 Welcome to module 15. Last time we studied co-hereditaryness of certain properties, psychological properties, especially our concern was about host-doubtsness, which is a part and parcel of assumptions in algebraic topology. And we saw that host-doubtsness is not co-hereditary and it is not all that easy a hypothesis also we need to have so that the quotient is host-doubtsness. Therefore, quite often case by case we will have to check whether something is host-doubts or not. Luckily what happens is the kind of quotient constructions we do in algebraic topology such as cones, mapping cylinders and so on as we will see later they all have the host-doubtsness going into the quotient spaces. The one single fact which can be attributed to this phenomenon is that equivalence classes are separate by open sets which are themselves union or equivalent classes. Most of one we will be dealing with maps which are co-fibrations which will ensure the situation. So when it comes to we will discuss this in a little more detail. So let us start. I saw things about quotient spaces like we observed that if you restrict the quotient to some smaller subspace in the domain in the mother space such that the map is still surjection and if that turns out to be quotient map again then that will be of quite help for us. But we may ask when such a thing is possible. Once again there is no easy criteria for that. We can say if this one, if that one, if that one and so on. So you have to verify it. For example of practical importance suppose you have an open quotient map or a closed quotient map then restricting it to a closed open subspace or a closed subspace will continue to be an open map as well as a closed map whichever the case may be. So all that you have to ensure is that the subspace is large enough namely suppose you take an open subset of X, Q restricted to that open subset must be also surjective. So that is all you would like to know. So why it happens is very easy because if Q is an open map and say A is an open subset then Q restricted to A will be also an open map. Similarly if A is closed set and Q is closed map restriction map will also be closed map that is all. And open surjection or closed surjections are automatically quotient maps. We have already noted this one. I will now recall that namely if you take the projective space which is defined as a quotient of Rn minus Rn plus 1 minus 0 you can restrict it to just the unit sphere there. The original map P from Rn minus 1 to Pn is actually both open as well as closed map and the sphere is a closed subset. So this entire thing will be a closed subset that is our idea. So it will be restricted to Sn will be also a quotient map. Why this is important? The first thing is that you immediately know that being quotient of Sn Pn is compact. Further restrictions you can do that namely only on the upper hemisphere that will give you a better structure of the projective space itself. If you take the upper hemisphere the lower hemispheres are all represented by elements on upper hemisphere. Therefore the identifications are only on the equator but the equator is one dimensional over sphere and the identification is again antipodal. Therefore what you get on the equator is Pn minus 1. The rest of the open cell which is the upper hemisphere strictly upper hemisphere actually remains as it is attached to the Pn minus 1. So that is an open cell because there the map is injected. So this description will be very very helpful in understanding the projective space inductively. In the case of n equal to 1 this already tells you that P1 is nothing but S1 again. Now let us consider slightly different kind of quotients could be quite weird. That is what I wanted to tell you. Namely there are spaces which are self quotients. Here is a simple example but there are many such examples. You cannot go on discussing all the examples. For n equal to 2 look at the map from S1 to S1 namely eta n of z is at power n. If n is 1 this is identity map. It is not much interesting. But if it takes z square, z cube, etc. Then you know that the kernel of this map is precisely the nth roots of unity. So it follows that the fibers are nothing but nth roots of unity multiplied by some you know translated by some z. These will be all the fibers. When you take z power k where zeta power k zeta is nth root of unity times z raised to n is same thing as raised to n. So all of them go to the same point raised to n. But now what I have proved is that what Vandham Theram Fallibhira says is all these maps are objective. So from a compact space to a hostile space. So this will be automatically a quotient map. It is automatically a quotient map. It will be equally a quotient map. Only thing is not injective. So in various ways, several infinite many ways S1 is a quotient of itself. This is a picture that we have. S1 person should take the quotient by action of say nth roots of unity here. So this action of z by z z by n z. So again what you get is again isomorphic to S1. Now let us come to another important question. Namely if you take two quotient maps will the product be a quotient map? It looks such a nice things to have but this is something which is not true in general. You should not be surprised because product is quite