 Today, we begin the subject, the CW complexes. In this chapter, we shall first introduce the most important class of the ploical spaces for algebraic topology, namely CW complexes, which are studied some fundamental points at the topological properties of these spaces related to the simple shell complexes that you have studied in part one. And also in subsequent chapters, we shall study its relation with maybe with fundamental groups and various other things also, homology and so on as and when time permits. CW complexes are built up out of from no error perhaps. So, by a sequence of operations called attaching cells, the attaching cells itself was studied in part one, but I will recall it so that to just to refresh your memory. So, fix an integer k and an indexing set lambda. For each index lambda, choose a copy of the unit disk in R and Rk. So, these all these copies are supposed to be disjoint now, they are copies of the same disk dk. Now, suppose x and y are topological spaces, we say x is obtained from y by attaching k cells from this indexing set. If there exists, so given two topological spaces x and y, we say x is obtained by attaching a k cell to y. If there exists a family of functions f alpha from the boundary of this disk which is sk minus 1, the sphere to y, for each alpha you must have such a function such that x is the quotient space of the disjoint union of y and all the copies of the disk dk by the relation x is equivalent to f alpha x, whenever x is in the boundary of dk of alpha. Remember boundary of dk of alpha is sk minus 1 and then there is a map f alpha, this is a quotient space. Let us denote the quotient map itself by q and then remember what is the meaning of the quotient space, the topology on x. What is it? A subset of x is open, if you find only if it is inverse image under q is open in the disjoint union of these things, the disjoint union of these spaces is given the disjoint topology, we have to remember that. The maps f alpha are called attaching maps for the cell. Let us restrict the quotient map q here to each dk, dk of alpha and then let us call them as phi alpha. These phi alphas are called characteristic maps of the cell, they are the homeomorphisms in the interior of dk and they are equal to f alpha on the bound. The image of dk of alpha is a compact subspace of x because dk of alpha are compact and f alpha 3 alphas are continuous. So, we call them closed k cells in this pair x y. Notice that why I am writing this x y is y is given space and x is obtained by this operation. The image of all these dk of alpha under phi alpha will be denoted by corresponding e k of alpha and they are called closed k cells. The y is an injective map on q, there is no identification within y. Therefore, we can use to simplify the notation, we can use the same y for the image of q y, image under q y. Any subset A y q A can be identified with A and it is a closed subset in q y because if you q A is closed in q y, if you know this q A is closed in x, that is the definition. But q y, if A is closed in y as such, its inverse image will be closed inside y again because under q it is just A itself and in all other dk of alphas, it is f alpha inverse of this A, A being closed as f alpha has been continuous, they will be closed subsets of sk minus 1. Therefore, they will be closed subsets of dk also. So, with this identification, it is not just synthetic identification, it is a topological identification. So, to reduce the number of notations and so on, we can now think of y as a subspace of x and that is why we can use this notation x y. This is a usually notation for whenever A comma B is used, B is a subspace of A, so that is the standard notation for pairs of topological spaces. So here is a basic lemma out of these quotient space topology, all these things are consequences of just taking the quotient space topology on the disjoint union of y and dk of alphas. So, first one is each ek is a closed subset of x. So, that is where we have to assume that y is an overdraft space, ek of alpha being an image of a compact space, it is a compact subset. The boundary, boundary of ek of alpha is subset of y, that will be closed. So, we have to see that, that will help to see that ek of alpha itself is closed in x. Each characteristic map, it is a quotient map, a subset A of x is closed in x, if we don't leave A intersection y is closed in y and A intersection ek of alpha is closed in ek of alpha. So, this characterization C is now exactly copy of the characterization of the quotient topology here. Here the characterization is in the top space here, on the mother space, whereas this part C here transfers the whole thing inside the quotient x itself, you don't have to go above. It is same thing, but you