 Today's session I have called the harvest because whatever hard work we did for the past few days, we are going to repeat now. A lot of results will be proved today using whatever hard work we have done for the past few days. One of the landmark results will be the invariance of the adjunction space. The details we will see in the work follows now. So the first thing is three results which relate the weak deformation retract, the deformation retract and strong deformation retract. We know that these things are one, you know, stronger than the other, but when they can be equal, that kind of thing we want to see. So the first thing says is you take a co-fibration from A to X. A is a subspace which is closed and A to X is a co-fibration. Then the first thing is that if A is a weak deformation retract, then it is a strong deformation retract. The only part is what we need. If it is a deformation retract, it is weak deformation retract is obvious. So if A is a weak deformation retract of X, if we are down lift, it is a deformation retract of X. The next thing is namely B, if A happens to be contractible, then as if the entire subspace A can be thought of as a single point, homotopically. So the correct thing is to go modulo A, namely I identify the entire space to A and take the quotient space. The quotient map itself becomes a homotopy equivalence. So this you would expect always, but it is not always true. It is true under A to X co-fibration. So a nice space is this is true. Third one is tightly a different variation of the same phenomena. Instead of quotienting out A, you look at A as it is, but take homotopies relative to A. Then can you say that contractibility of A will imply, you know, anything which is homotopic will be relative homotopic to A. That is again not true. When you need this time, you need even stronger hypothesis, namely a point A naught, wherever you want to concentrate the whole of A should be a strong deformation retract. If A is contractible, A naught is a deformation retract of A, but may not be strong deformation retract. If it is strong deformation retract, then the inclusion map of X A naught to X A is a homotopy equivalence. So it is like all homotopy information now inside A wherever you want to control A can be obtained by just controlling point A naught. So these things are themselves very useful in homotopy theory. We are developing the homotopy theory here, you see from smaller chunks to larger chunks. So let us go through the proofs of this which are now more or less one line proofs because hard work is already done. So how to say something is a deformation retract when it is a weak deformation retract? Given a homotopy inverse R naught from X to A, remember the inclusion map is a homotopy equivalence. It is the same thing as saying that A is a weak deformation retract. So R naught A is the inverse of R naught, the homotopy inverse of R naught for the inclusion map eta. The eta is a homotopy equivalence. It is same thing as saying A is a weak deformation retract by definition. So take R naught as its inverse. Let there be a homotopy from R naught composite theta to identity because these two are homotopy because this is inverse of that. So let F be such a homotopy from A cross I to A between R naught composite theta and identity of A. With this much of thing we can get a deformation retract. The key is use that A to X is a O vibration. So go back to your definition of a homotopy extension property. Namely that is what introduction three or four. So I will reconstruct it here below. So what I have is A cross 0 to X cross 0 inclusion map. Here I have put instead of G in the original thing I have put R naught here now. This is the inverse of, homotopy inverse of eta. And F is the corresponding homotopy from A cross I to I. So we take this commutative diagram. This is automatic. That implies because this is a co-fibration that is homotopy extension property implies there is a homotopy H from X cross I to A which fits this diagram. Namely restricted to A cross I it is F. Restricted to X cross 0 it is R naught. So that is what I am going to use now. Hello sir. Would you explain the diagram one more time to say why it is homotopy extension property it shows? A is a homotopy extension property is the assumption beta. You see that this is co-fibration is the blanket assumption on the entire theorem. Okay. Co-fibration means what? Homotopy extension property is there for every space and every map and so on. This is this you can choose as soon as you choose this such that this diagram commutative there is a map here that is the conclusion. So F is extended to H that is a homotopy extension property. Okay so I am going to use that. So we have H from X cross I to A such that H restricted to this A cross I through eta cross identity is F restricted to X cross 0 it is R naught. But then H defines homotopy between R naught and identity of A because H X 0 is R naught. H of X 1 will be F of X 1 that is will be R naught. Okay. So you have this also implies that if you take eta composite R naught okay eta composite R 1 we know is homotopy to eta composite R naught. But eta composite R naught is eta composite identity is identity of X. So we have got that R naught is a deformation retract. Okay. So to beat deformation retract became a deformation retract as soon as A is a A to X is a co-fibration. All right. So that is the key here. Similar thing will be done for the second part also. What is the second part here? If A is contractable, quotient map X to X by A will be a homotopy equivalence. Okay. So let us go through this one. F is a map from A cross I to A is homotopy of the identity map with a constant map A naught. All point A going to A naught, A naught is a point of A. That is the meaning of A is