 Welcome to module 35 of points anthropology part one, we shall continue our study of connectedness. So far things were not all that difficult, but now we are trying to take you a little deeper into connectivity. Here is a result that can be used to prove connectivity of a big class of interesting spaces. So, let us formally make a definition here. Once again, this is for namesake only. Take any set theoretic function f from x to y. For every y inside y, look at all points which are coming to y, all points of x is coming to y, namely f inverse of y. That will be called the fiber of f over y. What are the fibers of f? They are all inverse images of some point in y. For example, if you take a point y, there is no point which is going to that point f inverse of y will be empty. So, then I will say fiber of y at that point is empty. So, that is also allowed. It is purely set theoretic notion. There is no other condition on f. The word fiber is used in many higher mathematics. I am used to that. That is why I am putting that terminology here. So, everywhere fiber occurs, you can just write f inverse of y inside the core domain. Then you do not have to use this fiber word. Let f from x to y be any quotient map now, in particular is subjective. Suppose y is connected and all the fibers of f are also connected, namely I am writing it just for your sake, f inverse of y for every y is connected. They are all connected. Then conclusion is x is connected. You see, what we have seen earlier is that if x is connected, then f y is connected. Therefore, in a quotient map f of x is whole of y. So, if x is connected, f of x is connected, f x equal to y. So, y is connected. We are looking at a converse here. y is connected. You do not immediately conclude x is connected. But if the fibers are all connected, there is a extra condition, then you can do the other way around x is connected. So, it is quite a profound result, but the proof is very easy here. All that I do is take a separation instead of a, b I have called a 1, a 2, does not matter. Take a separation of x. Look at all points b i, all points of y below. So, there are points of y such that the fiber over that point is contained inside a i. b i, I am just defining it. That is my b i. This b i is subset of y. So, this I do for both i equal to 1 and 2. As such, b i could be empty, they could be anything. But here I am using two things. Namely, f is surjective and then it is actually a quotient map and the fibers are all connected. So, together these things will imply first of all, all these b i's are non-empty. b 1 union and b 2 is the whole space x that comes from surjectivity. And sorry, before that each f inverse of each point, namely the fiber being connected must be inside a 1 or a 2. Therefore, a 1 union a 2 will be the whole of x. What we claim is this b 1 b 2, b 1 b 2 union is the whole of x because each f inverse of y is contained inside a 1 and a 2. So, union of all f inverse of y is the whole of x. Therefore, y will be union of b 1 and a 2. What we want to show is this is, you know, we started with y as a connected space. This is a partition. This is a separation. That will be contradiction. If a 1 is non-empty, f of a 1 will come to something, it must be either b 1 or b 2, it has to be inside b 1. Similarly, f of a point in a 2 will have to be inside b 2. So, b 1 and b 2 are non-empty. Now, why b 1 is closed? b 1 inverse, f inverse of b 1, sorry, f inverse of b 1 is nothing but the entire of a 1. And to begin with, we start a 1 and a 2 are closed subsets. Inverse image of a set is closed implies the corresponding set is closed in the quotient topology, in this quotient map. Therefore, both b 1 and b 2 are closed. Clearly, a fiber cannot be both b 1 and b 2. It has to be in one of them, right? So, b 1 intersection b 2 has to be empty. So, all this show that b 1, b 2 is a separation for y and that is a contradiction. Why a contradiction? Because we started with a wrong assumption that x has a separation, okay? As an immediate corollary, we get a big theorem here. Take any finitely many connected spaces and then take their product that is connected. How do we do this? Do it for 2 at a time, then by induction it will follow for any finite product, right? For 2 at a time, how do you do? Look at x connected, y connected. Look at x cross y to y, the projection map or you can take x cross y to x also, no problem. That projection map is an open map. Any open surjective map is a quotient map. So, you can apply the previous theorem to this quotient map. Y is connected. What are the fibers of y? Fix a y, all points x comma y coming to y, nothing but x cross singleton y, which are all homeomorphic to x. Therefore, each fiber is connected because I start with the assumption that both x and y are