have to prove this one. So, what is the proof is using again and again the ausdorveness and it is a quotient map, how to perform and what is the definition of a open subset in x, or a closed subset in x. So, first of all, y is ausdorff, sk minus 1 is compact. For each alpha, f alpha of sk minus 1, which we denote by ek dot, ek dot denoting the boundary of ek of alpha, ok. All the interior points are taken away from this cell. This is a closed subset. Now, fix a beta, ok. So, see that ek beta is closed. For every beta you want to say this is a closed subset, right, this is a claim. So, all that you have to do is go back and check the criteria. Q inverse of ek beta must be closed in the disjoint union. In y, what is it? portion of y, it is just Q inverse of the pole which comes from the boundary part, ok. Q inverse of ek beta intersection y is nothing but ek beta dot. And this is closed. Q inverse of ek beta inside dk beta, it is identity map here, this is just dk beta. Q inverse of ek beta, beta not equal to alpha. What is it? It is just the boundary in f inverse of the boundary, ok. Since this is closed, this is closed and this is obviously dk beta is closed in dk beta, ok. So, this proves A. Once A is proved, now the criteria for ek of alpha being closed. A can be proved and ek of alpha is such that phi inverse of A, sorry, this is now what we are proving. We are proving that the quotient map restricted to each characteristic map, each dk of alpha, each cell that is closed, right. That is a quotient map. So, take a subset of ek of alpha such that its inverse image under phi alpha which is same thing as q alpha inverse but inside dk of alpha, only one alpha is taken. Pr, that is phi alpha inverse of A is closed in dk. Suppose this is true. To show that A is closed in dk of alpha, you have to verify that in all the full inverse in a, q inverse in A, it is a closed subset. That is what you have to show, ok. To show that A is closed in dk of alpha, same thing as saying that it is closed in x. Why? Because dk of alpha itself is closed. So, once again to show that means q inverse of A itself is closed in here, ok. By the hypothesis, f alpha inverse of A is closed in sk of alpha because you have started A inside this one, this is closed. Intersect with the boundary, it will be closed in there. Hence, A intersection boundary is ek of alpha, that is closed. So, once you have this, this first one here, the other two will follow just the same way here, ok. Now, once you have A and B, C is now directed consequence of B because q is a quotient typology. What is the situation here? C, C says subset of x is closed, you should know that A intersection y is closed in y and A intersection ek of, if it is closed, these things are closed because they are subspaces. Converse is what you have to see. Suppose A intersection y is closed, then that is q inverse of A intersection y, right. So, for that part it is fine. A intersection ek of alpha is closed, then by q the phi alpha inverse A, q inverse of A intersection dk of alpha is to be closed because this is a quotient map. So, that will complete the proof of that A, A itself is a closed subset. In the definition of the attaching, we could have taken the indexing set to be empty also, just for logical reason. And we should include this one, we should not assume that lambda is non-empty. Of course, what you get is y itself, when you do not attach anything, ok, then x will be y itself, the family could have been empty. If a family has just one member, ok, then this is nothing but x is nothing but the mapping cone of f, f from sk minus 1 to y is a function and then you are filling up this sk minus 1 with a, with the cell, with the dk, it is the same thing as taking the cone over that part, ok. So, this will have been, if you know, we have to know the map, what is the definition of mapping cone here, I am not recalling that here. So, that is a special case and in that special case we have this simple notation also, y union over f of the cell ek, there is only one cell, there is no need for writing alpha. In general, without the assumption of austereness, ek of alpha may not be closed. Very simple examples can be given, there is no problem about that, ok. However, even if you do not assume y is austere for anything, y will be always a closed subset of x in the attaching process. Also, unless on the boundary namely alpha's attaching maps are 1, 1, you will not have ek of alpha, homeomorph will be k, ok. So, here in the figure I have given you an example of attaching one cell only, k equal to 1 in this picture, the picture can be always one or two dimension anyway. So, this is your y, which looks like y anyway, and the first cell is attached to one cell, one cell is from d1, d1 is