 Today's topic is Existence of Simply Connected Coverings. First we shall investigate how the existence of Simply Connected Covering will actually solve the entire problem of existence of all coverings. So then we will come to the Simply Connected Covering itself, okay. So that is the plan. So first let us see once you have a Simply Connected Covering over a given space it will allow us to construct all other coverings. So start with a connected locally part connected space admitting a Simply Connected Covering space. Let us denote it by p from x bar to x. Then for every subgroup k of the fundamental group there corresponds a covering projection which we will denote by z to x q from z to x such that if you take q check of pi 1 of z it will be equal to k strictly speaking there will be a point also z bar such that q z bar equal to x etc. So you have to mention the base points here okay so that freedom is there. So this will completely solve the problem now for each subgroup you will have a corresponding covering okay. And then we know how they are related and so on namely conjugates up to conjugation they will be all isomorphic up to equivalence and so on okay. So the existence for arbitrary subgroup now. So first we have this bijection phi from group of covering transformations to pi 1 of x x naught remember p is a universal covering Simply Connected Covering therefore this itself is an isomorphism okay put k prime equal to phi inverse of k under this isomorphism that will give you a subgroup of the group of covering transformations the group of covering transformations is identified with the fundamental group. So any subgroup k of the fundamental group can be thought of as a subgroup of g p under the isomorphism so k prime is p inverse of k. We can take z to be the quotient space of x bar by the relation namely z 1 is equal to z 2 it is written only if z 2 is phi of z 1 for some phi inside k prime which is nothing but under the action of k prime on x bar we take the quotient space of x bar the orbits of this action okay so that is going to be our result alright. So claim is there is a commutative diagram with q prime as a quotient map q prime is from where to where it is from x bar to z okay so what is that commutative diagram it is this one p from x bar to x respects this identification right what is z 2 z 2 it is phi of z 1 right if you apply p to this one p of z 2 p of phi is just p itself so it is p of z 1 p z 1 is what p z 2 is still true if z 1 and z 2 are identified therefore this p factors down to the quotient namely equals classes of this relation so that is denoted by that one by q this one by q prime claim is this map okay she is obtained by p itself from z 2 this is a covering projection this is connected so this is connected no problem you have to prove that q is a covering projection you have to also prove that phi 1 of z take a base point above this one does not matter is actualized to this k prime okay then when you take q result of that if q check of that q check of that k prime it will come to the fundamental group phi 1 of z whatever we have taken because q b if we is covering projection then q check will be injective on to the image whatever we have taken okay so there are work to do here there is some work to do all right suppose v is a connected open subset evenly covered by p p will cover projection the entire axis covered by evenly covered open sets being locally connected I can take smaller neighborhoods which are connected at each point so they will cover and they will be even also because any subset which is even already of a any subset of an even the cover of that is automatically even okay so start with a connected even evenly covered open set okay we claim that it is evenly covered by q also so that will all settle that q is a covering projection this follows from the fact that p inverse of v is the distant union of ui just just writing the meaning of evenly covered okay write it as this way then any covering transformation from x bar to x bar okay maps each ui homomorphically on to another uj because every covering transformation respects p so p inverse of v has to go to p inverse of v itself but when you have written the distant union each ui may not go to ui it will have to get some shuffle that's all so homomorphic to another uj okay each of ui is now connected also because I have assumed v is connected so it is like the connected components of this p inverse of v i p inverse of v are getting what getting shuffled so that is the action of action of the any covering transformation all the covering transformation okay incidentally it turns out that q prime is also covering projection by this proof you know whatever we wanted to prove this is a covering projection now when we are arguing we will be able to prove that this is also a covering projection okay we just watch out that is a sort of extra bonus for us okay but we don't need it here so so you just just be watching out if you cannot get it leaving given omega an element in pi 1 of that we want to show that q check of omega is in k this is what we want to show okay q check is a such injective mapping okay so first you want to show that it's going inside k and then you are sure it is onto k okay so we take a lift omega bar of q composite omega start with a loop in z okay but take it q of that omega that will be a path in x at the point x naught all right okay take a lift lift omega bar of q composite in x bar everything can be literally x bar x bar is a connected covering it's actually simply connected covering so at the point x naught bar put f equal to put let f equal to f belong to gp the unique element such that f of x naught bar is the end point of omega bar we have seen that there is a one one correspondence here so take one element which corresponds with that such an f will be unique then by the definition of phi remember we have phi of f is nothing but q check of this class okay this is the definition of under this right concepts okay so since q composite q prime of omega bar is nothing but p of omega bar okay q composite q prime is p right p of omega bar by definition is q of omega so q of this path is q of omega okay and what is omega 0 omega is a path in z right omega 0 z naught which is q prime of