contractable. Identity map of A is homotopy of the constant map. This time instead of taking A here, I take this space to be X here. Actually in the original general for any space Y, any map G like this and so on is there. So instead of that Y, I am putting this time here, I am putting Y to X itself. And this map will be taken to be identity map. So that is what I am talking bigger for. But I am reconstructing it here. Again you get a capital H here to fit the diagram. Here again I am using that A is a co-fibration. So this time it is this map. Here it became X and I. Instead of R, it became identity map. Restricted to A, it is homotopy to a constant map. F, A comma 1 will be a constant map here. So that is the homotopy, F. This homotopy data, this data co-fibration for co-fibration gives rise to this conclusion. Namely there is an H here which fits in the diagram. So what is the meaning of that now? Restricted to H restricted to 0 or H restricted to X cross 0 is identity because it is identity. Difficulted to F cross I, it is A cross I, it is F. So what are the properties of this H? H1 of XA which is H of X1, which is X1 here has the property that F of, this is F of A comma 1, which is a constant map A0. Which means all of the points, that is if A belongs to A, then they are going to a single point. Therefore this map H1 factors down through the map Q, constant map X, X to X by A. It gives you a map from X by A to X. So that is what I am calling as a G1. Because under this H1, all the points of A are going to a single point which is the single point identity X by A or this is one single class. So this is a G1 is X by 8X because H1 is X to X. Which is just, I am writing what is the property of G1? G1 is H1 is equal to G1 composite Q. Also the homotopy 2 cross H, now if you take, first take now see H is from X cross I to X. Then follow it by Q. That means I have quotient it out by all the points here. This also factors down to define a homotopy from X by A cross I to X by A. The points of A which have been identified here, they are identified here also. That is what you have is Q cross identity composite with G is the same thing as Q cross H. So what is this map G? G is again now identity map X cross A to X to X becomes identity map from X by A to X by A. That is the starting point. And the end is Q cross G1, Q composite G1. So it is a homotopy between identity and Q composite. Which means Q has G1 as its right inverse, a homotopy inverse, right? And that is enough to conclude that this is a homotopy 2 balance. The G1 composite Q is H1, which is homotopy 3. The last one is a special case. What is the definition of last one? If you have instead of arbitrary contractability, in arbitrary contractability you have just a homotopy from identity to the constant map. It is strong deformation rectangle means this homotopy is relative to A naught. The same argument will give you same argument, same G and so on. All these things work in the same case. But now I have the homotopy being constant on A naught. So that is extra thing here. So that will tell you that H1 is a relative homotopy. That means the homotopy from H1 to the inverse is constant on A naught. That is all. Whatever homotopy you have got there is a relative homotopy on A naught. It is the A naught. Which will conclude that XA to XA, you don't have to actually, you don't have to, in this case you don't have to go down to X by A at all. As soon as I, you know, halfway here, in this here, before going down, as soon as you got this one, this H1 is homotopy inverse relative to A. So stronger hypothesis gives you a easier conclusion, a stronger conclusion. Next, so now you know, many books just say contractual therefore you can collapse it and so on without giving any proof. I have given you a proof and the proof is correct only, only A to X is a co-fibration. In fact, we will have easy examples when this is not a co-fibration, this is not a fact. You can't just take X by A and say that it is a homotopic warrant or original X. So, next result, take a map from X to Y. It is a homotopic balance if filled only if X is a deformation retract of MF. Remember, MF is the quotient of X cross I and disjoint union Y by the identification of X comma 1 is identified with its image Fx inside Y. We also knew that Y itself is a strong deformation retract of MF. Now this result says that if F is a homotopic balance then X is a deformation retract of MF. Suppose we assume X is a deformation retract of MF. X to MF you have deformation in a homotopic balance. MF to Y you have a homotopic balance. If the composite will be homotopic balance then that is precisely F. Inclusion map followed by whatever you have what F at is identity is equal to F. We have seen that right. So, one way is clear that it is a deformation retract then F must be a homotopic balance. The converse, how do you prove the converse? I have already given you here the proof of one way. Namely, inclusion map is a homotopic balance. It is the same thing as it is a deformation retract. In inclusion map followed by F at is F. What is F at? F at is a strong deformation retract from retraction from MF to Y. So, let us say a composite will be homotopic balance. So, let us prove the converse. If F is a homotopic balance then again from this part of the theorem A. In inclusion map 4.3 will be homotopic balance. Because F is I composite what? F at the other way around. F at composite I right whatever. So, if two of them are homotopic balance the third one is a homotopic balance. I think this is one of the exercises. The very beginning. Hopefully you have done that exercise by now. Now you combine part of D in 4.3 with theorem 4.4 one which we have just seen to conclude that extended deformation retract of MF. Inclusion map is a homotopic