 So, last time we introduced the notion of excision and we stated the theorem of Mayer Vitres namely excision theorem and we gave some two easy examples of excision namely the disk and the entire Rn minus 0 that is one pair, another pair was the sphere minus the north pole and the upper hemisphere. In practical situations often X1, X2 are closed of spaces and they may not satisfy that the interiors cover the wall of space. So, you cannot expect the excision theorem to be directly used. So, we have to do a little bit of circus here. So, there are a few of situations which can be taken care by the following lemma which is quite an important thing namely suppose A to X, A is a closed subsets and A to X is a cof operation the inclusion map is a cof operation. Then the quotient map Q from X to X by A induces isomorphism in singular homology, H star of XA to H star of X by A star. So, the role of putting star is as you have anticipated is same thing as taking the reduced homology here H star tilde of X by A. Taking the reduced or un-reduced on this part is does not matter. So, if you do not put that relative X by A star what is star? Star is the subset A identified a single point here is X by A. So, that point I doubt it. So, I can write it as the whole singleton A. A the entire A is a single point there. So, without the cofibration condition such a thing may not be true. So, the relative homology is not the same thing as the homology of the corresponding quotient in general. We need to recall some important facts from part one to prove this one. This is not quite an easy theorem or anything in this lemma. The first thing was which is easy and it was left as an exercise in part one. So, I will leave this also an exercise here just for your refreshing your memory you work it out as an exercise. Namely if A to X is a cofibration then C A to X union C A is a cofibration where C A denotes the cone over A. A is already subset of X only on the part of A you take a cone ok. So, X remains same not the whole of X E dot taking the cone but over A you take the cone. So, this is then you get X union C A C A container X union C A this is a cofibration. The second thing is if A is a cofibration A to X is a cofibration and if A is contractible X to X by A is a homotopy equivalence ok. So, this is theorem 3.8 part B there it was some other theorem perhaps but here I have a quotient based entire theorem I need part B of this one we will have to use here. So, here you see there is no hypothesis on A like contract that it is contractible and so on ok. Yet you can get homology isomorphism the trick is to use this fact. Now C A which is a cone over A that is contractible ok. So, we can put this exercise and this this theorem part B and get something ok. So, that is what we are going to do now so proof of the lemma 3.9 this lemma we are going to prove now ok h star q star from h star of x A to h star of x by A is a isomorphism. So, I will recall what is the the cone over A? Cone over A the quotient of A cross 0 1 where A cross 1 is identified to a single point A cross 0 is identified with a subspace of A as it is A cross 0 goes to A ok. By theorem 3.8 B ok this one combined with exercise 3.2 ok what we get we get that x union C A x union C A mod out by C A C A collapse it now to a single point this is a amount of equality because C A is contractible and C A inclusion to x union C A is a co-fibration ok. So, this when you collapse the contractible part this becomes a homotubic value ok. Next to apply Mary Mitre sequence now what you want to do is take x 1 as x union the half part of the cone namely 0 to half the rest of them you take x 2 equal to C A now C A and x 1 C A is x 2 x 1 and x 2 will be the whole of x union C A ok and their interiors cover the whole thing. If you take the interior of C A inside x union C A it will be the open cone the image of A cross 0 open 1 all right and the interior of x union A x x union A cross 0 half will be just this half open. So, they cover the whole space so we are fine but there is actually a strong deformation interaction x 1 to x ok sorry what I have here is yeah x 1 to x x c you have cone over A but you have not taken the full cone you have taken only half half part of it the the the apex of the cone is left out. So, you can push the whole thing back to x so x 1 to x there is a deformation retract which is identity on x 1 intersection x 2 or x 1 intersection goes inside A. So, the deformation retraction is simultaneous x 1 instead of x 2 would go inside A whereas x 1 itself is pushed to into the whole of x. So, there is a deformation retract of the pairs ok. Therefore, the inclusion map here from S star of x A to S star of x 1 comma x 1 intersection this is an isomorphism ok. So, step by step we have to build up this isomorphism namely final isomorphism that the q star has to be built up step by step. So, now, with all these background let us consider this commutative diagram the top arrow here is what we want to prove that it is an isomorphism this just now we proved that S star of x A to H star of x 