connected. So, theorem says that the product is connected. Once you have proved it for 2 by induction, it will be true for all finitely many product. The above theorem, together with this corollary, it is an easy corollary. Sometimes you have to directly use corollary. Sometimes you may have to go back to the original theorem. That will be very useful in proving connectivity of a large number of tucaloid spaces. So, that is what we will do now, namely what are called as the general linear groups, the orthogonal groups, the unitary groups and so on. They are all inside n cross n matrices. So, let us make a formal introduction to these things which are very important in mathematics, in the central mathematics. Look at, of course, my k is any field you will be to begin with, but finally when you talk about biology and so on, I am taking k as either real numbers or complex numbers. Remember that. So, this notation m n cross m k denote the set of all n cross m matrices with entries in k. Clearly, under entry-wise addition and multiplication, it is a vector space of dimension n m over k. This is part of your linear algebra. Therefore, by choosing an appropriate and convenient basis, we can identify this vector space with k power n m, the Cartesian coordinate space because dimension is n m. There are many different ways of doing it. For example, you may write the rows of A as, see there are n rows and m columns. So, each n row, the one single row is a vector inside c power m, k power m. Write it the next row to the, you know, on the side of, on the right of that and so on, instead of writing the below. So, write them side by side so that it will look like a n m order to pull. For example, if you have 2 by 2, then you will get a 4 vector, okay, a 2 vector followed by a 2 vector. So, you get a 4 vector and so on. So, that will give you an identification of this vector space with k power n m, okay. It just depends upon which way you want to write. Any of them will be as good as any other one. So, you better choose according to your convenience. That is why convenient and appropriate. Appropriate to the context. But you should keep in mind that your original space has a matrix structure for many other purposes. So, in order to rely, in order to remind you that instead of writing k m n here, k n m here, n m m n, once you write 5 into 3 as 15, it is lost. 5 into 15 could be 1 into 15 also. So, you write it as n cross m. So, this is just a reminder that it is an n cross m matrix, okay. This notation is used. It is a very clever notation. It is not mine. It is there in the literature. Instead of k m n, you write k m n. As a Euclidean space, they will be both same. As a vector space, they are the same, okay. As matrices, this will be different than k 1 cross n m or n m cross 1 and many other possibilities. Various subsets of m n k are of interest in higher mathematics. Now, if this notation m n k stands for n cross n, okay. So, instead of writing 2 of them, I am just writing n. That means they are square matrices. Here are a few of them. I am giving you definition of that. Look at all square matrices of size n such that the determinant is not equal to 0. It is the same thing as all those which are invertible. The S o n is inside now R n, okay. R n cross R n cross R n and so on. R n cross n. N cross n matrices with real entries such that A A transposes identity and determinant of A is 1. If you do not put this condition determinant of A, then this is called O n orthogonal group. So, this is a special orthogonal group, okay. U n is all those complex n cross n matrices such that A A star is identity, okay. Here also you can have another one S u n wherein you put one extra condition determinant of A equal to 1 here, okay. I am trying to give you only a few of them here, but I just told you two more also here. Note that the first one g l n k is an open subset of k n cross n or all the matrices. Why? It is given by determinant of A not equal to 0. It just means that inverse image of the function determinant which is a polynomial function, okay. So, therefore, all those points which are you know determinant is a function into k, right. k is either c or r. So, not equal to 0 means that is an open set. So, it is the inverse image of that open set, okay. So, this is an open set whereas for similar reasons S o n, this is an equation, that is an equation. It is not an equation, okay. S o n and u n, they are closed subsets of m n, m n k either c or r. This notation is A transpose and this is A star is conjugate transpose, okay. So, I claimed that our previous theorem along with the corollary about finite products can be used to prove that all these three here at least and then many more are all connected spaces. So, let me present a proof of one of them, the first