balance. What are these? This just means that inclusion map of X in MF is a co-fibration. This was one of the one of the results part D of theorem 4.3. And the previous theorem says if it is a co-fibration and a weak deformation retract then it is a deformation retract. Weak deformation retract same thing as inclusion map is a homotopic balance. I repeat once inclusion map is homotopic balance and it is a co-fibration then it is a deformation retract. So, this was the theorem that we just proved here ABC 3 parts of the theorem. So, use that. So, we get the corners here. So, corollary is proved namely if something is homotopic to balance between two spaces the same thing as saying that the domain is deformation retract of the mapping cylinder. Now, we come to another. All landmark results in homotopic theory. Two spaces are homotopic equivalent if when downlifts there is another space W which contains both of them as deformation retracts. This is not just for fun. There is no other way to show that a space two given spaces are homotopic equivalent. This theorem is used deep in differential topology namely in cobaltism theory. Lot of hypothesis is there with that you want to construct a homotopic between X and Y they are manifolds. So, do that inclusion maps you construct a cobaltism between them then you show that inclusion maps are deformation retracts homotopic equivalences plus you know that in the case of manifolds they are co-fibration outs. Therefore, because this theorem X and Y will be homotopic equivalent. So, one way is obvious but here we are insisting that you are not losing anything. You are not doing anything great. It is if and only if this will always happen. So, that is if and only if part is what you are asking. If X to Y is homotopic equivalence you take W to be the mapping cylinder and we have already seen. Okay inclusion map followed by FFAT composite inclusion map is F. Now F is homotopic equivalence FFAT is homotopic equivalence I will be homotopic equivalence. That is all. I will prove the converse here now. What is the statement here? What is the converse? Converse is obvious I have not written down here. The converse is obvious. Why? Because X to W is homotopic equivalence and Y to W is homotopic equivalence means there is a inverse map from W to Y you composite to that is all. Okay. So, this W could be anything but to get to one you have to take if this is homotopic equivalence you take W to be MF. It need not be MF always. It could be something bigger also something smaller also does not matter. If it is a property that inclusion maps are homotopic equivalence is from both X and Y then F will be homotopic equivalence that is the whole idea. The X and Y will be homotopic equivalence there is no F given you have to construct one but if F is given then you take MFS the space W that is all. Now we will get some more you know some more results take a co-vibration X to Z okay. So, I have changed the notation usually A to X I have taken now X to Z is co-vibration every time I am assuming it says close subset okay. X to Z is a close subset then for every continuous map F from X to Y the inclusion map of Y into the adjunction space okay is a co-vibration what is adjunction space of F F is from X to Y so it is Z disjoint union Y points of X are identified with points of Y to F F and X of X and F of X are identified that is the adjunction space right the adjunction space of F always contains Y as a subspace that may not be subspace X may not be subspace because F may not be injecting right. Y to F it is inclusion map that becomes a co-vibration under X to Z is co-vibration okay this is like now suppose X is X to Y is an inclusion map okay then extending the co-vibration here from X to Y okay so X to Z was a co-vibration Y to A F the larger one gets becomes a co-vibration there is more general than that this X need not be subspace of Y it is for any map any map so that is the importance of homotopy theory okay so again co-vibration what we have to do you can't go on verifying it for homotopy exchange property for all sort of spaces you have a ready made proposition there 4.1 namely show that A cross I union X cross 0 is a retract of X cross I in this case what you have to show A F cross I sorry Y cross I union A F cross 0 is a retract of A F cross I that is what you have to show so it is very easy from corresponding statement corresponding hypothesis that X to co-vibration what I get Z cross I sorry Z cross 0 union X cross I is a retract of Z cross I okay so this proposition says first we get a deformation okay retract we just retract is good enough then it will become a deformation retract is what we have seen already so just retract is enough retract X Z cross I to Z cross 0 in X cross I retract means what on this subspace it is identity now you just define Z disjoint union Y cross I remember the adjunction space was defined as quotient of Z disjoint union Y right so if you want to construct some map you go back to the mother space here with certain extra property you have to define a map so that it goes down to the quotient space so here I am using the fact that the product I am taking is a compact space means a interval therefore whether you take the first the quotient here then take product or first take the product and then quotient they are same this was one of the fundamental result we have proved and we are using it several times okay so I am defining it on Z cross Z disjoint union Y to I Z cross 0 union X cross I disjoint union Y cross I okay here later on points of X will be identified with FX same thing will be I done here the second factor I there is no problem there this must be the Euler front I here okay so R restricted to Z cross I the Z disjoint in Y cross I Z cross I disjoint in Y