 Last time we proved Venkampan's theorem and then derived that if you take the fundamental group of the wedge of circles then it is a free group. The rank was equal to number of circles involved there. We shall use that and try to do now the computation of fundamental group of any one-dimensional simplicial complex. Indeed, we would like to include the case like wedge of circles also. So we would like to extend the notion of this one-dimensional complex slightly a little more general. So such things are called pseudographs. By a graph I will mean a one-dimension simplicial complex. So a more general thing, a pseudograph, some people may call it just graph itself but for me a pseudograph means that you can have a single vertex and then a loop around that or two vertices with many edges between them which these things are not allowed in a simplicial complex, one-dimension simplicial complex, remember that. So this instead of defining like this ad hocly, I will do it systematically in a slightly different way which can be used full for us in doing more rigorous mathematics. So here is the concept of attaching cells back to the first chapter there wherein we have defined adjunction spaces. So this is a special case of adjunction spaces now. So fix an integer k greater than or equal to 0, 0 is also allowed. Let y be a topological space and f alpha indexed family of continuous functions from sk to y. So sk is the sphere of dimension k, okay. So I have indexed family of functions k is fixed by the way, okay. Alpha is varying, there are a number of them. It may be 1, 2, 3 and any number of them, okay. Take a as disjoint union of all these the disks of one-dimension higher indexed again over the same set, okay. The space x obtained by attaching k plus one-cells to y, y are the map of alpha. This is what I am going to define, okay, this entire phrase. So this is by definition in the quotient space of A union y by the identifications on the boundary of these each of these disjoint disks namely sk, I have a map here. So identify x with f alpha of x, x is an element of the boundary which is sk, f alpha is an element of y. So identify this. This you do for all x in the boundary of s alpha and for all alpha, okay. If you have just one, one thing or thing here, this is what is you would have called adjunction space. But here is the same adjunction space, this is A and this is y and this is map f. Disjoint union s alpha, you can call one single f as the union of all these f alpha. So it is adjunction space, nothing else, okay. But this is a very special adjunction space wherein all the, all the disks on same dimension are coming extra from y. So y, we have attached these disks, x is the resulting space, okay. Now a pseudograph I mean a space x as in the above definition where I start with y a discrete space just a collection of points and discrete topology, okay. And the k which I take here is 0. That means what I have this s naught which is minus 1 plus 1 to y discrete space. So minus 1 plus 1 is also discrete. So any function from a two element set to y is continuous automatically. Either the two elements here will go to the same element or they may go to different element that is the only two cases, okay. Then I am attaching a one cell k plus 1, k is 0, k plus 1 is one cell. That means what d alpha, d alpha is minus 1 to plus 1 the closed interval, okay. The minus 1 goes to some point of y, the plus 1 goes to some other point of y, maybe same point of y does not matter, take the coefficient space, okay. That will be called as a pseudograph, okay. So we will have this terminology y instead of writing y, I will denote y as x naught. That means the zero skeleton just like in the case of temperature complex. It is just the set of points that is y to begin with and call it the zero skeleton, okay or those points will be called as vertices also. This is similar to what we have done for simply one dimensional complex. Then q which is a union x naught to x, okay. This is the quotient map, okay. Let q be the quotient map. That means what q restricted to each each one cell, okay. That is either the end points are identified or the end points are going to different points. So depending upon that either it is actually a homeomorphism or in any case the interior it is a homeomorphism. The end points may be the same in which case the image will be a circle otherwise it will be an arc. It will be an edge, ordinary edge, okay. So x naught is actually which is y is the space of x under this quotient map. So again points of x naught are zero cells or you can call them as vertices or zero simplexes and so on. So simplex word we will not use because this is not a simplex complex, okay. Indeed a simplex complex one dimension can also be described by this one. Only thing is then you have to put a lot of other conditions, okay. A pseudograph is in this sense a slight generalization of a one-dimensional complex. Clearly it is locally path connected. It is connected if q a the quotient map, okay itself is x. All the points of y are covered by the alphas, f alphas then it is automatically connected otherwise it is not connected obviously, okay. q restricted to a itself is a quotient map in that case. You do not need y at all, okay. Why is there because there may be some extra vertices hanging. So if it is connected then you can define on this whole thing this quotient space just on a itself. How? Whenever alphas are going to same point you identify them otherwise do not identify anything that is all. Wherever it goes to I fall for something why some other every time I also come to that one you identify them. That is the way it has to be done. So this y helps to describe those relations, okay. Otherwise it is more complicated within a but it is done inside a. If you use y then it is easy to see what is happening, okay. So here is a picture of a pseudograph how it could be different from a symbolical complex. To look at all these heavy lines and heavy lines that is a symbolical complex. But I have attached a loop here a one cell here of which the end points have gone to the same point. Here also the same thing here also same thing. Here they are going to different points but there is already another edge here. So these are violating these things are violating the symbolical complex structure, alright. So