one, okay. When k is c, I am taking the gl and c, okay. If you take gl and r, then you have to be dangerous, then you have to put determinant of A equal to 1 or minus 1. That will have two different components, okay. Okay. The space gl and c, what is the space? It is now subspace, open subspace of the Euclidean space is connected for all n greater than equal to 1, gl runs, it starts with. So, just a temporary short notation g n depends upon n, right. So, I am going to prove this inductively. So, let us first prove c, g1, g1 is just c star. What is determinant of a 1 cross 1 matrix? It is just the element itself and that should not be 0. So, it is just c star which is c minus 0, okay. We know that c minus 0 is connected. Dropping out a finitely many points from r2, r3, rn, they are all connected that we have seen already, okay. So, now assume that the result is true for n minus 1 and n is the determinant of 2. That means, g n minus 1 is connected we are assuming, okay. Now, let us use this symbol En to this column, this is a column vector 0001 inside c power n. So, to write a column vector I will 0001 and put a transpose here that is all. Consider the map phi from n cross n matrices to c power n given by phi of a equal to a operating upon En. What does it give you? It would give you the last column of this matrix A, okay. This formula is nothing but the last column of the matrix A. Writing the last column you know is a bunch of coordinate functions. It is one of the coordinate functions. So, therefore phi is open surjective mapping, right. This is an open surjective mapping but we are not interested in the whole space here. We are interested in only an open subset of that g l and c, okay. So, all this I have told you that it is just a quotient map here because it is a projection map to the last n columns, okay. In particular, phi is an open mapping. Now, g l and c is an open subset of that map. So, it follows that phi restricted to g n is also an open mapping. But what is the image? If you take a non-zero, it is an invertible matrix that is determinant of A is not equal to 0. Each column is a non-zero vector that is a minimum thing. Therefore, the last column which is A of n, it is a non-zero vector. Therefore, the value is taking inside c minus c n minus 0 0 0. So, I will write it as c n star. What we want to say is that this is also surjective namely from g l and c namely g n to c n star. So, this is surjective is what I want to say. Why? Take a non-zero vector inside c n. Use your linear algebra to complete it to a what? A basis. How many elements will be there in the basis? Exactly n elements. The first vector you write it as a last one. V is equal to the last vector. Then write the other n minus 1 vectors, column vectors they are all in 3 to the minus column vectors. You will get an n cross n matrix. That matrix because each column together they span the whole thing. So, that must be a basis. This is what we have assumed. So, the matrix will be invertible. Determinant is not equal to 0, which is the same thing as a matrix invertible. So, what we have? What is an element A inside g n whose nth column is v? It just means that if you take phi of A, it is v. Therefore, phi restricted to g n to c n star is surjective. We are exactly in the situation of theorem 3.34 that we have just proved. It is a surjective open mapping. So, it is a quotient map. c n star, what is it? c n cross c n cross c n. So, sorry, c cross c cross c minus 0 0 0, which we know is connected. What are the fibers of this phi? If we show they are connected, then you are in a good shape. You can immediately conclude that g n is connected by this theorem. Okay. Therefore, what we need to show is that each fiber here, phi inverse of v, this is connected. By definition, what is phi inverse of v? All those n cross n matrices inside g n, such that when evaluated and last on a n, which is the last column will be equal to v. This v is fixed. First thing I want to show is that all the fibers are homomorphic to each other. What is the idea? Finally, I want to prove that they are connected. So, if I prove one of them is connected, all others will be connected. So, let me prove that they are homomorphic to each other, which is much easier to prove. So, what you have to do? Start with any non-zero vector. Choose a in g n. This we have done earlier, such that a of e n equal to v because we know it is surjective now. Okay. It then follows that the left multiplication by a defines a homomorphism. Take an element here, multiply it by a on the left. What is the inverse map? Multiply by a inverse. So, this is a homomorphism of the whole space to itself, g n to g n to g n. But I want to say that this fiber goes to that fiber. What is an element of this fiber? All those b such that b of e