1 x 1 intersection x 2 is alpha is an isomorphism this is what just we proved now here ok. This is actually a homeomorphism start with x A is a closed surface you collapse it to a single point is the same thing as putting a cone and then taking the whole cone to a single point is same as just doing this one the inclusion map itself gives you a homeomorphism here of the coefficients ok. So, this is very easy to verify that this is homeomorphism all right. So, this is an isomorphism in homology this is an isomorphism in homology. So, now you can come to the second row here x 1 comma x 1 intersection x 2 x union c A to c A this is the inclusion map this is by accession. So, this is where we have to show that eta is an isomorphism by accession axiom ok. Now, what is this one beta x union c A point star here and c A the whole thing is taken the c A is a contractible space. Therefore, if you use the homology exact sequence of the pairs alternatively you will get you know c A s are all contractible 0 maps you will get. Therefore, this is isomorphism this one homology exact sequence of the pair we would give you this one in the last day you have to use this star till then you do not have to use that for the last part you will get only that the reduced homology. What is this one x star x union c A star to x union c A by x A this is a quotient map right you have collapse c A we have verified that if this is homotopic equivalence is part of that what we have seen because c A is c A to x union c A is co-fibration and c A is contractible. So, this is homotopic equivalence. So, this this p star induces an isomorphism if this is an isomorphism this is an isomorphism p prime star whatever map here this again a quotient map in some sense okay c A has been collapsed to a single point this is also an isomorphism. So, this is an isomorphism this is an isomorphism this is isomorphism therefore this is an isomorphism q star. So, it takes so much of efforts to show that the q star from x A to x by A is an isomorphism accession is used co-fibration is used and that tricky this this theorem is used when you have contractible then x x by A homotopic equivalence okay provided A to x is a co-fibration okay. So, what finally we have is a ready made thing now namely sorry this this lemma namely whenever A to x is a co-fibration then you can collapse A x to x by A okay that quotient map from x A relative to x by A star this will be a isomorphism in homology. Now, let us see how we can proceed further. Now, come back to the situation of accession x is A in and B where A and B are closed subspaces if the inclusion map one of them A to x or B to x is a co-fibration then A B is an x is a co-capal do you understand how now we are going to we do not know whether x whether interior of A and interior of B cover x if that is the case then you do not need this. So, I have already illustrated this situation namely when you have written sphere as the union of upper hemisphere and lower hemisphere if you take the interior they will not cover okay. So, there are many other situations like this. So, an alternative approach is that their co-subspaces the inclusion map of one of them is a co-fibration okay you must remember that co-fibration implies that the subspace A is a strong differentiated track of a small neighborhood of A inside x. So, that is the key here which works because of because of this one. But we are going to do it in a slightly different fashion we shall prove condition E of lemma 3.8 remember there are 5 different conditions for for the excision okay. So, you can use one of them any one of them. So, condition E you will prove this is same as showing that h star of B comma B intersection A h star of x A is an isomorphism. So, this is what we will show under this this assumption okay not that A to A union B is a co-fibration implies B intersection A to B is a co-fibration okay this similar to the exercise here this exercise A to X is a co-fibration the cones are co-fibration which is similar to that okay may be easier than that. So, we have A to A union B is a co-fibration B to A union B may not be but B intersection A to B is a co-fibration. We now consider the quotient map X to X by A we want to use this lemma okay since B is a closer space of X it follows that the restriction map B to X by A which is B you can take this as because of the excision part okay B modulo B intersection A you can restrict the whole map instead of X take B itself so B and then its image B modulo B intersection A because what what you collapse is only B B intersection A part this is a quotient map okay therefore from the above lemma both Q and Q prime induce isomorphisms in the relative homology okay the isomorphism X X so B to A sorry B to B intersection A and X to X intersection A X A to X by this one they are isomorphisms okay B to B intersection A B comma B intersection A to to the quotient namely B by B intersection A you can go but that is same thing as X by A but that is given isomorphism to H star of X A so you have a diagram like this and come back so this is