 Before we take up the central problem in this chapter of classification of G-coverings, I would like to present a method of obtaining new G-coverings out of the old one. This time, change in the base itself, since this method involves a new concept which is important on its own, not only in algebraic topology but elsewhere also. So, let us study this one a little more carefully and in the real more generality, not fully generality. We start with base space B and take two functions FI from EI to B, their fiber product is defined to be a triple topological space and two functions PI, PI is from Z to EI, I equal to 1 and 2, with the condition that F1 pi 1 is equal to F2 pi 2. So, composite map from Z to E1 to B or Z to E2 to B, they are the same and this entire thing satisfying the so-called universal property. So, this is the picture, Z is here, E1, E2 are here, B is here. So, F1 and F2 are functions of B. Then Z is a function from pi 1, has a function pi 1 and pi 2 respectively to E1 and E2. If you have another commutative diagram like this, Z prime, pi 1 prime, pi 2 prime, respectively maps from E1 to E2 such that when you compose F2 here and F1 there, they are the same, F1 pi 1 prime, F2 pi 2 prime, they are the same. Then, this is the time we will have a map into Z fitting this dotted arrow. This is the conclusion of the definition. There exists a map G, unique map G such that the whole diagram is commutative, namely pi 1g equal to pi 1 prime, pi 2 equal to pi 2 prime. For every Z prime and pairs of maps which satisfy this property, there must be a unique map length. So, this is the universal property of this commutative square here. Then, this Z together with these two maps is called the fiber product of F1 and F2. So, this is the definition of the fiber product of two functions taking values in the same space. The uniqueness of such a pair Z E1, Z pi 1 and Z pi 2 like this up to homeomorphism follows easily by this universal property. Namely, suppose Z prime is another such pair which also has a universal property. Then, it will admit a map from Z to Z prime in the reverse direction here such that again the same diagram all these diagrams are commutative. Then, what you have is G prime, let us say G prime going Z to Z prime coming back pi G that will be another map from Z to Z itself fitting this diagram. Identity map fits this diagram from Z to Z. Therefore, there will be two of them. By the uniqueness of uniqueness property of this map that composite G followed by G prime must be identity of Z. That means what by the same argument that this G prime is a inverse of G, that G is a homeomorphism. So, this is typical of all universal properties, definition given by universal properties. Existence of such objects is not guaranteed but uniqueness is guaranteed. So, let us look at the existence. The existence here is very easy compared to many such existence theorems. So, for the existence what I do is I define Z to be subspace of even cross E2 consisting of pairs even E2 such that under F1, F1 of even is F2 of E2. Take the product and take this subspace. On this subspace there is already projection maps given by this product. Product E1 E2 has projection maps pi 1 pi 2 respectively to E1 E2. Just take the restriction of that on Z. So, these are now I am denoting them by pi i but now I am thinking them as maps from Z to EI. Obviously, if you project E1 E2 you get E1 then F1 of E1 is same thing as F1 pi 1 of E1 E2. So, F2 of E2 is same thing as F2 pi 2 of E2. So, the commutativity of this diagram is obvious. You have got commutative diagram like this. All right. That we have constructed. But now I have to show that this Z as universal property. That is also easy for if Z prime F1 prime F2 prime is another triple with the same property then we just define G of Z prime. I am taking G of Z prime here to be pi 1 prime of G, pi 1 prime of Z prime comma pi 2 prime of Z prime as an element of E1 cross E2. But when you go here they agree that is why by the very definition of Z that pair pi 1 prime Z prime comma pi 2 prime Z prime is inside Z. So, that gives you a map into Z. So, that is what I am doing here F1 prime Z1 and F2 prime Z2. This is not this definition in this notation I have put pi 1 prime here. So, I should take pi 1 prime pi 2 prime not F1 prime F2 prime. So, that G fits the diagram. This is the only way a map can fit there because any map into a product first of all is determined by the two projection maps pi 2 and pi 1. So, two coordinates you have to define. So, therefore if there is a map it has to be this map. The first projection first coordinate has to be this one this map second coordinate has to be this map and that defines an element of Z because Z is taken as those points wherein when you go to F1 they agree and F1 and F2 they agree. So, that is the construction