 So, today's topic is some kind of general discussion about covering spaces, namely what are the properties shared by both the base space and covering space? By properties I mean topological properties. Since a covering projection is a local homeomorphism, automatically all local properties are shared such as local compactness, local contractability, local connectedness, local path connectedness, being first countable and so on. But it also shares a lot of what are some kind of local global properties and some global properties also ok, manifold, being analytic manifold, C infinity manifold or being a topological group. So, many of these properties are shared by both X and Y. So, it is not possible to discuss all of them. So, here I will give you a sample of that one simple thing because we are just doing lot of simplicial complexes and so on. So, let me just discuss that being a simplicial complex, being triangulable that property is shared by base space and top space of a covering projection. For that I need another topological concept here, recall that a space is called compactly generated if a set is open, if we do not leave it is open in its intersection with a compact set is open inside compact sets for every compact set. So, the compact sets of a topological space determine all other open sets as well as closed sets. It is the same thing k intersection a is closed in k for every k, then and then only a contained inside x will be closed either open or closed this two can be one from the other you can go back to back and forth. Simplicial complex says the width topology is compactly generated why because we have proved that any compact set is first of all you know covered by compactly many as a finitely many closed complexes and each closed simplex is a compact thing. And finally, week topology is nothing but the topology controlled by each closed simplexes a set is closed if and only if intersection with each closed simplex is closed in the simplex ok that is the relevance of compact generated when you are dealing with simple shell complexes. So, the first thing that we proved a prove here is compactly generatedness is a local property. What is the meaning of local property? I have been using this word, but you may not know what the meaning I have in mind. What I mean is each point it is a neighborhood such that the given property holds like local compactness and local connectedness the same thing as you have an open cover such that on each member of this open cover the property is true ok. So, this can be taken as definition of local property right now suppose you have an open cover you such that the subspace topology on each you belonging to you is compactly generated then I want to say that x is compactly generated. In fact, if x is compactly generated every subspace you is compactly generated that is easy ok. So, now you are taking use an open cover then I want show that you can come back ok. So, this is a proof take any subset G such that G cap G intersection K is open in K for each compact subset K, K is fixed set and G intersection K is open in K. If this happens for every K we have to show that G is open ok for this because U is an open cover it is enough to show that G cap U is open in U if G cap is open in U then G will be union of G cap U and G cap is open in U so G cap is open in X because U is open in X. So, union of G cap U over U will be G which will be open ok. So, now at K we any compact subset of U then it is compact in X as well as because compactness is not a relative notion it is independent of where X is ok. Therefore, G cap U cap K is nothing but G cap K because K is subset of U already this is open in K. If K is a compact subset of U then G cap U cap K is nothing but G cap K. So, G cap K is open in K therefore, G cap U is open in U ok. So, this is if and only if ok it is no property then for each open subset this will happen and if this is the case then it is it is happens for the whole thing ok. Now, take any covering X bar to X ok instead of covering I am proving a more general result here namely take a surjective local homeomorphism then apply it for covering also. Any surjective local homeomorphism then X is compactly generated if it known this X bar is compactly generated. To see that what we have to take start with an open covering for X bar such that on each U inside U P from U to P U is a homeomorphism ok. This you can do because P is a local homeomorphism ok and of course, local homeomorphism means P U is an open subset in X. Then P U U belong to U is an open cover for X. U is an open cover for X bar the P U will be open cover for X and U to P U is a homeomorphism. Now, you can use the previous lemma to go up and down suppose X is compactly generated what does that mean that the topology on U for each U is compactly generated. Therefore, the topology on U instead of P U because P is a homeomorphism topology on U is compactly generated but U is an open cover therefore, X bar is compactly generated. So, when coming back if X bar is compactly generated each U on each open subset is compactly generated therefore, P U is compactly generated the P U is an open cover for X is compactly generated ok. So, you just need a local homeomorphism which is surject of course, you should cover the whole thing that P U should be cover. So, you shall surject your homeomorphism local homeomorphism is good enough. So, now, we come to the main topic of today's agenda. Let X be a covering projection X and X