n equal to e n. What is this one? All those say b prime of e n is equal to v. So, if you take an element such here b, a times b will have this property and go on us. Okay. So, homomorphism is very easier to verify here. Now comes, I have to just prove that one of the fibers namely phi inverse of e n, this is connected. You can see why I have done this one in the next step. Why the choice of e n is done here? Namely, this allows us to apply the induction, the induction hypothesis. Okay. So, the claim is that phi inverse of e n is connected. So, but what is it? I am just repeating it here. What is it? All those b in g n, b of e n is e n. This is the definition. I want to say that this thing is homeomorphic to g n minus 1 cross c n minus 1. Suppose I prove this one, then the theorem is over. Okay. This was only thing that was needed. Once one fiber is connected, all the fibers are connected. So, I can apply the previous theorem to conclude that g n is connected along with the induction hypothesis, it follows that g n is connected. So, all that I have to prove is that this set, this space is homeomorphic to this space. Look at what happens. b is in phi inverse of e n, means the last column is 0 0 0 0 1. Okay. So, this b has the last column. This is a big 0. This is a block matrix, by the way. This is n minus 1 cross n minus 1. This is n minus 1 cross 1. This is 1 cross n minus 1. And this is 1 cross 1. It is a block matrix. Okay. So, this whole column is just 0 0 0 1. That is the meaning of that it is e n. Right? This is some block b 1, which is an n minus 1 cross n minus 1 matrix. And this is another just row here. u n, u 2, u n, u n may be some z 1, z 2, z n, whatever, z n minus 1. Okay. So, every b which has this form, namely inside phi inverse of e n has this form. But determinant of this one is nothing but determinant of b 1 times 1. Therefore, this must be non-zero. The moment this is non-zero, it is element of g n minus 1. And what is this one? This is just a element of c n minus 1. Projection map here is to first n minus 1 coordinates here, n minus 1 cross n minus 1 coordinates. That is continuous. Projection to these coordinates also continuous. So, breaking b into two parts like b 1 comma u, that is a continuous function. This is what I am putting here. The compliment, I mean the inverse is also continuous from the whole six to this one. Therefore, this becomes a homeomorphism. This proves that g n c is connected. So, here little bit of many interesting topology is involved here. Okay. So, the method employed here is educative again. Not only the information that you have got now that g l n c is connected, okay. If you have learnt this proof nicely, it will help you in long way for many other things. Now I will have to go to another important thing here, okay. This time I may take a little more time. The topologist signed a card. Remember, I wanted you to, I want to, I promise that I will show you why the closure of a path connected that may not be path connected. That is what we are coming to here. This example will serve that purpose but pay attention to this example because it will serve as a many, many counter examples for us, okay. For us as well as for other people also. So, this thing, something you have to study properly. What you said? What is the sine curve going to do? It is starting with the graph of a sine function. Only thing is instead of sine x, we are taking sine pi by x. The x is inverted here. Therefore, the domain should exclude 0. At 0, it is not defined. So, I do not want to take 0 to infinity. I just take 0 to 1. That is enough for me. You can take 0 to infinity for other purpose. 0 excluded, 1 closed, no problem. Take the function sine pi by x and look at its graph. The graph is x comma sine pi by x, point like that, which will be subset for R2. Okay. So, the whole topology sine curve is going to be subset of R2. Okay. You can just look at this curve, but now we want to do something more here. Namely, we want to put the y axis part between minus and plus 1. So, 0 comma y between minus 1 and plus 1. This is the lines, closed line segment, okay, lying on the y axis, 0 comma y. So, you take the union of this sine curve, namely the graph of sine pi by x, union with b. That is my x. Okay. So, this is by definition the topologist's sine curve. Okay. Being the graph of a function, continuous function defined on an open or on some interval which is connected. Okay. So, the graph A is connected. Actually, it is path connected. The whole thing is a path. It is also path connected. All right. So, A is connected. A bar is the whole of x. So, this is what one has to see here. I will show you the diagram here, then it will be very clear to you. So, x is connected. A is actually path connected, but I am not going to use