 So, we are in the process of establishing classification of surfaces and last time we began with the canonical polygons and carried out two of the steps in the reduction process. One was to get pairs of type 1 which are adjacent eliminated. The second one was to see that there is only one vertex class. All the vertices on the canonical polygon should be identified at a single point. So, this we have achieved. Now, the next step, next there are three more steps I will say. So, step 3 here now to make any pair of edges of type 2 adjacent. So, that means that we are assuming there is a pair of adjacent type 2. There is a pair of type 2. It may not be adjacent. It is adjacent, there is nothing to do. If it is not adjacent, what is the meaning of that? There will be an edge A here and then there will be some sequence and then the edge A occurs again A followed by A and capital A followed by A and again another gap here. If B or A is empty, then A and this A will be adjacent. So, essentially you must assume that B and A are non-empty. So, this may be any length, this may be any length. So, what I do? I take the tip of first A, the tip of the second A. So, it is a convex polygon. So, you can join them by a straight line to take that one. So, name it as C and I am going to cut it along that. So, this polygon will be cut along this line. The top portion here is beta, alpha and beta. So, two parts. Beta is kept as it is. The top portion is turned around. If you have turned it around through 5.5, 4. So, this C which is directed like this is now turning around to come directed like this. That is not the object. The object is to bring this A which is downward arrow to match this A which is upward arrow. So, match them and identify. Identify these two areas. So, these parts of the polygon are identified now. To another thing, which may not be a convex polygon. That is not a problem plus. All that we need is the sequence and then we rewrite the convex polygon. That is what we do. So, what has happened? A has disappeared. A needs to have disappeared but a new edge has come which is of type C under C. So, this is the step 3. So, we have brought this together. Now, again we look at the rest of the sequence whether there is any type 2 which is not adjacent. Again we do the same thing. When we do that, this whole thing will be in one part either in A part or in B part of this one. So, whatever is I have, we have brought it that will not get disturbed. So, that is our point here. So, this way repeating this process again and again all the edges of type 2 will come adjacent. They may not be together. All of them may not, each of them will be adjacent. C C, then some gap, B B, some gap, A A, like that. So, that is the idea of this step 3 and I have explained how to do that. So, this is what we do. What is the corresponding relation code? The relation code is look at this one A capital A, A capital B. That is the sequence gets converted into, you can start from wherever you like C, C, B, A inverse, the other way around. This B is A, that is precisely what we have done. See A capital A, A, B is C, C, B, A inverse. A inverse means if A is say A 1, A 2, A n, then you have to write it as A n inverse, A 2 inverse, A n inverse, A n minus 1, you have to say A n inverse. So, that is the inverse sequence. Exactly the opposite direction you have taken. Note that such an operation does not disturb some of the other pairs B, B, C, C, etc., which is already present. Therefore, repeated application of this will make all the pairs of type 2 adjacent. Of course, it may be necessary to perform step 1 again, namely a new A A inverse, the type 1 may might have appeared which are adjacent. You just do the same thing, process of folding it up and eliminating them. On the other hand, what happens is the number of vertex class will never be more. This is what you have to observe. Number of vertex class will never be more. Look at here, for example. So, all these vertices were in the same class, one single class. What I have done? I will cut it here. So, this will take end of this one and beginning of A, wherein maybe they may have gone into different class. It will not happen because this beginning of A is beginning of C. Beginning of C was the point here. So, that is not in a different class. Now, the other one is end of A here. They have come together already. So, only one vertex class will be there in the new thing also. Check that there is only one vertex class. So, that is what they have to satisfy. At this stage, so, stage step 3 is also done. So, let us take a look what is happening. At this stage, there may be no pair of adjacent of type 1 left. Then what is the meaning of this? Everything is A, A, B, B, C, C and so on. That means that we have the list 3, A1, A1, A2, A2, A3. This is the list 3. So, this is a canonical, this is a normal form. So, they are done. So, what is that, what is the other case left out? The other case is precisely there is a, there is at least one pair of type 1 and they are not adjacent. So, what to do with that one? That is the point. So, this is the step 4 handling a pair of edges of type 1. Suppose there is at least one pair of edges of type 1, say labeled A and then other one somewhere A inverse. Definitely, if it is adjacent, we have already eliminated it. So, it is not adjacent. But A, some capital A, then A inverse, some capital B and then that. So, A and B, A capital B may not be, should not be, this is what the meaning of this. So, P has the form as shown in figure 36. A something non-empty, A inverse and