 So, at last we are with cypher 1 campan chair. So, let us begin with a another small preparation here, patching up of covering spaces. Start with a group G, let U and V be two path connected open subspaces of X, where X itself is U union V. Let P and Q be coverings, G coverings over U and V respectively. Suppose W which is U intersection V is path connected and there is a G map between the two coverings restricted over W, namely P inverse W, P, W, T inverse W, Q, W, there is a map V. That is what P is a homeomorphism of the top space, so it commutes with the projection maps and indeed recall that V of GX must be G of VX, this is the definition of G map. Then there is a G covering tau on the whole of X. So, tau is such that tau restricted to U twiddle will be P and tau restricted to V twiddle will be Q and the total space is exactly the union of U twiddle and V twiddle. That is the meaning of these two coverings on these two open subspaces have been patched up to give a covering on the whole spaces. Let us see how it is done. So X twiddle is the quotient of the disjoint union U twiddle will be twiddle. It is not just the disjoint union because below U and V have intersection. So on those intersections we must have some identification, namely Z is identified phi Z for every Z in P inverse of W. Remember phi is a homeomorphism from P inverse of W to Q inverse of W. For all points in P inverse of W, I identify with its image under phi. So that is the quotient map. That is the quotient space. So by definition X twiddle is the quotient of U twiddle disjoint union V twiddle but see that within U twiddle there is no identification. Similarly within V twiddle there is no identification. Therefore under the quotient map U twiddle and V twiddle go injectively and we identify these subspaces of X twiddle to be U twiddle and V twiddle. In that sense X twiddle will be union of U twiddle V twiddle. So because phi is a homeomorphism of an open subset of U twiddle to an open subset of V twiddle. Remember these things are W is an open subset because W is U intersection V. U and V are themselves open subsets in X. So their universities are also open subsets inside U twiddle and V twiddle. When you identify along an open subset by homeomorphism, it is very easily checked back the quotient map itself in open mapping. So it is a open surjective, open quotient. Moreover there is no identification within U twiddle and V twiddle. So lambda U twiddle can be identified with U twiddle and that is what we do. We identify U twiddle with the respective image under lambda. So X twiddle is defined properly. Now on the disjoint union, you can take P disjoint union in Q from U twiddle disjoint union in U twiddle to U union V which is X. But now because the P and Q are respected by this phi. See Q composite phi is P. Therefore the disjoint union factors down to give you a map. Let us say, well I have here, I have defined it as tau, distributed tau from U twiddle disjoint union to X twiddle. And this itself will be a covering projection because you can just see that if U and if P and Q are covered evenly, say that is P is covered by evenly by an open subset, the same thing will happen for lambda over that open subset. Because everything will be inside U twiddle. Same thing will be happening for under Q also if some open subset is evenly covered by Q and X is covered by U union V, X is U union V. So this follows that the union P, Q and Q defines a covering projection. Here in the statement I have put tau, here I have put lambda, it can change this lambda to tau. So that tau is an extension of P as well as Q. Same map collapsing to, I mean factoring down to the quotient map. So this was a fairly easy picture. The hypothesis is very important and it does not come freely. What you have to assume, intersection is path connected and suppose there is a homeomorphism of the respective things here, of the restricted coverings. Restricted coverings or G coverings, they must be isomorphic. That means there is a G map between them, that is all. So that you can patch it up along with that. The beauty of this one is I have never actually used that there are only two open sets. You could have had X as union of a family of open sets and assumed this condition for each U i intersection U j and perform this quotient space construction, it will work exactly same way for arbitrary things also. This hypothesis should be for each, instead of U and V, for each U i intersection U j, they must be path connected and then there must be homeomorphisms like this and under these homeomorphisms you identify them, then you get patched up for covering projection. The same proof will work. So this is what the remark here says that if you can replace the two open subset U V by a family of open subsets and W with the family U intersection U j, U i intersection U j, then the proof is identical, it will work the same way. But the problem is getting those phi ij's which are isomorphisms on intersections, that problem is not, that is a tricky hypothesis there. So that extra hypothesis, if you have some extra hypothesis, then only that will come. So let us see in the usage, what is the text hypothesis we need. So here is what we want to apply this patching up. Now let X be a union of open subsets, indexed over lambda where each U i is path connected, semi locally is simply connected so that we will have the simply connect coverings for them, that is all. Further assume that intersection U i, U j for all of them is the same W, this is the extra hypothesis, for every i not equal to j, okay, one single W which is path connected. I allow U i's to be not just one, two member