and the construction and uniqueness are over alright. So, this is a very simple minded thing to begin with but it has wonderful properties. The fiber product by very definition is symmetric in F1, F2. You can say you can write F1 cross F2 or F2 cross F1. So, I just omit having a notation for this one. It has some wonderful properties namely if F1 is surjective then I will use this diagram. If F1 is surjective then pi 2 will be surjective. If F1 is injective then pi 2 is injective. If F1 is a homeomorphism then pi 2 is a homeomorphism. If F1 is a covering projection pi 2 is a covering projection. If f 1 is a G covering, pi 2 is a G covering. So many technological properties of f 1 will be reflected here. Of course there are some which we do not. So for example, if f 1 is a cofibration then this will be a vibration. If f 1 is a cofibration there is no guarantee this will be a cofibration. The cofibration will happen if we reverse all these arrows and then take the cofibred product. Then we say something like co okay dual to all this when arrows are all reversed okay. So cofibration will not fit here. Fibration is vibration this will be a also vibration. So let us verify a few of them which I am going to use them also. Meanwhile we will get familiar with this definition and the construction also okay. Let us verify surjectivity. F 1 is surjective I am assuming. I want to say that pi 2 is what do I do? Take a point in E 2 okay. Pick up a point in E 1 which sits over f 2 of E 2. f 2 of E 2 is an element of B. f 1 is surjective therefore f 1 of E 1 will be equal to that point such a E 1 will be there okay. Now by the very construction E 1 comma E 2 is an element of Z and the second projection of this pi 2 is E 2. So starting with any E 2 I have produced an element in Z such that pi 2 of that element is E 2. So pi 2 is surjective alright. Let us verify the injectivity which is slightly little length there one line length there. Suppose f is injective f 1 I want to show that pi 2 is injective. So pick up two points in Z how do they look like E 1, E 2, E 1 prime, E 2 prime okay. Suppose pi 2 of them are equal what is the meaning of that? The second coordinate E 2 must be equal to E 2 prime okay. Then I have to show that E 1 equal to E 1 prime. If the second coordinates are equal I have to show that E 1 is equal to E 2 prime then that will complete that pi 2 is injective right. But what is the meaning of these points are inside Z they are not arbitrary elements of E 1 cross E 2. This means that f 1 of E 1 is f 2 of E 2. f 1 of E 1 prime is f 2 of E 2 prime. Therefore if you start with f 1 of E 1 which is f 2 of E 2 but E 2 is E 2 prime so it is f 2 of E 2 prime but f 2 of E 2 prime is f 1 of E 1 prime. But f 1 is injective is our assumption therefore E 1 must be equal to E 1 prime. So we are done okay. In particular if f 1 is a bijection then pi 2 will be a bijection okay. So let us go a little further. Suppose f 1 is a local homeomorphism then I have to show that pi 2 is a local homeomorphism okay. So start with a point in Z I must produce an open subset of Z for this pi 2 restricted to that must be a homeomorphism onto its image in E 1 in E 2 okay. I have to find such a neighborhood. First of all I should use that f 1 is a local homeomorphism. So I look at E 1 even as a neighborhood E 1 in E 1 such that f 1 restricted to E 1 to f 1 of E 1 let us call it as V E. This is a homeomorphism so that is a definition of local homeomorphism and V is an open subset of V then put U 2 equal to f inverse of this V f 2 inverse of this V that will be open in E 2. Now I have got two open subsets one U 1 another U 2 right U 1 open in E 1 U 2 open in E 2 therefore U 1 cross U 2 will be open inside even cross E 2 the product space intersection with Z that will be an open subset of Z and it contains the point even E 2 why because if you look at f 1 of E 1 that will be a point of U 1 this V f 1 of sorry that is the point of f 1 of U 1 that is V therefore every point here is f 2 of that is inside V. So I have taken a point wherever I have taken a point I have taken a point in U 1 cross U 2 intersect with Z that means if I take a pair here what is the property of that f 1 of a point here is equal to f 2 a point there. So even E 2 belongs to that so this is an open subset so even E 2 is a has a neighborhood that much we have formed inside that claim is this W has this required property namely pi 2 from W to U 2 below U 2 is this one is a local homomorphism okay. Now first thing is the bijectivity of this from W to U 2 follows from the bijectivity of this f 1 exactly as in the previous two steps okay f 1 was assumed to be bijective on the whole space U there whole space E 1. So restrict to the whole thing instead of E 1 and E 2 to U 1 and this U 2 instead U 1 cross U 2 you have this you have the same