 Having introduced the singular homology verified a number of its interesting properties and proved that computed the homology of a single point and proved that the homology, the singular homology is a direct sum of the single homology of its path components etc. Yet at this stage we still do not have a powerful tool to compute the homology. So now we should discuss one of the most important concept namely excision which will yield a powerful tool. You can say this is called Mary Wittere's principle in real terms which will give you a long homology exact sequence and that will help little more, much more than whatever we have done so far. So let us start with something namely start with two open start with any two subspace is subject of a logical space X okay and denote the inclusion maps by eta i. They will induce inclusion maps again from s of xi to s of x singular chains okay. Now consider the chain map from the direct sum to sx s dot of x namely a comma a1 comma a2 it should be a1 comma a2 so the direct sum eta1 comma minus eta2 defined on s dot x1 direct sum s dot x2 taking a1 comma a2 to a1 minus a2 take this map this will be clearly a chain map the linear combination of chain maps is a chain map and the inclusion maps are chain maps okay. What is the kernel of this map remember s dot x, s dot x1, s dot x2 they are all free abelian groups so the direct sum is a free abelian group okay. So an element is zero means all the coefficients of each singular simple x is zero that is the meaning of that is the only way alright. So kernel means what now if a1 and a2 are chains in s1 of x and s2 of x then corresponding coefficients of the singular chains must be identical for a1 and a2 that is the meaning of a1 minus a2 is zero okay this just means that a1 is identically equal to a2 in sx but they are inside one is inside x1 another inside x2 okay therefore a1 equal to a2 must be inside s of x1 intersection so you can easily check that s of s dot of x intersection x2 is actually the kernel the entire kernel now what is the image of this one that is easy to check it is just like two you know two element one element from here one element from here whether you take the sum or the difference is same thing sum of elements here okay so that will be a sub module here right and that sub module is usually written as the sum just without the direct sum without the direct sum notation ordinary sum so s dot of x1 plus s dot of x2 this is the image okay therefore we have a short exact sequence s of x1 intersection x2 the direct sum and the ordinary sum to 0 this is the image so this is this is this is exact because I can put 0 here that is a kernel of this map namely this map is a1 comma a2 equal to a1 minus 8 okay so you know exactly what map is though I have not written here so this will then give a exact sequence of long exact sequence of homology modules from which we hope and of course we have experienced this namely there will be information on h star of the the central chain complex here sorry h star of the sum here in terms of h star of the direct sum means h star of direct sum means direct sum of itself so this we know if you know h star of x1 and h star of x2 and something we know about h star of the intersection the two subspaces we should know what is happening there we should know what is happening in the intersection then we may be able to say something about the union so this theme was there exactly the same theme in mayor in when compounds stay around about fundamental okay so this through this this principle is near on here goes under the name may I raise two Austrian mathematicians of early last century so the question is can we replace h star of sx1 plus sx2 with h star of x why we hope to do that because topologically x1 union x2 is the whole of x that is the starting point but can you do this algebra h star of s1 x2 replace it by this one are they equal this is a this is some kind of a sub of this one though it is not a sub only the chain complex it is above the corresponding s of x okay so is there some kind of isomorphism so this is the question okay we shall now proceed towards some affirmative answers to this question the first technical part for the first thing was the we have we are making a lemma here the positive answer finally beautiful answer came by may wait a sequence that will that will that is our aim okay so this algebraic preliminary for that how to deal with this one namely when can we replace this by h star of x itself so as usual take x equal to a union we have changed the notation here instead of x1 x2 deliberately for because this is applicable in a larger context then the following statements are equivalent the sum of s of a and s of b to s of x induces isomorphism in homology so this is a very clear statement namely the inclusion induced homomorphism this inclusion into homomorphism must be itself isomorphism that is what is demanded okay the second thing is another condition sa plus sb go module sb on both sides that will induce an isomorphism on the homology past the homology then these are chain complex in these are chain maps when you pass the homology it must be isomorphism the third condition is you know symmetric in a and b after all sa sb instead of go modulo sb on both sides go module sa these two are obviously just symmetric condition because a and b there are no other conditions here x is a union b that is all the fourth condition is slightly different now just take sa and s of a intersection b to sx modulo sb okay so you are going down sb sb is not here is sa so I have to take s of a intersection b here there is a obviously there is a map here homomorphism at the chain complex level this induces