but any member, but all of them it should intersect in the same set, same open set, pair wise intersection must be same W, okay. Choose a base point inside W to X naught and let this be the base point for all the spaces involved in computer fundamental groups so that I will not write it down because one single same X naught is there, X naught is in W, X naught is in U i, U j and X naught is in X. So everywhere I am taking the base point X naught, okay, it just saves me of writing down that element, okay. Let eta i and phi i be the corresponding homomorphisms corresponding what induced by the inclusion maps, W is included in U i, look at the corresponding homomorphism in pi 1. Similarly here, their inclusion induced maps do not confuse them for monomorphisms, they may not be monomorphisms, that is the crux of the matter, okay. The inclusion induced homomorphism, then look at this collection pi 1 of X, that is a group, eta i phi i, okay, this becomes a diagram. What we want to show is that this is a push out diagram, this is a statement. In other words, once you know pi 1 of U i's and the intersection pi 1 of W and their monomorphisms etcetera, pi 1 of X can be completely determined, this has to be the push out. So this is the van Kampans theorem. Taking the fundamental group of the union by knowing them on each individual open subsets, so that is vaguely but this is the full rigorous statement here, okay. So let us go to our first proof. Take any group G, okay, take alpha j from pi 1 of U j X naught to G homomorphisms such that alpha j eta j equal to alpha i eta i naught equal to j. So this is the data for push out, this is another push out, this is another diagram, okay. If the original diagram is a push out diagram, then this is what it should admit a unique homomorphism gamma here such that these triangles are commutative. So this is the meaning of pi 1 of X is push out. So given such a thing, we must produce gamma. So that is our task now, okay. So what is our tool? The tool is the classification for G coverings. All G coverings are interpreted in terms of homomorphisms from the fundamental group into whatever G that we have taken. The correspondence is canonical and this is what we are going to use again and very strongly now, okay. So to get this gamma, so we are heavily using the classification theorem 8.2 along with the canonical property of meandu, all right. So what are these covering factor? On each Uj, there is a homomorphism from pi 1 of Uj to G. This will give you a unique G covering, unique in the sense isomorphism class, isomorphism class of a G covering. Every homomorphism gives you G covering. Let me take that one as Pj. For each J, I take this one, okay. Now look at alpha J from pi 1 of Uj X0 to G is the given map here, right. If you composite it with eta J, what do I have? I have the same thing as alpha I and then eta I composite alpha I. These two homomorphisms are the same means if I take this covering and pull it back here. If I take this covering and pull it back here, they are isomorphic because the corresponding homomorphisms are the same, okay. So this is the hypothesis that we get. This is connected thing here. You have to come here. To come here, you need a homomorphism from pi 1. To do that, you need a base point here, common base point, okay. So we have done not only just a common base point but a small neighborhood and that neighborhood is the same for all Uj. That was a crucial fact here, okay, alright. So now I get a covering, two covering transformations here and there will be a Pi, Pi J from a G map from one to the other. They need up and get a covering over X which restricts through these coverings when you pull back through these inclusion maps, okay. That is the extension of this covering. That covering corresponds to a homomorphism into G. Take that as gamma. Automatically, if you take Phi composite that, it is discovering therefore that covering is given by this map, this homomorphism. So it must be homomorphism, okay. So we have produced a gamma with the corresponding property. Why gamma is unique? Suppose there is another gamma prime which also fits this diagram. Then what happens? If you pull that, use this composite, this one, okay. The covering given on Ui will be the same thing as covering given on by this alpha i. Therefore, restricted to each Ui, the coverings are the same but Ui is forming an open covering for the entire thing. Therefore, the two coverings here are the same. Therefore, the map is the same. Same means what up to isomorphism. Therefore, the map, the homomorphism might be the same. So you see that is the end of the proof. So the proof of Wenkampans theorem has become complete tautology here. No problems at all, okay. I repeat here. So what we have done? Start with such a diagram. We have to produce gamma and we have to show that this gamma is unique, namely when it commutes with these diagrams. So what we will do? We will take these homomorphisms and construct coverings corresponding to that. Construct means we have this mu of this alpha, mu of that one I can take, okay. When I pull them back here, because the two homomorphisms are the same, they must be isomorphic here. So take those isomorphisms and patch up these, all these coverings. Start with this joint union of these and then identify them to get a covering on X. Covering on X, now you have to go back, reverse gives you homomorphism. That homomorphism when you compose with this phi i, it is eta alpha i which just means that if you pull back this covering over this one, it must be this alpha. So therefore this must be alpha i. I mean if this is alpha i, same thing as saying