross S1 is pi 1 of S1 cross pi 1 of S1 again which is Z cross Z this we this also we know so the point is when you want to learn something new concept you should check it or you should learn it with something which you already know very well so that you will be sure of your ground so that is one case here returning to the general study let us be sure that we whatever you have described as a CW structure on X cross Y is so given any point X comma Y in X cross Y as a set there is a unique P cell sigma in X and a unique Q cell tau in Y such that X is in the interior of sigma Y is interior of tau why because every CW complex is a disjoint union of its open cells each element will be in one of the open cells so it takes a sigma and tau that would mean that X cross Y is precisely a disjoint union of interior sigma across interior of tau these are subsets of X and Y respectively so X cross the product will be subset of that take the disjoint union these things are all disjoint because in the beginning interior sigma and interior sigma prime two different cells they are disjoint inside X similarly tau tau prime inside Y they are disjoint you take all possibilities then you get the entire X cross Y and not only that what happens to K skeleton K skeleton consists of all cells of dimension less than or equal to K you have to put them together so that is why you have to take union of all P plus Q equal to K X power P cross Y power Q for each zero cell X naught of X if you take X naught cross Y that will be some complex because all those cells will come and previously their boundaries would have been inside that if you take a K skeleton here X naught cross Y power K the K plus one cell which looks like X naught cross some cell will have boundary in this way so these will be sub complexes of X cross Y similarly for every zero cell Y I cannot take arbitrary point here ok remember that you have to take cell zero cell Y naught of Y then X naught cross X cross Y naught will be sub complex the product cell can be thought of as obtained by attaching JP cross JPQ with characteristic maps as products of the two characteristic maps where P and Q range over all possibilities P plus Q equal to K in particular all closed cells of X cross Y are closed and compact subsets of the product space because they are just products of compact sets closed subsets of compact sets closed subsets and compact sets ok so the only major concern is when X and Y are in final dimensional or one of them is in final dimensional ok so let us resolve that then we will be very happy with the product of CW complexes so let X and Y be any two complexes we have the first thing is X cross Y W whatever CW complex structure is there the same thing as take just the the product topology and then retopology is it by using compactly generated subsets so that is the meaning of writing W on the outside remember for any space say Z what is the W Z W is the coindistopology through all the compact subsets of Z I want to say that these two are the same so that by chance if you do not have to write here you can write here then also it is correct do not write X cross Y W without bracket that will be different ok the identity map once you have this X cross Y bracket W to X cross Y is continuous so this is we already know so B is automatic ok once you know why this is true then this is automatic identity map is homeomorphism if you know this product is compactly generated this also we know ok identity map X cross Y W to X cross Y defines the bijection of compact subsets this also we know these all these things generally we have studied so we can apply them to the situation of CW complexes the only thing we have to see is why this is true ok now look at the family F of product cells sigma cross tau a sigma and tau range over all close subsets of X and Y that is a cover for X cross Y and these subsets are compact if you have a sub family of compact sets covering the whole space and give the induced topology from that that will be the same thing as the induced topology from all subsets ok so therefore the co-induced topology X cross Y W from this family is actually finer than the weak topology on X cross Y X cross Y W ok because we have to verify only for a chosen family smaller family if we have smaller family the co-induced topology will be finer we will have more open sets it has to satisfy less number of conditions but equality follows from you have to know why it is equal equality follows from the observation that take any compact subset of X cross Y it is covered by finitely many members of this family there will be finitely many close subsets sigma cross tau for any compact set you know that is covered by finitely many cells and what are the cells they will look like sigma cross tau therefore these two topologies are the same and we have seen that BCD other three things are automatically satisfied in the general case so only this was this has to be seen ok so we still have not come to the case when things are infinite what happens we still do not know whether X cross Y W is equal to X cross Y so that is our next concern it will never happen always so what are the best possibilities what we are going to study so here is a definition which will be a nice condition which will guarantee some kind of equality amongst X cross Y W in X cross Y so what is the condition a CW complex is said to be locally countable if every open cell meets only countably many closed cells let me elaborate on this one take a zero cell it is both open and closed what does it mean it means that all the cells which will hit this one how do they hit by using attaching maps all those cells if you count them it will be finitely many in particular the edges or one cells which are coming out of this zero cell there will be finitely many not only that the two cells may be far away I do not care if they have one boundary point in the given zero cell that will be finitely many so this will happen for every every vertex it will happen for every edge if it happens for every case in Plex that is the meaning of this one so the first remark here is if you place open cell by closed cell then this condition is still true suppose I take this condition every open cell meets only countably many closed cells take a closure of that that is a closed cell it is a larger one it may intersect too many of them no it will also meet only countably many why that is so because the closure of each cell being a compact thing covered by finitely many open cells this is one of the properties that we have seen each finitely many open cells will meet again finitely many sorry countably many closed cells so countable union finitely enough countable sets will be again countable ok whether I put every open or closed here does not matter they are the definitions will be the identical now here is a lemma take any locally countable CW complex ok take every locally countable CW complex every point X is contained in a countable sub complex which is a neighborhood of X in X that it is contained in a countable sub complex is obvious you can just take you know one of the cells containing that that is not a point but that will not be a neighborhood in general you have to have a neighborhood of X in X that means the set should contain an open subset of X and contain the point little X ok so this seems to be some more nontrivial thing we shall see this one next time thank you