that. Just connectivity is enough to conclude that x is connected. Let us look at this curve. So, this is the sine curve. This is 1. That is 0. At 0, it is not defined. But what happens to this? This is 1 by 2. Okay. This is not 1 by 2. This is what? Yeah. 2, 1 by 2, see pi x by pi. So, yeah. So, you see for, let me see, this pi by x is there. Put x equal to 1 by 2, then sine 2 pi you get. That is 0. Okay. Like this. So, at all these 1 by n or 2 by n, 4 by n and so on, it will take either this point or this point or this point or this point or this point. It depends upon even and odd. So, it will keep going like this. This part is b. This line segment on the y axis. This is the y axis from minus 1 to plus 1. All the, the entire curve is between this y equal to minus 1 to y equal to plus 1 in that strip because sine x does not take values other than minus 1 to plus 1 in the interval minus 1 plus 1. So, this is like. So, now look at any point here. There are all these points coming closer and closer to this point. So, every point on the y axis from minus 1 to plus 1 is in the closure of this a. This is a. Therefore, a bar is the whole of x. Is that clear? So, every point on the y axis from minus 1 to plus 1 is a closure point of this a. Therefore, our theorem says that this space is connected. A itself is path connected because it is actually, itself is a path. Except that in the definition of path you must have a closed interval. So, we do not have a closed interval. So, what you do? Suppose you take a point here, from any other point here to any other point here, you can take the restriction of the function sine pi by x that will give you a path here. So, this is path connected. So, since a bar is x follows that a is not closed. Of course, it is easily seen that b being a line segment that is closed. We claim that both a and b are path components of x. That will show that x is not connected, x is not path connected. So, this is the final thing we have improved. That a and b are both path components. They are path connected because one is a closed interval, homeomorphic closed interval. Other one is the image of of a continuous function defined on an interval, open one side open as a cyclone. No matter, it is an interval. So, what we have to show is that take any point on the b part and any point on the a part. Two points to take show that there is no path from one to the other. That will show that these two are what path components. See, if you have taken a, what you have to do bigger than a, you have to take a point of b. So, you must be able to join a point of b to a point of a. So, I say there is no such path. So, if I prove this, what the claim that a and b are path components follows. Therefore, x is not path connected. They are exactly two path components. Suppose you have a path. Path means now what? Some continuous function defined on a closed interval. I can change the closed interval to any 01. There is no problem by re-parameterization. So, you can assume that there is a path from 01 to x such that omega 0 is inside b and omega 1 is inside a. So, remember x is subspace of r, r2. Look at the projection map r2 to r, the x axis, x projection, first projection x, y going to x. Then, pi composite omega 0, this is what? This is the x coordinate of omega 0. Omega 0 is inside b. So, the x coordinate will be 0 because the whole of b is just 0 cross minus 1 plus 1. So, pi of omega 0 is 0. Pi of omega 1 will be something on the x axis between open interval 0 to 1. So, it is positive. Of course, it is less than 1, less than or equal to 1, but it is not equal to 1. It is some positive number. By intermediate value theorem, see this map is from where? From x to r. So, the codomain is r. So, intermediate value theorem for this map 01 to r, not from x to r, but I have taken 01 to x and then composite with pi. So, 01 to r, it follows that for every k inside n, there exists a tk belong to 0 comma 1 by k. Why I want to take this one? Because I want to make this tk convert to 0 as k tends to infinity. That is why I have made this one. But these points are where there is some point here. So, what is the statement? Statement is look at omega 0, that pi composite omega 0 and pi composite omega 1 by k. Between these two, there must be some point, take that point conveniently, tk will go to that point, that is the intermediate theorem. So, such that pi of omega tk is equal to this point, 2 divided by sum 2 nk plus 1. All that I want to say is look at omega 0 is 0 and omega 1 by k instead of omega 1 here, sorry pi of omega 1 by k is some positive number. Between them, there will be a point like this. So, that is what I want to say. So, it can be that both of them are 0 and this I have to choose is that is not possible, that is the point. So, 2 divided