that pull back, this covering must be the same thing, okay. As an immediate corollary, we will write down this one in a simpler case, namely when X is union of only two open sets, connected, path connected, okay. And the intersection is simply connected. This is one of the cases that we are interested in, okay. Assume that X is locally contractible or locally, semi locally, simply connected and so on, so that covering spaces makes sense. Then the inclusion induced homomorphism, I check from pi 1 of u to pi 1 of x, J check from pi 1 of u to pi 1 of x. Why I am not writing the base point? I am assuming that the common base point in the intersection which is connected is taken, okay. Same common base point for all of them, that is why I am not writing. These are injective and pi 1 of x is the free product of their images. This is the consequence of the push-out diagram when here, when this is trivial group, it is simply connected. In this trivial group, the push-out diagram is same free product of this. And we know that these are monomorphisms. So you can identify each group G i with the free product of them. So here I am only applying two of them at a time, okay, in this theorem. You could have had any number of them, alright. So that is a simple version. So all these are, there are several versions of Encompass theorem. Unfortunately, some of them are very, very complicated. They cannot be handled directly by our classification theorem. However, whatever we have proved itself is of tremendous potential, tremendous application potential, okay. The general form of this theorem without local contractability or etc., but slightly other kinds of assumptions are there, okay. You can read an excellent book of Masses, an algebraic topology, but even this one do not give you all the versions. There is another prototype version available in TIFR lecture notes by Dharam, algebraic topology lecture notes. These things are available on our website also, downloadable. So you can read them. And if you are working in three-dimensional topology, you need to have many other versions also, okay. But once we have, this training you have received, the fundamental relations and how these are working here and what is the relation between covering spaces and fundamental group and so on. Those things when you read they will become easy for you, that is all, okay. Just for instance, you just look at the circle S1. You can, circle S1, you can write it as disjoint union of, sorry, overlapping union of say S1-1 and S1-1, okay. But the intersection is not connected. It will contain two open arcs. All of them are trivial now. So you can think of these are u and v, they are arcs. So they are simply connected, pi1u is trivial, pi2u is trivial, but the intersection is not simply connected. Intersection consists of two components, okay, and both of them are simply connected. So what is the push-out diagram? Everything is identity here, everything is identity group. But you know the fundamental group of S1 is not trivial. So where did it come from? So if you cleverly answer this, which is what the RAM does, you get a different kind of push-out diagrams, okay. So lack of, because lack of time I cannot do all that now because it will go beyond course, okay. However, let us, whatever we have done, let us use that in a as cleverly as possible and carry quite a few interesting results here. So let X be one point union of copies of S1 index over a set lambda, one point union, okay, what you call as wedge of circles. Then the fundamental group is a free group over the set lambda. In particular, its rank will be equal to the cardinality of the set lambda, how many circles you have taken, okay. So this is an immediate consequence, all that you have to do is take the common point wherein you have identified all the circles, one point from each circle, let us take that as X naught, the base point, okay. Pick up a point Xi in the copy S1i different from X naught, throw away all these points from X. X is the union of all these circles, throw away all these points, that is an open set, call that as W. Put U i equal to W union one of the element, one of the point that you have thrown out, okay, call this as U i. If you put Xi, call this U i, that is also an open set, okay. So for each i, I have defined these sets. Clearly union of all the Uis will be definitely equal to X. And if you take U i, okay, all of them will contain this W. So their common intersection, all U i will intersection W, okay. So that is an open set. So now what is W? W, you have removed all the points, one point from each of these S1. When you remove that point, you are left with two open hours, which you can collapse to a single point X naught. So this you can do for each circle, which just means that W is contractible. In particular, W is simply connected, okay. So you have the nice picture, W to Uis, various Uis, in the singleton, the W pi 1 of W is 0. So that is a trivial thing. So the pushup diagram becomes a free product, okay. So what are pi 1 of U i? In U i, what you have done, all, instead of all the Ws which are, which is contractible, in one of the circles, you have put back this point. So that circle appears. The rest of them is contracts to single point. So that is a homotopy type of a single circle. Therefore, pi 1 of U i is infinite cyclic. So what is the conclusion? Conclusion is that the free product of this infinite cyclic group is the pi 1 of X, which is the pushup, okay. So we get a free product of infinite cyclic groups, one for each i in lambda. That is why you can think of this as free group or lambda, okay. So more applications we will see next time.