this is the general picture of a pseudograph. Immediately you can see that if I put two more vertices here on this loop this part becomes a symbolical complex on this part. Here what should I do? Here also I should put two more vertices, okay. Even if I put one it is enough because this point and this point would be now a single edge this is another edge so this will be like a triangle. So here I can do with just one vertex putting two more vertex no problem. What is the meaning of that? You are as if we are subdividing this pseudocomplex we have not defined anything like this subdividing the bad loops a bad edge here and so on to get a symbolical complex. So the in a hidden way the one-dimensional symbolical complex theory can be applied to pseudographs also. Okay I will use this remark again, okay. Now I will make a one more definition here namely a connected pseudograph is called a tree which is a definition if it is contractible as at a political space it must be contracted, okay. Then automatically it is connected of course such a thing will be called a tree by a subtree of G we mean a subcomplex T of G such that it is a tree it must be a pseudograph on its own but it must be a sub, okay. In a for example in this picture you can remove this one and look at the rest of the picture that is about subtree so there is a subgraph pseudosubgraph, okay. If I delete this one then also it is true you can I can just delete this edge and keep this vertex, okay but I should not delete this vertex and then I cannot keep this edge, okay. So edge has to have you know a complete vertex where it is attached that is the question. So both end points must be somewhere so here it is okay the both end points have gone here, okay. So you can remove this edge but you cannot just remove a vertex, you can remove a vertex only if it is isolated vertex. If you remove a vertex here all the edges which are emanating from there has to be removed that is the meaning of subgraph, subgraph, subseudograph, okay. A connected pseudograph is a tree if and only if it is simply connected so once it is simply connected it will be actually contractively so what I have to show. If it is contractively simply connected is obvious so I have to prove only the if part, okay. So start with x simply connected pseudograph given a vertex v0 belonging to x we shall define a homotopy x cross i to x such that x 0 is always this v0 and h of x1 is identity map. For this we first notice because of the connectivity of x that the quotient map okay restricted to a itself is a quotient map this I have already remarked earlier. Therefore constructing the map from x cross i to x is same thing as constructing it on all these d alphas or j alphas whatever okay the interval minus 1 to plus 1 copies of that cross i in a compatible way construct a quotient map on a quotient is always you can go back to the original space and then do that okay so j alphas I am denoting minus 1 plus 1 you can have a d alpha does not matter and let f alpha from is the q restricted to j alpha the q is the quotient map these f alphas we can say as characteristic maps this is actually earlier notation it was only defined on the boundary but I can extend it to this f alpha is extension of this one to the interior we can put a f alpha hat if you want because it is a castic maps okay given any vertex v now in x since x is path connected there is a path omega v starting at v not and ending at v fix those paths okay there may be many paths I do not care take some paths like this now you define h of v t equal to omega v t okay where I have defined h of v t omega v t for each v I have defined this one for each vertex cross i I have defined this one that is the meaning of this one now let f from f alpha from d 1 to x again I am writing d 1 or j okay minus 1 plus 1 this year be the characteristic map from one cell of x for each of them okay there is one cell in each one cell d alpha to cross i I have defined the map okay this map g is defined by g of s 0 is v not s 1 and the top is f s f s is this map f alpha every time I am not writing f f f f r 5 that is all g of minus 1 t okay is look at this omega by an f of minus f of minus 1 is a vertex right f of minus 1 is a vertex you take omega v t what is omega v omega is a path from v not to v okay and at one at one comma t you take its f omega f 1 okay so look at all these what I have done is on the boundary of d 1 cross i I have defined a loop okay on the bottom they are namely 0 cross i cross 0 a minus 1 plus 1 i minus 1 plus 1 to 0 I have taken it as v not then from v not I go by a path v f minus 1 omega f minus 1 on this part omega f plus 1 okay on that part these endpoints are joined by f alpha namely by the by the simplex there one simplex d alpha so you get a loop inside x and x is simply connected therefore this loop can be this function can be extended to d 1 cross i a function which is defined on the boundary of boundary of d 1 cross i which is d 2 in century can be extended to inside because it is not longer topic okay so x is simply connected it is used here okay g is continuous so you can extend it since x is simply connected you have an extension g alpha for each alpha okay now capital G you take the disjoint union over this one so that on each restriction it is g alpha all that you have to observe is that wherever you have identified it is the same old thing for each of them they it is compatible therefore this factors down to a continuous map as from x cross i corrects okay by the very choice when s is 0 it is v not when s is 1 it is the map surface okay so therefore it is identity map all right so let t be a subtree of a pseudograph g then a quotient map g to g by t is a homotopy to balance if it is a tree okay then it is contractible when we have done long back a a result when you can collapse the contractible subspace and yet the quotient map will be a homotopy to balance so i i would like to recall this namely when the inclusion map of this contractible space into the whole space must be a co-fibration remember that theorem so use that theorem to conclude that g to g by t the quotient map is a homotopy to balance namely when you collapse a tree tree means it is contractible thing okay like a vertex can be collapsed union of two sorry an edge can be collapsed union of two edges if