by 2 n plus k, this is some integer, this is r integer is what I want, 2 n plus 1, 2 divided by 2 n plus 1 for some nk and such n. Once you have chosen this way, what happens is omega tk will be what? It is x coordinate is this point. So, I have to write the x coordinate here. Once x coordinate is written, this omega tk is some point inside the graph. So, it will look like this. See, what I am trying to do is if the point is already on the x y axis, I throw them out. After all, the curve starts from omega 0 and omega 1. So, between them, you have lots of parts of this path a, this is the graph of a. So, it must be somewhere in this graph, that is what I want. In other words, all these points, see this point is omega 1, right, pi kappa 3 omega 1 or something here does not matter. Now, when you come here, this must be because intermediate value theorem, it must be omega of something. This must be omega of something. This must be omega of something because they are all, once I have chosen this one, the point must be between these two. Point must be between these two like this. I keep choosing them closer and closer, that is the whole idea. So, once you have chosen that way, this will be sign of pi divided by this number, pi divided by this number 2 nk plus 1 by 2. So, what is sign of this one? It will be plus minus 1 and x coordinate is this one. So, that is precisely what I have been pointing out. x coordinate will be somewhere here or here or between these two points and y coordinate will be precisely that. x coordinate will be somewhere here, the y coordinate will be that. This, that, this, that does not matter, all of them may be this point. But I will always choose something after that here, one here which I can always choose them alternately. That is why I have taken it as odd numbers. All these things must be there. All that happens is by continuity, if you take the limit of this one, it must converge to some point in B. What is that point? Actually, it must be omega 0, this 0 it must be, right? Omega 0 is 0, 0, 0. Omega 0 is 0, it must come to that one because this goes to 0. But this point is plus minus 1, plus minus 1. This is not a convergent sequence. So, this contradicts the continuity of omega. So, it proves that there is no path, all right? So, this proves that this theorem 3.29 namely the closure of a path connected space is connected, that is not true. Closure of a connected space is connected. If you want to put path connectivity, it is not true. It is not valid for path connected space, okay? Let us show that this same example gives you a counter example for this also for path connectivity, okay? So, it does not take much time. So, we show that one also. Now put x equal to the topology sine curve and y equal to the closure interval is 0, 1. And f from x to y, the surjection projection map to the x coordinate, which is a quotient map which we know because it is surjective open mapping to begin with, restricted to x also it is surjective, all that I have to say that is all, okay? We know that x is not path connected, right? But y is path connected, y is just 0, 1, okay? What happens to the fibres? The fibres of f0 take any point between f inverse of 0 is nothing but sorry, f inverse of 0 is nothing but the b, the component b, 0 cross minus 1 plus 1. So, this is path connected. What is f inverse of any other point x where x is not equal to 0? Look at this picture. There the only part is the sine curve, which is a graph. The graph means for each point is only one point here. Each point is only one point. What is it? x comma sine pi by x. Therefore, the inverse image under the projection is precisely one point here. The fibres of each point between 0 open to 1, there are only one points. So, they are also path connected. So, the hypothesis is satisfied. Everything is satisfied yet the conclusion is wrong, okay? Conclusion says that this is path connected but that is wrong. Therefore, the theorem is not valid for path connectivity, okay? Okay, so this example, this example, you know, has a very peculiar property. You take a small neighborhood of this point on the y axis here, okay? A small neighborhood. Then look at the intersection of this entire thing there. It will have lots of segments of this curve, infinitely many of them. They are all disjoint with each other. Every point on the y axis, every neighborhood, no matter how small the neighborhood is, has this property. It is union of all segments. Disjoint, first segment will be on the x axis, that is a straight line. But otherwise, all other segments are part of this sine curve, right? So, this phenomenon we want to study and make it into a property namely locally disconnectivity or locally connectivity. That will be the next topic. We will do it next time. Thank you.