 So, in the previous lecture, we ended our lecture with an example from group theory. So, we saw that given a group and given a normal subgroup, we can look at the equivalence classes. So, the normal subgroup defines an equivalence relation on our group and we can look at the equivalence classes. We can give the set of equivalence classes a group structure such that some nice things happen and today in this lecture, we will see that we can do something similar in topology. So, let us address the question first what the question is. So, let x be a set, suppose we have a equivalence relation. So, let x mod equivalence be the set of equivalence classes. So, then we have this obvious map from x to x mod equivalence. What does it do? It takes x and sends it to its equivalence class. So, right now everything is only happening for sets. Now, suppose we have a map of sets f from x to y such that x equal to y implies f of x equal to f of y or this property we can phrase it as follows simply stated f is constant on equivalence classes. So, then this function this map of sets it factors uniquely as c v of f or equivalence is pi. So, it factors like f naught. We got a map of sets from the set of equivalence classes to y right and what is f naught? So, f naught is defined as follows. So, let alpha be an equivalence class right and let x in x be such that the equivalence class of x is this alpha right. So, we define f naught of alpha to be equal to f of x. Now, this is well defined because f is constant on equivalence classes right. So, clearly this definition makes sense as f is constant on equivalence classes. So, now we are going to bring in topologies and we are going to complicate this problem. So, now further assume that or rather now consider the situation where x is a topological space we have this equivalence relation right. So, is this an equivalence relation on the underlying set ok. So, what we want to do is we want to put. So, the question is can we put a topology x mod equivalence. So, this is just like can we put a group structure on g mod n such that the following two things happen one the map the natural map pi from x to x mod equivalence is continuous ok. Suppose we are given a continuous map continuous map f from x to y. So, y is another topological space and we are given a continuous map such that f is constant on equivalence classes. So, then as a map of sets we are going to get a factorization like this f is this we have pi we have x mod equivalence and we get this map of sets f naught then we get a map of sets f naught from x mod equivalence to y right. So, the second condition we want is that f naught must be continuous. So, we do not. So, we get a unique map of sets in fact right even here the f naught is unique it is completely determined by f. So, let me make a remark. So, if you lean pose condition 1. So, then this problem has a very easy solution easy solution namely we can just say that this give x mod this equivalence relation the equivalence class is the trivial topology right. So, since there are only two open subsets the empty set and the entire set the inverse image of both these will be open and therefore, pi will be continuous here, but if you give this the trivial topology then the second condition the second property may not be satisfied ok. So, the main result the main theorem we are going to prove in this lecture there is a unique topology x mod equivalence which satisfies the above two conditions. So, let us prove this theorem. So, let us first show there is such a topology topology tau on x mod this equivalence which satisfies satisfying the two conditions. So, we are going to define tau first. So, let tau be the collection of subsets to contain in x mod equivalence such that this pi inverse u is open in x. So, it is easily checked and I will leave it as an exercise satisfies the conditions to define the topology ok that is a very easy check and I will leave it as an exercise yeah. So, obviously so, now let us prove that both the these two conditions are satisfied. So, first we have to show that this map pi from x to x mod equivalence is continuous here. So, x mod equivalence for this we have to show that given an open subset v in x mod equivalence pi inverse v is open in x yeah, but this is clear from the very definition of the topology on x mod equivalence right v is in tau if and only if pi inverse v is an open subset of x ok. The second condition we want to check is suppose we are given a continuous map f from x to y which is constant on equivalence classes. So, then we get this resulting map f naught of sets a map of sets determined by f. So, we need to show that we need to show that f naught is continuous for this we need to show that. So, let v contained in y be an open subset. So, we need to show that f naught inverse of v is open x mod equivalence yeah. So, for this applying the definition of tau this happens if and only if pi inverse of f naught inverse of v is open in x right, but simple set theory tells us that pi inverse of f naught inverse of v is equal to f inverse of v which is open as f is continuous. So, this shows that tau satisfies next let us prove that this topology is unique. So, suppose there is another topology tau prime on x mod equivalence such that conditions 1 and 2 hold for tau prime ok. So, let us make this diagram. So, we have x we have this map pi this map is to x mod equivalence comma tau prime yeah. So, the first condition is that the obvious map x goes to the class of x this should be continuous if x mod equivalence has the topology has the had should be continuous in our topology. So, we are giving x mod equivalence the topology tau prime and by assumption since tau prime satisfies condition 1 this map is continuous right. So, but we also have this map x mod equivalence comma tau right now as tau satisfies condition 2 we get this dotted arrow like this right. So, let us call this pi naught yeah ok. So, the dotted arrow we will get anyway. So, note that the map of sets pi naught pi naught is forced to be be the identity as this right. So, this map of sets pi naught that we get that is forced to be the identity right. So, since tau satisfies condition 2 this implies that pi naught is continuous. So, thus if v is in tau prime v is an open subset in tau prime then pi inverse pi naught inverse of v which is equal to v is open in tau. So, this implies that tau prime is contained in tau ok. So, similarly we can switch the roles of tau and tau prime. So, we have this right. So, we get this pi naught right. So, as tau prime satisfies the second condition this implies that pi naught is continuous. So, once again this implies that tau is contained in tau right. So, thus tau is equal to tau ok. So, this completes the proof of the theorem. So, this topology the above topology x mark equivalence is called the quotient topology. So, we will end this lecture here.