they are not a loop they do not form a loop that can be collapsed and so on okay so this is what i mean if g is a graph this follows from t contains a g is a co-fibration by the way this co-fibration business was done only for simple complexes but the situation can be converted easily to a simple complex because a pseudograph can always be subdivided by putting extra vertices into a simple complex now let x be a connected non-empty pseudograph let t0 be a subtree in it then there exists a subtree in x that is t containing this given t0 such that this t contains all the vertices of x this is one way of telling that that this is a maximal tree you cannot make it into a larger tree by putting extra edges because when you put x all the all the vertices are there as soon as you put one extra edge there will be a loop so let us prove this one a rigorous proof requires now johns lemma or some such thing or hand varying it can be done and so on but in the finite case you can actually do this there is no problem for finite case you don't need johns lemma okay so let tau be the collection of all subtrees in x which contain t0 order it by inclusion okay this in this collection i have an order partial order now we shall apply johns lemma and then conclude there is a maximal one all of them contain t0 the maximal will also contain t0 so let ti be a chain in tau chain in this collection chain means what totally or not set we claim that union of ti is a tree okay so how do we know that it is a tree given any point in this union it will be in one of the ti s because it is a union then there is a path in ti from x to some point t0 because a tree is a connected thing already okay which will also path in t and so t is connected okay each ti is a tree so ti s are connected but now i approve that t is connected okay now suppose omega is a loop a loop means what it is continuous function from i to t therefore the image must be compact okay therefore we know that once it is image is compact it is contained in a finite sub graph finite sub pseudo graph okay it is compact it follows that the image contained in the union of finitely many closed one cells right closed one cells are means what i have told you that it is just either edges or full circles which are all contained some ti because here there are only finitely many of them let us say all the vertices are in t1 t2 t3 and then take the maximum of them in that ti they will be all be there but then this loop is inside ti it is null homotopecant ti but ti is a subspace of t so it is null homotopecant ti okay therefore every chain has a upper bound so that is sufficient condition for John's lemma John's lemma will tell you that there is a maximal tree a maximal element which contains belonging to tau means it is a tree containing tau 0 all right a maximal element i have got now suppose there is another vertex which is not in this maximal tree maximal t that would mean what because x is connected from that extra vertex you must be having a path coming to this t that would mean that you will get a vertex somewhere such that the next edge from that vertex to this tree the next other edge will be inside the t okay so this is what it means but then you see really seen that this okay what i have done here there will be an edges which will have one end point in t and the other end is not in t from the end point which is not in t you can collapse the whole thing to t contract that means t union s deforms to t or t is a strong deformation detect of s therefore t union s is contractible therefore t union s is a larger tree that is a contradiction to t is maximal therefore there are no more vertices so that completes a proof that every every tree can be contained in a larger tree and the largest tree such that all the vertices of x are there okay in particular there is always such a thing we can start with a single point single point is a tree after all so every connected pseudograph is homotopic to a bouquet of circles so this is the the corollary of whatever you have done finally how how does you get it start the pseudograph take a point any point you want there will be tree containing that that tree i will take the maximal one which will have all the vertices in it all the vertices of x are there since it is a tree i can collapse it okay then g by t g to g by t is homotopic balance what will happen to g by t there will be all those edges which are not in t they would become circles every edge which is in t has become a single point at that single point i will have some vertices some what some circles what are these some circles only for those edges i did say that all the edges are inside t all the vertices are inside t all those edges which are outside t they will become circles now so g by t is a wedge of circles namely one point union now circles the number of circles will be precisely equal to the number of edges which have which we have missed from t okay so this is the gist of this thing we are starting with t not equal to v not a single any single vertex take t to be the maximal okay containing all the vertices a tree containing all the vertices then x to x by t is a homotopic balance all the vertices have been identified to a single vertex because t contains all the vertices okay so all the edges in t and also have become two single point so the edges in x minus t are the ones which survive and how they survive they become circles and two end points of these edges are identified single point therefore x by t is a bouquet of circles therefore what is the conclusion you take a pseudo graph which is connected take any point in it pi 1 of the pseudo graph with any point a is a free group that is the conclusion right so let us the next theorem given a connected pseudo graph pi 1 of x is a tree a free group of sorry pi 1 of x is a free group of rank equal to the number of edges outside any maximal tree in x therefore this number is independent of what we use to by the way the maximal tree may not be unique you can think about that but the number of edges outside because of this will be the same that is the beauty okay so that is something which you are down next time we will use this one to prove a big theorem in in group theory and some more topology later on so that will be the last module in the last lecture for this theory thank you