bar locally path connected and connected. This connectivity is not exactly necessary all the time anyway, we have told several time. Let H from mod k to X be a triangulation of a connected topological space X ok. Then for every covering X bar like this there is a triangulation of X bar namely H bar from mod k bar to k and a simplicial mass V from k bar to k such that this diagram is commutative. So, even the H bar here which is which is a triangulating map is not arbitrary. In other words, you can forget about X bar and k what we get is mod k bar to k mod k ok that map V bar itself will be a covering projection ok. In other words start with mod k take a covering then there will be a simplicial structure on this covering such that the map P the original covering map becomes a simplicial map. So, this is the gist of the first part here. Further moreover for each n simplex k the cardinality of the set of n simplex system k bar is equal to alpha where alpha is cardinality of the fiber any fiber. Remember if this is where connectivity is necessary that is all. If X is connected the cardinality of the fiber is a constant ok. So, this part says this every n simplex of k there will be alpha many simplex is in the X bar exactly same. So, look at the vertices how many vertices where for each vertex here there will be alpha many vertex is above right ok I will make it clear in the proof of this far. So, what I am going to do is V be the set of vertices S be the set of simplex is for k then we are going to define k bar V bar S bar S for all. V bar is P inverse of H of V remember H is a homeomorphism from mod k to X. So, I am picking up the vertices in X bar to be P inverse of the vertices in X that is the meaning of this H of V are the images of those points which are vertices vertices are just P inverse of that that is the vertex side. What are the simplexes? This S bar is collection of all H bar of F what is the H bar and what is F I am going to tell you where F is a simplex of S ok like vertices triangles edges all those things have to be coming from S. Then look at this H bar H bar is a function from mod X mod of F to X bar that means a lift of this H H restricted to F. Mod F is a subspace of mod k H is a homeomorphism mod k to X restricted to the mod F then you must you must get a lift of this one H bar is a lift because P is a covering projection take all possible lifts all of them look at this all these H bars then take H bar of F F is a finite set remember that. So, H bar of F will be a finite set and what are the these are the vertices of simply so they will be inside H of P inverse of H of F. So, they will be vertex of this one to declare the mass simplexes if you have a subset A of B which B is H bar of F then A is nothing but H bar inverse of some F prime where F prime is subset of F. Therefore, you can restrict the same H bar to this subset A mod A so every subset of B finite subsets there they will be also subsets of S bar by this definition. So, what we have seen is that this k bar is a simplicial complex that is easy to define but why modulus of k is homeomorphic to X bar in such a way that when you take composite with P it will be a simplicial map from k bar to k. So, that is the harder part to show already definition is over before proceeding further I will want to tell you some simple situation look at the circle easiest way to triangulate is take one omega and omega square three points join one omega and join omega to omega square omega square to omega by those arcs okay. So, it is a triangulation it is actually a triangle now take the covering let us take simplest covering namely r exponential function. So, what is the corresponding triangulation on the real line which will make this namely exponential map itself look like a linear map. So, what is that declare all the inverse image of these three points as vertices. The inverse image of one we are familiar already we have done this over time namely all the integers come inverse image of omega all one third integers will come one plus one third two plus one third three plus one third all of them will come then omega square one plus two third one plus two plus two third all those things will come. So, each interval music interval will divide it into three parts okay. So, now exponential map from each interval to another interval can be thought of as as a linear map of course it is arc there. So, when you take actual triangle to that one that homeomorphism is precisely the linear map what is that homeomorphism you look at the triangle and push it back to the circle that that is the homeomorphism right by the radial projection right. So, what are the triangles what are the what are the edges I have declared by this formula what I have declared is you take an edge here lift it lift it at various points okay if we have a lift then join the two points the end points by a simple x. So, that is precisely what this h bar of x means I cannot join just one third two some five by three or some other 25 by three that is not possible one third you have to lift there are two possibilities. So, one going backward one coming backward that is the only two lifts will come there. So, what you get is a real line okay. So, this condition is very important that we cannot take orbital subsets of v bar as simplices of this one of course that will be a simplification complex but it will not be giving you mod k bar will not be put x bar okay for that you have to use this x bar itself and