isomorphism in homology is what the fourth fifth one is similar now instead of sa i book sb sb to s of a intersection b to sx by sa so this is just usual kind of you know what you can term as noiter's isomorphism theorem first isomorphism second isomorphism and so on it is of that nature so let us quickly go through this one equivalence of a and b is a consequence of five lemma see we have opportunity to use this five lemmas which we have introduced just a couple of days back sa plus sb to sx there is an inclusion map this is the quotient of sa plus sb by sb and that is the kernel so this is an exact sequence sa plus sb when you quotient over sb you can quotient out the same thing by sb on this side also that will give you the same map inclusion gives rise to this map you can call this inclusion okay so this is inclusion induced map this is the restriction map only to sb and this identity map here the identity map okay gives an isomorphism therefore this is an isomorphism if it only if this is an isomorphism the five lemmas says that if this is actually much simpler than five lemmas because both these two are zeros here okay so this is an isomorphism and this is an isomorphism but if this is an isomorphism and this is identity isomorphism it's clearly is an isomorphism is obvious yeah so once you have this okay when you go to homology level this is an isomorphism would imply this is an isomorphism because now you have to use five lemmas when you let take the ladder so hi hi hi hi minus one hi minus one hi minus one so alternatively you will get this and these are isomorphism again they will repeat here so the middle one will be isomorphism so that is the five lemmas you have to apply to the long exact sequence above and below okay b and d let us look at b and d s of a plus s of b modulus b sb I have taken modulo sb here and I am taking modulo s intersection b so this is this is actually like the isomorphism theta okay b and d follow from the commutative diagram below in which the horizontal arrow is the isomorphism theorem given by Noether isomorphism theorem s a to s intersection b s a plus sb by sb this is noether isomorphism theorem you say this is an isomorphism if and only if this is an isomorphism so b implies b okay this this is b this is d okay interchanging a and b you will get c implies e a and b are equivalent okay by five by the five lemma and the long exact sequences b and d are equivalent because of noether isomorphism theorem interchanging a and b what do you get c and e are equivalent alright so once your b is there just interchange a and b roll away and b a will give you c also a and c are equivalent c and e are equivalent so all the five statements are equivalent equivalent to each other okay so now we shall make a definition take x at a political space and a and b be any two subspaces okay just for the sake of definiteness you can say x is a union b which is not necessary if you want just work with a union b let x be there but we want to work with a union b so I have cut down x itself which is the same thing as saying this is just to set the other thing x minus b is equal to a minus a intersection b okay we then say the inclusion map a to a comma a intersection b to x b is an excision map so I could have directly said this one namely this is a 1 a 2 to x n by the difference x minus b here saturated different is equal to the difference here and they are inclusion maps so say thing is excision map so this is purely saturated condition okay in the relative thing whatever extra space comes must be the same in both the cases and this whole thing is an inclusion map a pair a b of the political subspaces of x is called excessive couple so this definition is something different okay this is not trivial thing for singular homology excessive couple for singular homology okay excessive couple for some other homology or something else there may be different excessive couple excessive couple for singular homology so we are defining like this if the inclusion map s of a plus s of b to s of a union b at the singular chain complex level this must induce an isomorphism in homology that is the first condition of this lemma right s of a plus b to s of x x is a union b okay which is the same thing as or any of these five conditions so we are taking conveniently the first condition and that is what our aim was after all okay so we take that condition and make it into a definition so wherever we if this inclusion map is an isomorphism in homology or if any one of the other four equivalent conditions of the lemma is satisfied condition D of this lemma immediately gives us you can write it as a theorem inclusion map a 8 section b to a union b b this is always an excision map this is an excessive couple okay inclusion map of an excessive couple a b a b is an adhesive couple means what h h star of s a plus s b to h star of a b is an isomorphism so that is the hypothesis if that is the case okay I can write it as induce an isomorphism this is if and only if this is induce an isomorphism okay then h star of a a intersection b h star of a union b this is an isomorphism that is condition D here let us look at this one a plus b modulo h star of a is relative to a union b modulo a to sorry this is see this is a a intersection b to x p okay yeah so a a intersection b to x which is a union b to b okay so this is just a restatement of lemma the above lemma here and the definition so you you should remember that if I give this condition then this is true if I give this condition that is that is all you have to note that a pair a b is excessive or not inclusion map is always an excision map an excessive couple there must be some homology condition okay so this