minus 1 plus 1, the interval minus 1 plus 1, ok. So, minus 1 has gone here, plus 1 has gone here, f of minus 1 is this point and it is identified, f of minus 1 plus 1 is this point is identified here, that is the meaning of this identification space. This could be just anything homeomorphic to open interval here. Similarly, this is another open interval here, another cell. Remember the interior of this cell, interior of this cell, interior of this cell, they are already shown on the boundary what happens that is left to f alpha. Boundary should be always has mapped on to y somewhere inside y. In this d1, d1, 3, the third cell, both the points of the boundary have gone to a single point. So, corresponding to f3 is a one point map. So, now you see that this is not homeomorphic to d1 at all, these things are homeomorphic to d1, these two, but this is not, ok. So, this is what happens. If you draw a picture, suppose I draw a line like this, like this that is not considered that cannot be a attaching map, it is a, you know that is not part of attachment. On the other hand, I can just take a point here, just a point here that is a cell out. Then it will be attaching a zero cell, ok. It is not attaching one cell. So, just now we have defined k equal to a constant and k greater than or equal to 1. So, let me define what the meaning of attaching a zero cell also, alright. So, we say x is obtained by attaching a zero cell, I have defined k greater than or equal to 1 here, k cells, right. So, I want to include zero cell also. What is the meaning of attaching a zero cell? Namely, x is disjoint union of y and z, where z is a closed discrete subspace of x, discrete and closed subspace of x, y itself will be closed. Both y and z will be closed by this definition because it is a disjoint union of two tuple equal spaces. Why is a given space? z d is a set of points, the topology on those points is just discrete. So, this is completely justified because first of all you have to say what the meaning of zero cell or d zero, ok. D zero, zero cell r zero, what is it? Zero vector space. There is only one vector there, which is a zero vector. So, a zero cell is just a single term, ok. It loses all the other geometric properties here, it is a single term. And then what is the boundary? The boundary is empty. So, there is no map there. Therefore, no gluing takes place. So, this is a logical justification for defining like this. If you have difficulties in this logic, you can just take this as definition, no problem, ok. Attaching element that belong to z is then a zero cell of x. These are elements of z are called zero cells. Point, why is a black box? There is no, no, no, this is like a base space. There is no name for this one. When you attach a zero cell, they come from a disjoint set. Attach one set, they come from a disjoint set indexed by a family. That is what it is. And zero cell by the very definition by the topology and whatever you want to say, each point in z is closed as well as open in the whole space. So, it is zero cell is both an open cell as well as closed cell. The quotient topology retains a number of properties of the base space of the top space. There are quite a few of them like false purpose. They do not go to the quotient space. We have studied quotient space thoroughly last time, ok. So, what happens is out of away from y, you can expect a few properties of the Euclidean space in z x. I mean x minus y can have a number of topological properties inherent from the quotient space, from the Euclidean space. It is a decay I have been attached there, ok. The first attempt is here is, so what kind of things may happen inside dn and sn minus 1? That is an elementary thing here. So, I have put it as a lemma which is the starting point of topology here. Take a subset a contained as sn minus 1, ok. Fix an epsilon between 0 and 1, strictly between 0 and 1. Let us put n epsilon a, this is a notation. All x in dn is that norm x is bigger than epsilon and when you divide over to a modulus namely the unit vector is inside a. Remember a is inside a sn minus 1. So, it has to be unit vector, ok. Take all such elements as this x pi norm x varies over a. What are these things? This is just like take a vector v and multiply it by some number r between 0 and 1, ok. So, I can say it like that also. So, point and then a small line segment going towards 0, but 0 is strictly avoided because this 0 is less than epsilon x and 1, ok. All those points you take, all right. Then this n epsilon a, ok. See x, this norm x is bigger than epsilon. It will include 1 also. Only the epsilon is between 0 and 1. This norm x can be 1. Norm x will never be 0 because norm x is bigger than epsilon, that is all. So, n epsilon a intersection sn minus 1 will