omega bar okay therefore the uniqueness of lifts in z okay so in z i have two different lifts namely q prime composite omega bar and q composite omega these must be the same we have q prime composite omega must be equal to omega because we apply q to both of them this one and this one they are the same and they have starting points are the same so i took a path in omega path omega in z come down to x went back to lift it to all x bar the image of that path is the this picture start with a path here come here lift it here when you come here it is the same path that we started so this is follows by uniqueness of the path liftings okay so we have got this path q composite omega bar is something omega in particular q prime of the end point is the same thing as omega end point which is say z naught some some point which is nothing but q prime of x naught path by definition okay so this implies that f is inside k prime and hence phi f is inside k so this proves q check of the entire pi one of z is contained inside k now we have to show that k is contained inside this one that is similar to the proof is similar okay so therefore q check of pi one will be equal to k okay so i repeat how the covering transformation has got because the identification under subgroup is nothing but some uis have to be identified some ujs it is a full subgroup then all of them will be identified with a single one component right because then it will transitive remember all p inverse of v these uis this indexing set is nothing but the same thing as the group indexing set because there is a bijection between the fiber and the entire group what is this you can i can be chosen as group elements okay so there is some shuffling here and some of them are identified rest of them go to homeomorphically on to v that is what is happening after identification that is how q becomes a q becomes a covering transformation and if you fix one single thing there that identifications or become spread out to again some ui now this time the disjoint unit will be much smaller only on one single coset so that was showing that q prime is also a covering projection okay so this is much easier than we thought right getting getting coverings for arbitrary subgroups is easier once you admit a covering for the trivial group namely the simply connected cover okay so let us now proceed with existence of simply connected existence of simply connected covering does not come freely it itself requires some more conditions on the space x let us see why suppose p from x bar to x it is a covering projection x bar is simply connected take any point x okay and if u is an evenly covered neighborhood of x then there are copies of u copies of uis which are homeomorphic to u right of u in x bar that is we have open subsets v of x bar such that p from v to u is a homeomorphism the p itself restricted to v is homeomorphic because u is even recovered right so this is the starting point all the time we are doing now what happens let us see at the fundamental level we can write the inclusion map u to x as a composite of see v to u is a p which is homeomorphism okay the restriction map I can take a p inverse there okay this is like taking a branch of the logarithm function for exponential function okay this is not defined on the whole of x defined on u and there are many many p inverses here one of them goes to v so that p inverse followed by the inclusion of into x bar followed by p this is nothing but now the inclusion of x because p p inverse is identity so the inclusion map from u to x is broken up or as a decomposed into p inverse inclusion p inclusion p inverse that's the way therefore if you take take a point here start with x x belonging to you right then take pi 1 of ux to pi 1 of xx okay this goes through pi 1 of x bar x bar some point which is sitting over this one pi 1 of x bar x bar is a trivial group so if a homeomorphism going through trivial group the composite must be trivial so what we have proved u is to begin with a neighborhood of x some arbitrary point x but u is a neighborhood which has a property that it is evenly covered and connected the connectivity just passing in individual then the inclusion map induces trivial homomorphism why that should happen we have never bargained for that one right so far right so this is a must this will happen if there is a simply connected company okay so this is very important so we make a definition of all this far so we say start with a locally path connected space we say x is semi locally simply connected very strange kind of definition okay but wait a minute you would explain itself if each point of x let this point x little in x has a path connected open neighborhood such that the inclusion induced map is trivial so this was a condition which we observed which holds if x bar is if there is a simply connected covering for x so we make this as a definition okay as a condition then we put that this condition should be true okay just now just a definition if such a thing has for all the points then x is called semi locally simply connected you could have state only locally simply connected that would mean that pi 1 of u x itself is trivial you don't need it we need only the inclusion induced map the homomorphism must be trivial the inclusion map need not be inclusion in the fundamental group okay if that is also true then we would have what pi 1 of u would say be trivial but that need not happen for example you take s1 included in d the disk pi 1 of s1 is not trivial but the inclusion map to d is trivial because pi 1 of d is trivial okay so that can can happen so how to ensure that such things happen well there are stronger conditions like local contractability locally contractability means what at each point if you have a fundamental system of neighborhoods which are contractable which hence we are saying that given any neighborhood of a point there is an open set contained inside that and that open set is contractable containing the point okay the neighborhood should open set should contain the point okay similarly we can define just locally simply connected which I define just now instead of semi locally what is this definition given any point and a neighborhood there is a smaller neighborhood of that point which