something non-empty. All right. If P or A is empty, then I can immediately eliminate it by step 1. So, that is not a point. So, we can, if no edge in A is identified any edge in P, now let me examine what happens. So, there must be some identification. Each edge here is identified some other edge. But if suppose edges inside A are identified within themselves, namely inside A itself, B inside B itself, then what happens? Look at this endpoint here, the endpoint here. That goes to one single class which is coming from A. This endpoint, the beginning and beginning of A, they will be in a different class because A, when you identify A, they are not getting identified. They will be in a coming out of B. There is no identification from A to B. That means that there will be two vertex classes. That is not allowed now because we have already made one vertex class. Therefore, some edge here must have identified some edge here which will make finally all the vertex class, single vertex class, all the vertex is being identified. So, one of the edges here must be identified one of the edges here. Therefore, what we get is a better idea namely this figure one here, A, A find inside this upper part there will be some B which will be identified with a B here. So, A, B little B, capital B, capital A inverse this one is C and this is again B and B inverse. B is this way, B inverse this way, D and A. It cannot be B because B and B are all adjacent. So, A, A, B, B both of them are pairs of type one. Such a thing is called interlocked pair of type one. Interlocked means what they are not together. They are A to A that is something in between B and B something in between A something in between and so on. If these are empty then there is nothing to prove. So, you better assume they are non-empty otherwise whatever you are doing is already somewhat useless thing whatever it is already achieved what you want to do. So, let us see what we want to do now. So, this time so cut top of A which is top of two line from top of A to top of A. Earlier we did top of A to top here all time right top of A, top of A cut off, cut off this part all right. Whatever remains is drawn here is alpha this beta will be now brought back so that B and B are getting identified. You just slide it down and put it here identify the edges B and B. The new sequence will be what now we just read from wherever you take. So, it starts B and then C is the new edge then A. So, this B is inside now. So, this B you pick up from here D, A this is a new edge C, A and then C that is a new sequence. You redraw it as a convex polygon here in diagram 3. Same thing I have C I have now read A to D this is a this is a black box right we do not know what sequence it is some a sequence nonempty sequence. So, you write it as A to D A D. Okay this is A C A this will be what this is C B okay now we have this C and C which is again type 1 C C inverse right. So, from top to top I mean end point to end point draw line D cut it here cut it here okay now bring this A and A together okay that means what this portion alpha portion has to be brought here and turn it around okay so that A and A coincide. So, new phenomena here what it is C okay C D inverse C inverse A D C B D okay. So, you can redraw it and then you see that D C D inverse C inverse. So, this whole thing is now together and then it is really something D C D inverse C inverse okay. So, there is no in between things here now. So, this is the best thing you can do of course you cannot bring them together D D inverse will not come together there will be C D C D inverse C inverse. So, this was precisely the sequence for the torres. So, this portion of the surface will give you a torres and then this will be connected some with something more. So, we will transform the polygon so that the two interlocker pairs of edges become consecutive okay. So, the corresponding thing is starting with A A B B A inverse C B inverse D D finally got turned into this one A D B A D C B something D C D inverse C inverse. You can call it as X Y X inverse Y inverse if you want what label you do is immaterial okay. Now, look at A D C B do the same operation on this part by calling this something nice already you will never this never be disturbed. If there is a pair here which is not adjacent you perform the same operation here okay. So, keep doing this all the interlocker systems of type one they will come together okay. So, this is what we have by repeated application of this get the canonical polygon of the form A 1 B 1 A 1 inverse B 1 inverse A 2 B 2 A 2 inverse B 2 inverse and so on. And then all the other ones are type 2 C 1 C 1 C 2 C 2 C 3 C 3 C M C N if either M is 0 or N is 0 then M is 0 means it is type 3 list 3 and N is 0 means list 2 okay. We had in the classification we had three of them first list was just the A A inverse which corresponded to the sphere 2 S okay. Then list 2 was only this kind of things up to A A inverse B A inverse the list 3 was this one. This is combination this is neither in it is not in the list. So, this is not normal so we have to do one more hat trick here to make it a normal sequence okay. So, that will be the last step to be handling the case of mixed type unmixed type means what either only A A inverses are there or C 1 C 1 C 2 C 3 these are type one these are type two. If both of them are there what to do okay. So, now what we do is we concentrate on the middle portion is here A M B M A inverse B A inverse C 1 C 2 one of the type one pair here and the other one is type two of them together okay you have