 So, module 12, quotient spaces, the idea of a quotient map and the construction, open the floodgates of geometric topology to us. We can now study a large class of very interesting geometric objects via topology. Before proceeding with the topology, let me make it clear that all of you understand the meaning of a surjective map, a surjective function from one set to another set and what it means in terms of decomposition of x or equivalence classes, there are three different pictures of this which imply the same thing, same concept. Look at this diagram, the first portion here represents a surjective function, a onto function from one set x to another set y. When you have this, you can take f inverse of y as y ranges over capital Y that will give you disjoint union of x as I have written down here, f inverse of y means all points of x which come to a, the point y here, okay, so by the very definition of f they are disjoint and the entire x is a union of this thing. Because f is surjective, each f inverse of y will be what? Non-empty, therefore what we get is a non-empty disjoint decomposition of x, okay. So such a thing is called a decomposition or a partition. So what we have got here is a surjective function into another set implies a partition of x or a decomposition of x. Now suppose you have a partition of x, okay, then you can define an equivalence relation on the set x, namely in which the partition members will become equivalence classes, namely x is equivalent to y if and only if it is in the same subset. In particular here, what will be the relation? x1 will be x, it will be equivalent to x2 if and only if fx1 is equal to fx2. So a partition giving rise to a relation or a function directly giving a relation, this picture you can get from here directly or from directly from here. Finally, suppose you have an equivalence relation like this on x, look at the equivalence classes that is a set call that set as y, then what should be the function from x to y, take any x and take its equivalence class. So from equivalence relation, you can come to a surjective function. If you follow this procedure again, this cycle, what you get is whatever you have started with, namely suppose you have an equivalence relation and then you have defined f to be x going to the equivalence class of x, then what will be the disjoint union here, there will be equivalence classes and what is the equivalence relation induced by that, the same equivalence relation because these are the equivalence class correspond to this relation. So this is what I am saying, given a surjective function on a set is equivalent to giving a partition by f inverse of y and then giving a partition say x is disjoint union of ai, you can define x is equal to x2 if and only if both x1 and x2 are the same subset ai, so this is an equivalence relation. Finally, giving an equivalence relation, you can take y to be the set of equivalence classes and define f as f of x equal to the equivalence class of x. So verify that if you run the cycle of arguments again, what you get is wherever you started from, you come back to the same point, same concept. So when you have an equivalence relation or a surjective function or a decomposition, even if one of them is given, you should have all the three in your mind so that you can use them whichever one, whichever description you want you can use. Now let us come to the topology. Suppose x tau is a topological space, x is a set and tau is a topological space and q is a surjective function from x to y where y is any set. We want to make y into a topological space such that q becomes continuous and this we want to do in an optimum way. How many open sets can we put as much as possible? That is the whole idea. So I define tau prime which is going to be a topology on y as follows. Take a subset of y. If q inverse of that subset is inside tau, then put it here. That means a subset of b belongs to tau prime if filled only if q inverse of b belongs to tau. This tau prime will be automatically a topology on y. Very easy to check. Namely, q inverse of intersection of two sets b1 and b2 is nothing but q inverse of b1 intersection q inverse of b2. So since tau is a topology, that intersection belongs to tau prime. Similarly, q inverse of a collection bi, union of all bi say say arbitrary union is nothing but q inverse of bi and take the union. That will be inside tau because each bi is there. q inverse of bi is there. Therefore, union of bi should be inside. So tau prime is a topology. Automatically q is continuous. Why? Because take a set here in tau prime, its inverse image by very definition is in tau. So this topology does satisfy that q is continuous. This topology is called the quotient topology and the space y with this topology is a quotient space. The map q will be called quotient map. Some people call it identification space because surjective map, it may not be injective, identifies a certain number of points which are taken to the same point in y. It is also called decomposition space using the second description that we have there. Every surjective function gives you a decomposition of the set or partition of the set. So those names are also used by some authors. Often quotient map from x to y and the quotient topology are described as one of the four with two statements. So you should know various