the construction that is those points which are f 1 of U 1 equal to f 2 of U 2 that is why the step 1 and step 2 are valid here also okay. So bijectivity of pi 2 follows by bijectivity of f 1 alright now we have to show that it is a homeomorphism inverse is continuous is what you have to show. So look at S which is the inverse for this f 1 this is a homeomorphism from W to U 1 there is an inverse of f 1 which is continuous that is the meaning that is the homeomorphism using that define a inverse U 2 to W by the formula H of E 2 prime equal to S composite f 2 comma E 2 prime second coordinate E 2 prime automatically when you apply pi 2 to this one you will get E 2. So this H will be inverse of pi 2 why it is continuous because this is second coordinate is just identity here the first coordinate I S composite f 2 which is continuous therefore this is a continuous inverse okay a homeomorphism the inverse of f will produce an inverse here alright so it is a local homeomorphism actually this also proves that if f itself is a homeomorphism then pi 2 would be a homeomorphism okay so that proof is also here more generally local homeomorphism implies local homeomorphism is what we have seen now the next thing is if f 1 is a covering projection then so is pi 2 in fact we claim that you take an evenly covered open subset of B evenly covered by f 1 take its inverse image by f 2 inside E 2 okay claim is that open set is evenly covered by pi 2 so V is contained inside the open set which is evenly covered by f 1 the claim is W which is f 2 inverse of V is evenly covered by pi 2 suppose I prove this one by the very definition of covering projection you have an open covering of B consisting of evenly covered neighbourhoods evenly covered open subsets inverse image of those things will be an open cover for E 2 each of them I am going to prove that they are evenly covered here so so pi 2 will be a covering projection so it is enough to prove this statement okay namely W is evenly covered that means I have to look for pi 2 inverse of W write it as a disjoint union of open sets such that restricted to any one of them pi 2 is a homomorphism what is my source is similar statement for f 1 namely write U write V this must be V this is all typos write V f 1 inverse of V as disjoint union of U alphas U alphas are open in E 1 restricted to each U alpha f 1 is a homomorphism let us put S alpha to be the inverse of that that will depend upon alpha right so V 2 U alpha S alpha is the inverse of f 1 all right then pi 2 inverse of W remember this is same thing as what is W W is f 1 inverse of V here it is correct instead of from V to U a hogai either it this must be V pi 1 inverse of f 1 inverse of V okay because this is W is f 2 inverse of V f 2 inverse pi 2 inverse f 2 inverse is in the pi 1's f 1 inverse it is pi 1's f 1 inverse but f 1 inverse is disjoint union of U alpha therefore this is a disjoint union of pi 1 inverse of U alpha pi 1 being continuous these are open subsets of even cross A2 but I am restricting it to Z intersection with Z I am taking so they are they are open subsets disjointness is still there because they are disjoint inverse images of disjoint open subsets here okay pi 1 inverse of alpha U alphas disjoint so I have got a disjoint family okay I want to show that each one of them projects homeomorphically onto this W under pi 2 this part already follows why what we have seen in the local homeomorphism picture what is the inverse corresponding inverse use this S alpha to get the corresponding just the way I have done it here so H alpha here will be defined as S alpha composite f 2 comma identity okay so that is what you have to do here okay so just like what we have seen in the previous paragraph it follows that pi 2 from pi 1 inverse of U alpha W is a homeomorphism which is inverse given by H alpha of V2 is equal to S alpha composite f 2 of E2 comma E2 now we come to one of the most interesting part part now namely if f 1 is a G covering then pi 2 is a G covering okay if it is a covering this is a covering we saw already done so only thing is now we have to see that they are given by actually an even action of the group G okay suppose G acts on even evenly and the corresponding projection map from even to be is the resulting coefficient map which is a covering obviously so f 1 I am thinking of this as a quotient map now by the action of G we claim that there is an even action of G on Z such that this pi 2 becomes the corresponding quotient map you cannot change pi 2 by the way okay in Z the pi 2 pi 1 etcetera are already defined okay once even and E2 f 1 and f 2 are given that is defined but I want to claim that this results becomes a quotient map corresponding to the action of G and therefore it is a covering projection not only that okay the pi 1 from Z to E1 there are G actions