an isomorphism okay from B comma B intersection A you go to B modulo B intersection A but that is same thing as X by A and then you go back by H star of X A that is an isomorphism Q and Q prime isomorphism and bottom they are equal so the top thing here is an isomorphism so if one of them is a co-fibration then we have A union B A comma B is an excessive couple so here is a corollary take a pair X 1 X 2 of close subspaces of a space okay this is an X equal couple for singular homology under the following conditions either A or B whatever X I X I X 1 intersection X 2 is a relative CW complex X 1 comma X 1 intersection X 2 is a relative CW complex or similarly X 2 comma X they must be related to the complex then X 1 X 2 will be excessive couple means you can go to X 1 union X 2 okay similarly the other one is now simplicial complexes X I is modulus of K I where K I are simplicial sub complexes of a simplicial complex K with K go to union of suppose you have two sub complexes of a simplicial complex such that the union is all of K you have proved that inclusion map of any sub complex whether it is a simplicial sub complex or a CW sub complex the inclusion map is a co-fibration and of course sub complexes are always closed so this this corollary is direct corollary has as its name says it from this theorem okay they are co-fibrations therefore so this is a very very useful thing you take a sub complex okay so there is no way to get open subsets and so on you don't have to worry about that part errors we have done that kind of things also neighborhood retracts have been proved but co-fibration is you have to remember that is much stronger than anything else the close is you have close subsets of this true by the way this was the way one of them we who I think it was when campan he proved the his own version of of what is called as when compound theorem for fundamental group for for sub complexes for opens of sets it was the cipher who has proved that okay and they didn't know each other it is not a joint work in europe they they will call it cipher's theorem in america they called van compa's theorem okay so all that you have to use is 1.13 and 39 statement b is special case of a okay statement b a special case of a because they are sub sub simple shell sub complexes are simple shell complexes are simple c w complexes and simple shell sub complexes are c w sub complex for a part you have to use this theorem nine and then co-fibration 1.13 it just says that they are co-fibration and close subspace of course okay so finally we have produced we have we'll produce this ready made long exact sequence due to island work strain rod out of this theorem okay so this is just another step the it's not reaper the just putting it in this fashion it helps to helps to compute things very much so we shall concentrate on putting the excision theorem to good use by first deriving an important ready to use result so take start with x1 and x2 in excessive couple whether they are open or co-fibration somewhere excessive couple okay in the long exact sequence of homology groups namely of h star of x1 plus x2 h star of x1 direct sum h star of x2 and h star of the intersection this h star of s1 plus x2 can be replaced by h star of x1 union x2 that is the gist of the excessive couple the statement excessive couple so what we obtain is hi plus one of the union instead of hi plus one of sx1 plus sx2 you just put union of s2 then there is this correcting homomorphism to hi of x1 intersection x2 okay so one one index lower then it will come continue hi of this one this i star is the inclusion remember x1 x2 h1 x2 what is this one this is inclusion comma inclusion i star then j star is inclusion a minus a1 minus a2 so that is what you have to take for x1 union x2 so that is the way we have defined this exact sequence here and then again there will be a jump to hi minus one and so on to go on till we hit h0 of x1 in the last three terms are here h1 of x1 union x2 delta delta is the connecting homomorphism goes to h0 of intersection to h0 of the the direct sums then h0 of the union so this is what you have to use now this is called mayor vitreze sequence so so in honor of mayor vitreze who approved this one island works in rod i have called this one mayor vitreze okay so i have given you the formula here what what we call i star of z i star of z minus i star of z2 i have deliberately changed it here you can take plus here and minus here no problem at all okay you can minus here if you put you have to take plus here both the conventions are existing okay so the exactness if you take plus plus it will not be exact or minus minus if not that's all you have to remember okay so where what are these i1 is inclusion inclusion map i2 is inclusion here j1 is also inclusion and j2 also including correspondingly okay so mayor vitreze sequence is one of the major ready to use tool in computing homology it's a simple consequence of fair vitreze sequence we are going to prove this one but that way we'll do it the next time it's already a time now we will stop here