different definitions here. They are all, they could all be taken as definition of quotient space. The first one is u contained inside y is open if and only if q inverse u is open in x. The second condition is by taking a De Morgan law f contained inside y is closed if and only if q inverse infinity is closed in x. If and only if is important only if follows by q continuity, continuity of q. If this is there, we will tell you that everything that whatever satisfies as important. Therefore, I get a third one which describes the quotient topology namely this theorem which says that quotient topology on y is the largest topology with respect to which q is continuous. If suppose some lambda is a topology on y such that q is continuous with respect to that topology. X topology hasn't changed, topology on x hasn't changed. Then I want to say that every element of lambda, every open subset in this topology is in the quotient topology that we have defined namely the top prime. How? Take a certain say lambda, its inverse image is open in x because q is continuous but that is enough to put this one inside top prime. Therefore, lambda is contained inside top prime. Therefore, top prime is the largest topology. One more criteria. So, this will be the fourth one which is called in some parlour as the universal property of the quotient map. So, let us go through this carefully. You start with the quotient map. Now take any function, any continuous function from x to z which as a function factors down from y to z. Factoring down from y to z just means that there is an f total here such that you start with q and then composite trace, f total composite q is equal to f. It is the same thing as saying that whenever f takes two points to the same point here, sorry, whenever q takes two points of x to same point here, f must take the same points, same point to the same point here, these two points, same point here. f of x1 equal to f of x2, whenever q of x1 equal to q of x2, then you get a map like this, function like this. This is a satiatic fact. But what this theorem says is that this f total is automatically continuous. That is if you need continuous map, no matter what f is, what that is. Once they satisfy the satiatic property correctly to get a map, get a function like this, continuity of f will imply continuity of f. You see continuity of f, it automatically implies continuity of f because f is a fact composite q, composite of two continuous functions is continuous. Here what this theorem says is that f is continuous and f hat is continuous. So, this is the so-called universal property of q because it is true for all f and all such. And there is no other function, which is continuous like this, we satisfy this one. Such maps are called quotient maps. You can take this as definition. Let us go through this one, why this is so. Securitically, at condition, we will tell you there is an f period. By why f total is continuous is what you have to show. The continuity of f total follows, start with an open subset in set. Its inverse image must be open here. But what is the condition for a set to be open here? Its inverse image under q must be open in x. So, I have to take the open subset u here, f truly inverse then q inverse. So, that must be open in x. But f truly inverse q inverse nothing but f inverse because f is nothing but q composite, f true in composite q. So, if f is continuous, f inverse of q is open here. So, we are done. The above theorem can also be taken as a definition of the quotient topology. If we have given a topology tau twiddle on y, I have some other topology y, and it satisfies this property, then tau twiddle will have to be tau prime. There is no other choice. That is the meaning of this. So, I once again repeat. So, such a property is called universal property. Here is an example, quotient maps occur at plenty in mathematics. For instance, take a subjective open map. It will be automatically a quotient map. Subjective, open and continuous. Automatically it is a quotient map. Why? Because take an open subset in y, take any subset in y. It is inverse image. I want to show its suppose I want to show inverse image is open. Then I want to show that u is, but that follows because f of the inverse image of f under the inverse image of u is u itself. f of f inverse u is u. f inverse u is open and f is open. Therefore, u is open. So, this is the only point. f is continuous. So, if u is open, f inverse u is open is easy because f is continuous. So, openness gives you that it is the quotient map also. Exactly same way by taking De Morgan law, we have a closed map. Closed, continuous and surjection that is also quotient. You may be under the impression that these are the only cases. No. Of course, quotient maps are much more general than open maps or closed maps. But open maps are plenty again. Namely all coordinate projections from any product space into any factor x cross y to x, x cross y to o, y, r2, r3, r4 to any of these coordinate spaces, they are all open maps, projection maps. But be careful projection maps are not closed. You may be knowing that the standard projection x y to x or x y to y inside r2 to r is not a closed map. Just look at the image of the hyperbola x y equal to 1. By the very defining x y equal to 1 is a closed set inside r2. Its image inside the x axis, what is it? It is all the real numbers minus to 0. That is not a closed set. So, I will give you some examples of maps which are quotient maps, but they are neither open nor closed. So, I have just hooked up this example. It is not very standard. It is not occurring naturally. There are natural things also. Here just for simplicity, in one single example, I will show you a quotient map is neither open nor closed. So, what I do? I take the whole of real line or some large interval, okay. So, I will just take the whole of real line. Then look at the open interval 0, 1. All the points in 0, 1, you identify to a single point. 