on both sides this pi 1 becomes a G map indeed this extra thing will tell you what you have to do okay so this is more or less like last part is more or less like a hint here for the construction of Z so we define the action of G on Z obvious action there is an action of G on E1 and we do not know anything about E2 E2 is just an arbitrary term like a space therefore take the action of this G on even cross E2 why are the first factor G of even E2 is G even E2 now suppose even E2 is already inside Z then when you act by this namely G even E2 this will be also inside Z why why it is inside Z I have to verify that f 1 of E1 is equal to f 2 of E2 this is the property for even E2 inside Z but what is f 1 of G E1 because f 1 is a quotient map under G action f 1 of G E1 is f 1 of E1 therefore f 1 of G E1 is also equal to f 2 of E2 okay so therefore the right hand side is inside Z if even E2 is inside Z so this action of G on even cross E2 actually gives you action of Z action on Z okay that is the meaning of that this is well defined now clearly each orbit under G action is mapped to the same point E2 second part is not changed at all in pi 2 therefore the fibers of pi 2 are orbits and they are precisely the orbits okay because even prime comma E2 is in Z is any element okay even prime must be inside f 1 inverse of f 2 of E2 by definition that it is inside that okay f 1 inverse of something whatever it is one single fiber one single G orbit and hence there must be G belong to G such that G of even must be equal to this element even prime because it is in the same orbit that just means that G of even E2 is even prime E2 okay so this shows that the fibers of pi 2 under Z inside Z they are precisely equal to the orbits of G C action G which just means that this map is a quotient map okay well it does mean that the whole thing is a quotient map but it is a covering projection already that we have already seen right a covering projection is an open quotient therefore it is open subjective therefore it is a quotient map the fibers are correct but why it is a quotient map requires some explanation but we have already seen that it is a covering projection okay there are slightly different ways of seeing that quotient that is map so what we have seen is that pi 2 from Z to E2 is actually corresponds to covering projection by G action look at the way we have defined defined this action here G of even E2 is G even E2 what happens to when you come to pi 1 pi 1 of this is pi 1 of G E1 okay it is G of just G E1 this this pi 1 is just G E1 with G of E1 on this side it is E1 so pi 1 of even E2 operated by G is the same thing as G E1 so this very definition shows that the projection map to pi 1 is a G action yeah G map okay so this was last part here projection is G map so I have taken that as a key as a Q to define this action and it works all right so these are the basic things that we need to explain our classification of G coverings so this is going to produce large number of G coverings if you have one G covering and any map into the base space you can you can construct the G covering on the other other space so this procedure is precisely called has a different name which is non-symmetric in the in the definition in the idea because the role of f1 you know properties of f1 are reflected in pi 2 f2 may not have any of any of this properties may be just a continuous function that is why the role is important though in the definition of the fiber product f1 and f2 does not matter which one does not matter that is the beauty of this so we introduce a non-symmetric wording here namely this is the point f1 f2 so the fiber product is symmetric in that the role of f1 f2 is different okay so we shall introduce a terminology which is not symmetric in that sense namely we shall call this map pi 2 from z to e2 as the pullback of the map f1 even to be via f2 and denote it by f2 star of f1 this is a categorical notation which will it does not depend on the construction here we have heavily used the construction but all these things can be done in a categorical language without appealing to the construction okay that will be taking you little more deeper we do not have any time for that okay here we use the easy set theory which gives you all these answers easily so this map will be called the pullback the corresponding covering projection will be called pullback covering projection so if you start with a g covering e2b I am going to use a different notation now e2b is a g covering there is a map okay there is a map f from b prime to b then I get a g covering on b prime which I am going to write as f star of p p is a g covering f star of p with g covering what we have seen is if p is surjective injective local memory mark them covering projection recovering etc the same is true for f star of okay thank you