misbehaved with many many particular properties. The Cartesian product that we take in standard quantum topology as well algebraic topology has this problem. So you have to very careful in extending results to product spaces. The product of two maps say x1 to z1, x2 to z2 they are quotient maps q1 cross q2 from x1 cross x2 to z1 cross z2. Is this a quotient map? The general answer is no. However one would not leave it like that. So you want to understand why this is so and what best we can do. We can decompose q1 cross q2 as q1 cross identity composed identity cross q2. This identity is first one this one is identity of x2 this is identity of z1. So identity maps are of different spaces. The problem reduces to we know the composite of q composite of quotient maps is quotient. Therefore if we answer either this or this which is symmetric so second one now q1 is a quotient map second one is identity map. If I can show that this is a quotient map then for symmetric view this will be also quotient map. So the composite will be quotient map. So we ask when is q cross identity is a quotient map. So here is a satisfactory answer. Suppose y is locally compact hostile okay and x to z is any quotient map. So what you are doing is taking product with a locally compact hostile space then q cross identity of y is a quotient map. So the new creature which comes here you could be any quotient map this one should be locally compact. Then no matter what q is q is some quotient map q cross identity is also quotient map okay this is the self is not so difficult thing. The point is why put such locally compact hostile space condition and so on may be easier and so on. The the stranger thing is nothing else will work as soon as y is not locally compact hostile space there will be some quotient here such that the product is not quotient product identity is not quotient. So that is the beauty of this hypothesis. So in some sense it is a full answer also all right let us go through this one. Now you see the function space theory that we have studied comes to help because we have locally compact hostile space okay q cross identity because q is a surjective map is surjective and continuous there is no problem about that. What we need to prove is the following universal property we have to prove for any space w and any functions at cross y to w if g composite q cross identity say f equal to this one is continuous then g must be continuous. So this is the property of z cross i being quotient map any quotient map has this property anything from quotient map quotient space to any other space is continuous if and only if composing with the quotient map it should be continuous here we are talking about q cross identity q cross identity will be quotient map if it has this property okay so this is the universal property of the quotient maps it will define the quotient map if this is true for all g as soon as f the composite is continuous g must be continuous okay now go back to the exponential correspondence that we have that we have established earlier okay a function like this is continuous is same thing as x to w raise to y you get a map that is continuous right namely the map f hat from x to w raise to y given by f hat x of y is equal to f x y remember f x y is a function from x cross i x cross y to w okay so if this is continuous then f hat will be continuous and conversely so we have passed down to this one so here we have used the fact that y is locally compact all okay now this map f hat factors down through q to give a continuous function g hat from z to w no w raise to y because q is q is a quotient map and the beginning the g itself is is such that you know f itself is g g composite q for that reason it factors down to a map from g hat to z z ratio z to w raise to y okay such that f hat is g hat composite q so that is the meaning of fact of down okay but then g which is e composite g hat composite identity okay therefore if g hat is continuous e composite is continuous so g is continuous okay and g is continuous so we wanted to prove that g is continuous all right so you pass to the you know the exponential by exponential correspondence you factor function space use the function space argument so this becomes such an easy thing you can directly write down the proof but then you will have to repeat the proofs of the exponential correspondence etc verbatim more or less exactly same way you have to work out instead of doing that you can use that one that is ready made ready made results for you so once we have this we have observed the corollary is that suppose you have two quotient maps q i from x i to z i okay such that x 1 and z 2 are locally compact or z 1 and x 2 are locally compact as the case may be depending upon that q 1 cross q 2 can be written as q 1 cross identity composite identity cross q 2 or the other way around therefore it will be a quotient map each time you have to have both the cases you have to have the identity factor that you're taking must be locally compact quotient map could be anything crossing with a space then you are taking identity that space must be locally compact okay now I repeat this one namely though I started saying that it's a partial answer this is a full answer there is no other way namely this result is