the lifts of lifts of this homeomorphism piece of homeomorphisms restricted to each simplex and then you get a simplification complex okay. So, now even if you do not understand all the details which are getting lost in notation you must be able to figure it out yourself okay. So, let us go through some of the proofs here clearly k bar is a simplification complex which we have seen since for each f in s mod f is a simply connected mod f is actually contractible it is homeomorphic to a disk right. So, lifts are always possible the cardinality of lifts is precisely equal to alpha at each fillet to one point you lift it you will get that many disjoint lifts in fact every under a covering space every contractible space every simply connected space we have seen the university may you know disjoint union of copies of this they are all evenly covered okay. So, look at this h bar in the h bar whatever you lifted and look at the endpoints declare those endpoints as union of those things as the simplex system okay and then you have you can fill it the filling you will be also done by h bar that is all okay so that is the trick. So, what happens is compatibility compatibility comes because uniqueness of the lifts okay. So, suppose f1 and f2 are inside s bar and intersection is f3 is non-empty if it is empty there is no question of compatibility suppose h1 bar h2 bar h3 bar are corresponding lifts of this edge from mod f1 to mod f2 and mod f3 into x we claim that h1 bar restricted to f3 is the same thing as h2 bar restricted to f3 that is nothing but h3 therefore compatibility is there why this is true this follows because as soon as the lift is defined at one point take one point here okay it is defined uniquely on the whole connected sector the f3 is non-empty so if you take one point here okay so wherever that goes because if it is non-empty here there is one point then whole h1 bar will be same on a whole of f3 which is empty then there is no question there is no so but that is same thing as h3 because at one point it coincides okay it follows that uniqueness of lift connectivity of f3 this we have to use okay. So, it follows that if you have this one all the various lifts will patch up on the intersection they agree so there is one single function h bar such that p composite h bar is x okay automatically on each this simplex is its continuous so it will be continuous also so this is where we have to use that mod f mod k has the what is this width biology a function is continuous if and only if it is continuous so on each simplex so now we have got a function h bar from mod k bar to x bar such that when you composite with p come back come come here okay and restrict it to each k here that is h so I have to define what is the map from k bar to k can you guess now what is the map it is very easy what are what are the simplex of k bar they are lifts of some f so you map them accordingly what is p inverse of what is v bar v bar is what v bar is p inverse of v so wherever it goes to you define phi to v that map so that is the simplex map that is the vertex map it will become simplex map by the very definition so vertex map is already there so that becomes simplex map because of the very definition okay so diagram is commutative is fine finally what we have to do is that show that h bar is 1 1 on 2 and homomorphism once again on each simplex it is a homomorphism by very definition the lifts are automatically homomorphism okay and they go into compatible things therefore the bijectivity automatically follows okay on toness follows because take any point in x bar come come down to x it is inside a simplex okay so that simplex will be can be lifted up so that is subjectivity all right so I have written little more details of this one here okay so h component phi equal to p component h bar I have written down complete detail here so what we can do is look at the star of a vertex the star of a vertex is an open subset of mod k under h it will be open subset of x the star of x star of v are all contractible spaces so it is in particular they are simply connected therefore the inverse image will be disjoint union of open subsets which are mapped on to each star v this star v is evenly covered okay so you can use that to to establish that there is such a homeomorphism diagram commutative diagram h is restricted to star v this extra star v p is a homeomorphism on each connected portion of the inverse image okay and that will become star of v prime okay so this will automatically show that this h prime which is restriction of h bar okay is both open as well as continuous okay this is another way of doing this one it's a homeomorphism at each each each open covering of this one okay so I have given you a heuristic argument why it's a bijection here is a straight forward proof to show that h bar is a bijection okay so there are many interesting properties shared by by covering projection under covering projection by top space in the bottom space but in the right in the beginning we remember we have we have cautioned you that half stars nest etc are not shared regular cover regularity or normality etc are not shared so you have to be careful about that well I have given you some exercises so there will be chance to look into these exercises and then discuss them also maybe next time we will discuss some more examples and then go to the next topic so we will discuss some more examples of covering projections then go to the next topic okay so thank you