be exactly a. For all points, take a, say v belong to a, that v itself will be there, ok. So, a is contained here, this one is clear, but intersection sn minus 1 is precisely a because norm x to be 1. The norm x to be 1, this is precisely could be a. N epsilon a is an open subset of a. If you know if a is open in sn minus 1, it is also clear. If this is an open subset, intersection is open subset is clear. If a is open, why this one is open? Because this will be like, just like then an open arc, you know, it is like a product of biology now, cross with 1 minus epsilon into 1, 1 minus epsilon open. That will be the topology of it. That is why this is n epsilon a is open. The last thing which is very important is that you look at this map which joins x and x by norm x, the line segment, 1 minus t times x plus t times 1 minus x, t times x by norm x, ok. If t is 0, this is x, if t is 1, it is x by norm x. This is strongly from the tract of n epsilon onto a. That is a picture. If you allowed epsilon to be 0 also, then there is no deformation because you do not know where it was under 0. 0 has to be not there. That is important. So, I want 0. Then you have this, these are called, you can call it a collar neighborhood of a inside dn, ok. So, from this one, now we want to extend this one to the attaching cells from y to x, ok. So, what we do? We take a family of dk alpha and suppose I have attached them to y, just like in the definition above. Now, start with a subset b of y, ok. Then look at, for each alpha, look at f alpha inverse of b. That is a subset of sk minus 1, the boundary of dk, right. So, treat this as a, choose an epsilon, call it as epsilon alpha, take n of that, n epsilon of that. So, you have chosen these open subsets. Let us say this is open, then that is open, otherwise it is not open, it is a very open collar. Inside dk of alpha for each alpha, take the union, that is a distant union, include b also, take the image q under q, q of that. So, call that as n epsilon b, ok. n epsilon a in the earlier thing was defined only for this picture. Now, I have defined this for when x is obtained by, obtained from y by attaching cells, ok. The first thing is to take n epsilon b intersect y, points which are inside y of this one. They are nothing but qb, ok. If q of, if you take this part, they will also coincide inside y. So, they will be just qb. n epsilon b is open, if you know only if b is open in y. So, it is a straight forward from previous lemma. If b is open in y, each of this is open. So, then n epsilon is open. These are the intersections of n epsilon b inside dk alpha. And this part is b, this is part of y. So, that follows from directly from the previous lemma and quotient topology definition. b which is equal to qb, you can write it as b or qb, because we have identified these things, they are suspects of y. It is a strong deformation type of n epsilon b. So, this follows from the third statement here, put together. So, how do you do that? All that you have to do is on each alpha h alpha, take this map for each h alpha, take this map, ok. Put them together in the disjoint union. What you get is h is restricted to b cross i, put it as identity. And in the rest of these things, put this is h alpha, ok. So, this patches up to define a big homotopy from this entire space to this space. q of that, q of that cross i have taken, q of that cross i identity, you will come back to h hat, ok. So, this will give you a homotopy from n epsilon b cross i to n epsilon b. Homotopy of the identity map with the, it was with the retraction there, ok, on to b. Everything will go inside b here, ok. So, this property we are going to use at the end again. So, you can recall that this notation t 3 space means what? A host of space t 2 plus regular. Similarly, t 4 means a host of space which is normal, ok. Now, this is a simple exercise, maybe it should take some time like you have to verify case by case and so on. But this is an exercise which you should be able to do by yourself, do it so that you get familiar with what is going on with this lemma, ok. Now, let me give you the definition of CW complex. A relative CW complex will define slightly more general things just like attaching maps from y to x, right. By the way, one of the thing is that y could have been empty set also, ok. Can it be, if y is empty set k is greater than equal to 1, what are the maps? There is no map. So, you will never get to attach any one cells, but you can attach 0 cells, ok. So, y is empty is allowed if you can attach 0 cells, ok. And that is why we want to take empty family as well as empty set as permitted, ok. So, let us take this relative CW complex. So, what is the definition? A is a space, some typical space and x is obtained by a