is simply connected then this should be locally simply connected what is easy here is is the following namely if you locally contractability that will imply local simply connectedness that will imply semi locally simply connected so what we are interested in is this weakest condition and it turns out to be that that is enough this is always true whenever x bar there is a sorry there is a simply connected covering x for x these two this condition is too strong this may not be true this also is stronger this may not be true but this will have to be true and if we put this condition then we can guarantee the existence of simply connected covering for x okay however how to check this one if local contractability okay you can go ahead simply connectivity locally so what are the things that guarantee you this one manifolds are such thing superficial complexes are such thing we have not proved it yet maybe we will prove it in the next part in a part two of this course okay there are CW complexes which have this property and so on there are lots of spaces which have this property so it is not hopeless okay so entire covering space theory is applicable to this manifold superficial complexes CW complexes and so on okay so let us prove now the following theorem over a connected locally path connected semi locally simply connected space there exists a simply connected covering space the connectivity part is optional locally path connectivity and semi locally path connectivity is not optional you understand if it is not connected what you can do is connected component then prove for each connected component and put them together that will give you the simply connected covering for all of them okay so that is why connectivity assumption is like a simplifying assumption that is all not mandatory the idea involved can be definitely traced back to function theory of complex variables one complex variable especially to what is called as analytic continuation and germs of homomorphic functions here we more or less imitate the construction of Riemann surfaces of a meromorphic function one may say that homotopy lifting property together with unique path lifting property there are two of them captures all the homotopy properties of covering projections okay though they fall a little short of characterizing the covering projections under some slightly stronger condition this is what it is and that is what happens in function theory of one variables does a covering projection covering space can be thought of as a suitable space of classes of path is in a given space and that is why you needed existence of sufficiently many path is in the given space X so that is given by locally path connected so there are plenty of path is in the space but semi locally simply connected will tell you that there are not too many of them so it brings some sobriety into the existence of path path is there is not too many of them mainly if you take many of them when you go to the larger space they will be all homotopy that kind of result is semi locally simply connected okay take a loop low in a in a local loop locally you may not be able to make it hominolomotopy but if you globally if you go into the whole space it will be homotopy so that will show that not too many funny kind of loops are there okay one on the one hand there are plenty of them on the other hand no no no no hold on there are you know some conditions on them not too many so that is what is happening for existence of simply connected coverings so there is quite a bit of work to do here so let us do it one by one so here is the definition path is out to be studied therefore we go to the function space x raised to i x raised to i what is it all path is from all path is in x since we want everything starting at a point you know base point is fixed and so on so we start x not the whole space x we start at the base point and we would like to have only path is starting at that point so this is the subspace pxx not contained inside x power i remember x power i we have to given compact open topology okay so same space subspace topology will be given namely compact open topology on pxx not also okay so this is called now path space actually this is a path space but which is too big we want this one okay this is a path space path is starting at x not and all of them are inside x okay so remember there is a evaluation map from x cross i x power i cross i to x right evaluation map at various point evaluation maps are nothing but coordinate projections restrict it to p that will be still continuous so what is this evaluation map i am taking here is e of a path omega take the end point why i am taking end point the initial point is always x not okay so other one is important the end point remember in fundamental groups and while studying fundamental groups you have to want two points you have to concentrate on initial point and end points in the homotopy also these two points have to be kept fixed throughout so that is why you see all these things are very natural they are not new i mean they are not strange why they are coming is obvious because in the study of homotopy classes of i mean loops and so on paths and so on if you don't keep both the end points fixed head everything you leave no longer there is nothing no theory okay so we want x not starting point is fixed head now we are looking at the end point so that map is very important and that will tell you the entire story so this and e is evaluation map its behavior is comparable with exponential map so it is just a chance of english in a language that the same e will denote both evaluation map as well as exponential map that is good for us anyway so x is path connected therefore e is surjective take any point that is a path as starting from x not ending at that point omega 1 will be that point so this means e is surjective e is continuous e is a projection map therefore e is also an open map right well openness is not clear because we are not taking the product topology we are taking even stronger topology there so you have to wait for openness of space space itself is contractible and the evaluation map is an open map openness is not easy not as easy as continuity with continuity is fine because we have this projection map and