due to Michael which is very recent actually in some sense 1968 in topology by the way in 1968 it's quite recent I would think that everything by sixties were completely proved topology was people were thinking that now which is correct in some sense because all major problems were solved and so on rest of them were very hard this is what has happened in around sixties so it gives you a complete answer to this problem namely so I will state it here this statement whatever I give may not be exactly as it is here because that is a paper it will write many other things so I have extracted something okay so what is it let x be a regular or hostile space then the following two conditions are equivalent namely x is locally compact and x is overall compared plus half dot we have every quotient map y to z the product with x now y and z have changed here huh y and x have changed identity cross q is a quotient map so wherever whatever identity affects you take this x must be locally compact then this is if this is true for all q then x must be locally compact so that is the theorem of Michael okay so I declare that I am not going to the details here we will not prove it here nor we have any use for this theorem as such that product in one way is is we need but the other way we will not need you are welcome to read the Munkres book for an example of a non-locally compact space and a quotient map view from y to z says that the product is not a quotient map but if you know this result then reading Munkres example is redundant okay the proof of this is interesting in some sense namely for each low non-locally compact space canonically it cooks up a quotient map the quotient map is such that product with identity of x will not be a quotient map okay it's not one example for each locally non-locally non-locally compact space there is a there's a nice canonical example all right so I would like to now discuss a few exercises here I am not going to give you the solutions as such okay but let me go through some of these one by one they are illustrations of quotient spaces okay and they will be helpful in understanding various constructions in in algebraic psychology to start with the unit disk in in R2 okay this is the closer unit disk in R2 all x1 x2 such that x1 square plus x2 square is less than equal to 1 take only the subspace in which the first coordinate is greater than equal to 0 so it's the half that is okay now make the identification namely 0 comma x2 0 comma x2 will be what x coordinate is 0 only y coordinate look right there x2 is identified minus x2 only on the on the this line x2 x2 or y yx whatever you want to take x1 is x2 is identified minus x2 if this is the case you have to show that the quotient space is again homeomorphic law that is okay the proofs are not hard so this is beginning of the kind of things I want to you see namely next one is take the right hemisphere this time instead of this I am taking the sphere then again I am taking only half of it namely all the from the first coordinate is greater than equal to 0 okay so it is a cup like this okay now it is in three dimension but the sphere itself is two dimensions this is like a cup here on this I want to make two different operations here the first one is the quotient space obtained by identifying similar to the first exercise this time I will not touch the x2 axis x2 coordinate 0 comma x2 as it is but x3 corresponding x3 will be added 5 to minus x3 okay so this this is the boundary which is actually circle on this circle I am identifying x3 with minus x3 so this is not anti polar action you have to be careful here show that y is quotient is homeomorphic to a closed thing now full x2 okay this was a the half sphere but the quotient is a full sphere it is homeomorphism you do not see in the quotient space they are not embedded subspace in the second example the same operation same space the quotient space is obtained by taking anti polar action okay so anti polar action this is p2 this is the projected space p2 which is what we have already discussed while discussing the projection space so I have already given you the solution here why it is a objective space okay so I think this screen is cutoff and it should be but that is all here namely the quotient space you have to show is p2 which is more your host there here which is cutoff I cannot help it now the third example is slightly more complicated but if you have worked out first two they will let you know how things have to be worked out okay so they are hints for exercise three what is this on s1 cross s1 you take x going to x inverse diagonal action so x y going to x inverse y inverse going to means what this is the action of g2 to go more below this action okay that means you have to identify x y for each x y should be identified with corresponding x inverse y inverse the quotient is homeomorphism s2 that is what we have to show so so the the two sphere can be thought of as a quotient of s1 cross s1 okay this is a stalled exercise this is only for people who are advanced quite sufficiently advanced with the point set topology they will go through munkres paper or maybe michael's paper and so on so then they will be able to to prove this you know solve this problem all right so let us stop here thank you