excision map is just to tell you that a union b the biggest thing minus b is a minus a intersection okay so instead of saying all that you take a pair of topological spaces and they then say this is an excision map that's all so excision theorem is precisely due to mayer witteries the present form is due to eilenberg and steenrod okay so between mayer witteries sequence invention of this and proper modern formulation by eilenberg it took almost 20-25 years okay so mayer witteries did not prove it in this way if x is the union of x1 and x2 in such a way that if you just take interior of x1 and interior of x2 that must cover the whole of x then x1 x2 is an excessive couple for singular homology so this is a statement which just means now that I can replace s of x1 plus s of x2 by s of x itself replacing means what passing to the homology after passing to homology so this condition looks strange but this suppose x1 and x2 are open subsets then interior of x1 is x1 itself interior of x2 is x2 okay this is the easiest way how the excision theorem can be applied whenever two subsets are there which are open and they cover the whole space x then you are in good shape then you can compute the homology of the whole space in some sense by knowing the homology of x1 homology of x2 and the homology of the intersection how you can do that is the long homology exact sequence so that part putting it into a long homology exact sequence is island work steenrods contribution okay the principle and the details of this theorem for example like this were there in work of mayor and writer is all right we shall postpone the proof of this theorem just like we have postponed the proof of homotopy axiom homotopy invariance of the homology okay for the present we shall only make a remark what kind of proof is there okay given a singular n simple x sigma which is which may not be inside x1 or may not be inside x2 okay so it is in the union of x1 x2 assume for the time being that x1 and x2 are themselves open okay then the typical thing to do just like in all analysis work you have to cut down cut down the simplex into finer pieces such that each part is inside x1 or x2 so one subdivides delta n in such a way that the image of each sub simplex of this subdivision is contained inside x1 or x2 one thinks of original singular simplex sigma as an appropriate sum of these little pieces it is just like integration theory if you have a interval of definition you can cut it down into two intervals integral on the entire thing is same thing as integral over the first part and then integral over second part the similar things are there for area and all that all that you need is that integration should not overlap then you have so this is the homology this is the motivation and this is how all this homology was developed through motivation and guidance from what happens in the integration theory in the analysis okay you cut it down into little pieces strictly speaking though they are different chains and most of the effort is to show that they represent the same element in homology groups so some somehow you have to show that there is no integration nothing so we have a different kind of definitions and we have to show that in homology they represent the same element in analysis it is the same thing as the integration is the same of course all this thing has to be done in a canonical fashion not exactly depending upon the actual nature of x1 and x2 once you have done the whole thing it should be true for when you are writing some other y1 union y2 also very y1 and y2 are open that such and so on okay now let us stop there first for more explanations etc for actual proof we will we have to wait but now let me give an example here is a few examples of x2 couples okay which occur in practice yes the first one is take the disk dn closed disk or open disk let us take open disk dn okay rn minus 0 that is also an open set the union with the whole of rn right I want to say this is an x2 couple because dn is open substrate and rn minus 0 is also an open substrate over but now the conclusion is that dn dn minus 0 which is the intersection of dn and rn minus 0 right to rn rn minus 0 this inclusion map induces isomorphisms in the singular homology groups and this is one of the very useful thing okay this is starting point of our discussion in in manifold theory and so on we will see that okay the second one take the sphere sn let n and s denote the north pole and south pole okay I am just looking at the north pole now and let u bar denote the closed upper hemisphere upper hemisphere okay then sn minus n sn minus 1 is what the top point namely the north pole is deleted that is all and u bar this is an x2 couple and hence the inclusion induced map should be an isomorphism in the homology what is that u dot u bar u bar minus n which is which is what you get when you take sn minus n intersection u bar so sn modulo sn minus singleton n so everything is everything is here in that one except the one point here this is anything u bar u bar minus n I have deliberately taken u bar here u bar is not an open set but interior of u bar and sn minus n they are open set and they cover that's enough so u bar is not open here yet see here also I told you dn could have been a closed discord okay the interior of dn and rn minus 0 they are open and then they cover it so that is enough so that was this kind of statement here in this area interior of x1 and interior of x2 they cover the whole space okay so these two things we will use again that they are excessive couple so there is an isomorphism like this in in the homology all right for more examples we will have to wait so today we will stop here thank you