the topology on x x not is finer than the finer than the topology which is given by the Cartesian called Cartesian product topology okay evaluation map continuity we have seen already in exponential correspondence so how to show that the path space is contractible that means I have to construct a homotopy from p cross i to p right p cross i to p which is homotopy of the identity map to constant map that is what I have to construct so I look at a very simple map here you know which is again given by what happens this in in i cross i so what I am taking is now instead of p cross i p cross i to x power i I have to give a map but I will bring it p cross i cross i to x then h of omega omega is a path here t comma s I am sending into omega of ts okay look at what happens if s is zero irrespective of what is it is omega zero that is x not if s is one then it is omega t which is the identity map so this is omega t is zero irrespective of what s is it is omega not okay so all this is important that will tell you that under the exponential correspondence this is continuous it is obvious right because t t into s is continuous omega is continuous which is continuous function under the exponential correspondence there is a map h hat from p cross i to x power i but where does the starting point of this one goes all the time for each path here star set x not because when s is zero this is always always omega zero therefore this h hat takes values inside p what is the definition of p p is p of x x not all paths must start with x not that's all so it is a map into p and of course when s is one you see it is the constant omega right therefore this is a homotopy of the identity map of p to the constant map so p is contractible so this is easy this came very easy okay to prove that e is an open memory we have to show that open subsets image is open what are open subsets this is the compact open topology that's what we have to do so far okay so what is compact open topology take a compact subset of the identity of the interval on zero one take any subset u of x open subset then take all paths which take the compact subset inside u so that is the the bracket k u right this is a sub base set of all such k use forms a sub base for the compact open topology collection of all such things intersection of finitely many such elements forms a base okay so it suffices to prove that v which is intersection of k i u i finitely many of them 1 2 up to n minus 1 then apply e the image of that under e all these things are open if all basic open sets this is basic open set the image is open then e will be an open memory because then every other thing is union of such elements e of the union is union of ease e of the intersection is not the intersection of the ease for any function f f of a intersection b may not be equal to f a intersection m b but the union is fine so you have to do this one not just e of k u is open that is not enough okay so where k i's are compact subsets of i and u i's are open subsets of x okay so let us complete this one so take x belong to be omega in the intersection and omega 1 will be equal to x this is the evaluation e of omega is this one okay take this such a point x belong to be such that say what do you mean x belong to be v is the v is the e of this one now that there is some omega so omega e of omega is omega only put it if one is in k i see one k i is subset of 0 1 interval that they are compact subsets if one is in one of the k i then x will be inside u i right because omega of this omega of that one would have been inside u i so let us say un be a path connected neighborhood of x containing contained in all the u i's for which one is in circuitry okay it should be intersection of a finite intersection of all those contained in all those u i for which one is inside k i by by continuity of omega that let us choose an epsilon between 0 and 1 such that omega of epsilon to 1 will be contained inside u i because omega 1 belongs to me okay omega epsilon 1 is contained in the un is possible such that if you take epsilon 1 intersection other k j's they must be empty the k j may not equal to one of the u i's here for which one belongs to okay that is possible so whenever one is not in k j you can choose epsilon to be epsilon intersection 1 is intersection this one is empty okay put k n as epsilon 1 now so I have got one more compact set along with k 1 k to k n k n minus 1 okay now you take w to be intersection of all these along with the bracket k n u n also one more so I am taking one more here that is all I have chosen this k n and chosen you very carefully and carefully okay so once you do that you are home so now we claim that the un is contained inside e of w then e of w is contained inside okay so I have formed a neighborhood of x x un is a neighborhood of x okay that neighborhood is in the image of w and this image is contained so every point is covered by an open subset in the image which means the image itself is open okay so this v itself will be open we wanted to show that v is open okay so given any point y belonging to uh un sorry not v n e of that one is open this whatever yeah given any point y belonging to so so let us prove this far take y inside un choose a path what do you mean by y is in un epsilon 1 to un okay you choose a path from epsilon omega epsilon to y omega epsilon is already inside un okay and each un's are path connected in a connected components as path connected subset actually to join epsilon omega epsilon to y okay if gamma is defined to be equal to omega up to epsilon okay so epsilon to 1 it is this path okay tau okay then one observes that gamma is in w because first part is already inside all this intersection the last part will satisfy that it is inside kn un because kn is nothing but epsilon 1 it goes inside so tau of this one is connected so it is an element of w now okay so so this so defines equal tau equal then one observes that gamma is such that is inside w and e gamma is y by definition so this completes that un is contains w therefore v is open okay so what we have proved here is v which is image of k u in your intersection of k i u i that is an open set therefore the evaluation map is an open map so we stop here and continue the construction tomorrow thank you