0 less than t less than 1, all of them will be identified to a single point. Similarly, look at the closed interval 2, 3. All the points in between, namely 2 and 3 included identify it to another single point. Different points, another single point. So, these two are different points, not the same point. So, these are all relation. Rest of the points are not identified with any other thing. Okay. Look at the space y obtained and let q from r to y be the resulting quotient map. So, y is a quotient space. Then I want to say that its map q is neither open nor closed. Okay. Why it is not open? The open interval 2, 3, the image of that q of 2, 3 is equal to but single point, namely q of closed interval 2, 3. The closed interval 2, 3 and open interval 2, 3 are the same image point. That point is not an open set. Why? Because its inverse image is closed interval 2, 3, which is not open inside r. Okay. So, therefore, q of open interval 2, 3 is not open. That means q is not an open mapping. Okay. It is not a closed mapping. You come to the first part. Take any point in 0, 1, any single point. It is a closed set in r. But the image which is the same thing as image of open interval 0, 1, one single point. Open interval 0, 1, one single point, q of that is same thing as q of 1 by 2. Right? That point is not a closed point because its inverse image is open interval 0, 1, which is not a closed subset of r. So, this map is neither closed nor open. So, here is a picturesque representation of what I have done. This is the open interval 0 to 1 and this is 2 to 3, the closed interval. This entire closed interval is going to a single point. Here the open interval is coming to this red point. But the end points of this interval, they have come just nearer to that one. Why is what they mean here? If you take the closure of this red point in the quotient topology, that will contain both the end points. Okay. So, in particular the red point, singleton red point is not a closed set. Okay. So, that way that you can just go up. If you open the subset, its closure is this entire 0, 1, 0 and 1. Okay. In particular, the quotient space is not a t 1 space because this image of 0 here, image of 1 here cannot be separated by open sets. Image of this open interval, this red point, if you take any neighborhood of the red point, it will contain both of them. So, it is not t 1 space. The singleton 0 is not open is enough to say that it is not t 1 space. Okay. Finally, here is an example. While discussing loops, we have used the fact that the space obtained by identifying the end points of the closed interval 0, 1 is homeomorphic to s 1. So, let us write down a neat proof of this one. By taking the exponential function on the interval 0, 1, q theta equal to a raise to 2 pi i theta. Then what happens? q of 0 and q of 1 go to the same point and no other point is identified with any different point. In other words, q is injective on the open interval 0, 1. Therefore, to look at the inverse image of a point in q, only the inverse image of 1 contains two distinct points. All other things are singletons. So, this is the same thing as saying that the identification set as a set is it is s 1. 0, 1, 0 and 1 are in the same equivalence class and rest of them are in different equivalence class. That is the meaning of that what we have done by identifying 0 and 1 in the interval closed interval 0, 1. So, set theoretically we are fine. So, what we have to verify is the subspace topology of s 1 coming from r 2 is the quotient topology on s 1 coming from i via this map q. This is what we have to verify. In other words, q is a continuous function is fine because we know that exponential function is continuous. It is surjectively also fine. What we want to say is q itself is a quotient map. If we know q is a quotient map then our justification, our claim that identifying 0 and 1 inside the closed interval 0, 1 to a single point whatever space you get is only 1 to s 1. So, what we have observed is now we have to verify that this exponential function is a quotient map. So, we know that this can be done if we verify that this is a closed map because we have observed that closed maps are quotient maps. So, that can be done very easily. Take any subset k of i which is closed. k itself is a bounded set because i is bounded. Bounded then goes subset of of a Euclidean space in r. It is a compact space. k being compact, the image of k under continuous function q, 2k must be compact. A compact subset of a metric space s 1, standard topology. So, it must be closed subset. So, fk is closed, qk is closed. So, closed set, closed set means the function q is closed and that is the end of this argument here. So, we will study more of quotient spaces next time. Thank you.