sequence of processes, namely each process is an attaching k cell where k equal to 0, 1, 2, 3 and so on. So, let me make an elaborate definition here. Consists of a topological space x and a closed subspace x, which is Hausdahl. This is a part of the definition. Some people do not assume it, but I am going to assume this, ok. Together with a sequence of closed subspaces xn, n greater than equal to 0. So, I am going to define x0, x1, x2, x3, ok, inductively. So, what is x0 going to do? A is a state of x0 and x0 minus a is a discrete space. It is precisely what is the meaning of we have already defined that x0 is obtained by a by attaching 0 cells. So, instead of saying that I have just said x0 minus a is a discrete space, ok. A is a closed subset. xk is obtained by attaching k cells to xk minus 1, for all k greater than equal to 1, ok. So, x1, what is x1? It is attaching 1 cells to xk minus 1, which is x0, ok. x0 could have been just a, because all that you need a, a is contained as x0. This is a discrete space. If it is empty set, it is also allowed, ok. So, do this for each k. Then what is x? x is just the union of all these spaces. Remember, by the very definition here, xk will be a subset of xk plus 1 and so on, each of them closed in the next one. So, you have sequence of closed spaces. A contained is at x0, contained is at x1 and so on. Take the union, that is your x. The last thing is, what is the topology on x? This topology is called the Witt topology, or what is known as the coherent topology or co-indistropology. All that names are of no use, unless you know what is the meaning. So, meaning is precisely a subset of x is closed, if and only if its intersection with each xk is closed inside xk. xk is the space is here which being obtained from xk minus 1 and so on by attaching k cells, ok. So, the topology on each xk is well defined as soon as you know what is a and you know what are the attaching maps. There is no ambiguity. But what is the topology on this infinite union that has to be defined and this 0.4 here takes care of that. It says it is a coherent topology, coherent with respect to each xk, ok. Automatically with this definition, each xk will be a closed subset of the whole space x. Not only xk is closed inside xk plus 1, it will be closed inside xk also. All these things we can see, but this is the definition first of all, ok. The interesting case I told you, a could have been empty. Then x naught minus a cannot be empty. Cannot be means what? If it is empty, then everything else will be empty. Because x1 have to be attached, you know one cells, one to attach one cells, you need maps. Maps into empty set, there are no maps empty, ok. So, then everything will be empty. But x naught may not be empty. There may be one point at least, two points, 50,000 points, no more, no problem, ok. It is a discrete space, then you can start attaching more, more and more. But in between, you may not attach, for example, one cells at all. You may not attach two cells at all. You may directly attach three cells. That is possible. In between, some xk may be equal to xk minus 1, ok. So, all these things are allowed. So, one of the interesting cases when a is empty, then we do not write x comma a at all, we just write x. And that is called a CW complex. So, I will redefine what is CW complex. A CW complex, if it is non-empty, it has to start with a some discrete set, x naught. It has to. Otherwise, it is empty, ok. So, there will be a set of points. You can call them as vertices, just like in the case of simplicial complex. So, this x naught, it can call them as zero cells or vertices solve no problem. So, it is a discrete space. Then it is a question of, I mean, what you want to do, what the space is. There may not be one cells, there may be one cells, there may be 50,002 cells and so on. The only thing that is needed is, if you take a k cell, the attaching map to that k cell, which is on the boundary of this k cell, namely sk minus 1, that has to be a map from sk minus 1 into the xk minus 1. So, these xk's are called the kth skeleton of x. It is just a name, and a useful name. So, that is what. It may happen that after certain stage, there are no k cells. No cells have been attached. So, x is stopped at xk. If that happens, we call this x k dimensional, less than or equal to k dimension. Suppose there are at least one, there is at least one k cell and no more cells of dimension bigger than that, then we call dimension of x equal to k. So, these are just definitions. So, open cells, characteristic maps, attaching maps, all these things make sense in the case of CW complex also. Whatever you have done for attaching maps. So, that is just the gist of definition. We shall take up this study next time. Thank you.