take you cannot take just one of them at the other one is type 2 pair. So, how what to do with this okay. You will convert this one into entirely type 2 okay that is the whole idea. Combining results of the two cut and paste done in figure we obtain the following relation okay. So, what I am calling is A here up to here A M minus 1 B M minus 1 A M inverse A minus 1 inverse B M minus 1 inverse you call it capital A then you take this part call it as A B A inverse B inverse then take the C 1 C 1 C 1 C 1 call it as just C C. So, this portion is getting converted into A into x y x z y z okay. So, let us see how this gets. So, this is what is A B A inverse B inverse okay then C C and then a big gap which I have not written down anything this capital A you can write as capital A here A B A inverse B inverse C okay look at the tip of A okay to tip of C. So, you have to make this guess where to cut and so on this is not at all obvious okay. So, there are two of them you have to break this C at this point. So, you must be really worried because you have brought this one with so much of effort now you are asking cut it okay. So, you have to do that cut it here okay and bring this part beta part is kept as it is bring this one on this side so that this A and this A are together okay this part of A and this part of these two things together identify them. So, this A disappears what is the new thing that appears this D will appear now okay D is this way and you have turned it around. So, this will become downward here okay D inverse B D C then this capital A here which was which was blank thing here okay after that what you have this part here now B inverse and C ending with D here okay rewrite this one okay just like this way now one more cut you have to do look at the starting part of B to the end of C okay B D C are there already cut it along this this is E now cutting along E okay this portion is brought back here so that the C here will match C here. So, you are actually destroying the old C the old A but you are getting some new of them okay. So, what you get is now E D inverse E again some gap okay then this is what this is B inverse D inverse B inverse okay you better start from here and keep reading and then capital A at the last okay D inverse D inverse B inverse okay then E D inverse E all right. So, we can now apply three successively to make this into A X X Y Y Z Z finally what we have what is this process we have got A into X Y X Z Y Z. So, what I am doing I have instead of writing like this this is A this is X right B inverse but you write it as X X Y okay again you are getting X here because B inverse X okay X Y X and new one Z okay so that is why this sequence is X Y X Z and then Y Z okay and then Y Z. So, this is this is not what we wanted okay it should be X X Y Y and Z but that you can always do by step three right whenever you have X Y X so this X and X can be brought together okay and anyway there are X X Y Y Z Z three of them all the three of them can be brought together X X Y Y Z Z Y step three okay so you have to repeat step three here again to make it what we want. So, what has happened finally is the following you started with one one A B A B inverse of this step and one of this one okay rest of this part and this part you could have taken all this part in single here and assumed that this is the at the end by cyclically permitting okay then that all that A remains as it is and this part becomes X X Y Y Z Z so it becomes three pairs of type two okay so one plus one here one type of this one one point of this one we do three pairs of type two whatever is remaining you repeat the same process again on that all right each time you see there are plenty of type two things to mix with that one till all the these AIs and BIs have disappeared one by one if there is only one by now it would have become one here one here it would have become three C C three C one seven C two C two three three there there are two of them again again three more okay each time with mixing which is one this will convert into two C two of the sequence okay by repeating this process so in the middle you keep changing that is what I have told in middle this will be replaced by X Y X Y Y Z Z and rewrite the sequence this part again take the next one and you will make it X X Y Y Z Z and rewrite okay finally you will be like with X 1 X 1 X 2 X K Y X K X K repeated okay this will be the sequence okay actually this K will be precisely equal to whatever n was there each of this m will be get into two of them so two n plus n okay so this completes the proof of part A of the theorem 6.48 starting with a polygon the first case is A A inverse which is S 2 we do not want to do anything that is it is declared as normal otherwise what should happen it will be of the of the future two let me just go back the list two is is like a repeated torus A1 B1 A1 inverse B1 A1 A1 A1 A1 A2 A2 etc projective space repeated projective space okay the second part is A2 distinct members of this list Q surfaces which are non-homomorphic so that is the part that will okay all the surfaces you can see here is one part which is already a big thing now the list is not a redundant list there is no unnecessary elements you know duplication it is not there this is what you have to assure so that part we will prove using fundamental group it will give a very easy proof then just to justify all these homologies which will use a simple homology and give a proof we shall also give a proof which will say cellular homology okay so today we will stop here thank you