cross I okay you take it as your R here this readymade function is here on Y cross I you take the identity over after that you have to check that it goes down to define a map from F cross I to F cross 0 is part of the F cross 0 union Y cross I on Y cross I will take an identity okay so if you try to prove this one directly you will see how difficult it is the thing proving the homotopexiation properties I am not saying that it is not possible alright but here the this proposition 4.1 makes the life very easy okay finally the homotope invariance of of what of a junction spaces take X to Z deco-fibration there is already say where X is a closed subset of Z two functions f and g from X to Y I am going to make a junction spaces if these two maps are homotopic then the extension spaces are homotopic valent indeed a stronger statement is true namely as relative pair the f comma y a g comma y why is a common subspace here okay f comma y a g comma y as a pair are homotopic valent this was done right in the beginning remember what is the meaning of this there is a homotope there is a map from f to a g okay which identity on Y similarly a g to a f which the identity on Y these two are homotopic valent relative to Y okay so that is a strong statement namely a junction spaces do not depend on exactly what f is but only depend on the homotopic class of f so this helps a lot in doing function in homotopic theory often you start picking up a map here but you do not pick up the map you you are actually interested in the homotopic class of that map and then you want to do something but then you want to say that whatever you have picked up you know it may induce something another one if you pick up something else no a junction space as far as concern the homotopic type has not changed so this is a very useful result so let us go to the proof of this one start with a homotopy x cross i to y okay from f to g we claim that now you can define a junction space of H of this H itself what is it it is z cross i disjoint union Y x cross i points of x cross i being identified with their image under H okay so that is H so that will also contain Y okay so what we are saying this a f y say a pair is a its topological pair and a g y another topological pair both are subspecies subspecies of the inclusion maps are homotopic violins is enough to say that they will be homotopic violins but what we are saying is they are deformation ray tracks hello sir yes sir can you say one more time how you made A H I mean yes how what is A H A H is a junction space of H yes say you a junction space is defined for every map yes I mean you used to say it cross i right yeah of course x cross i is not contained in z z this z cross i okay so that will come yeah I have just only claimed here I just had explanations will come now huh okay okay what is A H this is the picture see this is Y this Y here this is Y here X is a single point I chosen in the diagram this circle is your z okay where did X first goes to this point and then this Y is chosen as a path it goes to this point and it is homotopy means it is a path here right a homotopy of a point is nothing but a path in the picture I have to only take the simplest cases okay so the first point is F the end point is G when singleton axis is singleton map is nothing but the point right now homotopy is nothing but a path here here I have taken AG AF here this is why distant union z but X is identified with its image is it clear okay in this picture I have taken AG I want to say both of them are subspaces of A H here what is H it is it is from z cross i X cross i identified with its image here you see yes it's not a straight line now right it is you have to take the path whatever path the capital H is this is the edge junction space for H now is this picture is clear now what I have to show is this this part is subspace here and that is subspace there okay they are deformation retracts okay alright the argument is exactly symmetrical if I show one of them is deformation retract other one will be also deformation retract for the same reason in fact there is a t factor you have to reverse t to 1 minus t then the role of X cross 0 z cross 0 z cross 1 will be interchanged that's all so this part is Z cross 1 here this is Z cross 0 when you take the homotopy here you see we are actually doing much more than cowardism theory this looks like a cowardism when things are manifold here they are not even manifold but we have a word even kind of things here okay so these things will be these results will be used in a very deep differential topology so here is a proof finally you have to write down proof this is an idea again by this 4.1 what is the assumption assumption is X is a co-fibration X 2 Z is a co-fibration that is a standing assumption so I have a retraction Z cross i to Z cross 0 even X cross i okay this induces a different interaction R naught bar okay from A H which is the ejection space for H Y union over H Z cross 0 union X cross i okay let Q be Y distance union Z cross 0 union X cross i to Y disjoint union Z cross 0 and X cross i this is the quotient space this is a disjoint union this is the ordinary where this identification is taken place its restriction Q on to Z disjoint in Z cross 0 forget about this part that is a surjective map okay moreover for any F a closed subspace contained in the Y union this part check that Q inverse of this one is Q inverse of F union H inverse of Y intersection F this part I have written down only for people who have forgotten that product of quotients you know first take quotient and then product with identity map when I is compact is a quotient first is the first take the product and then the quotient this part is not needed for you now because I have actually proved that one but here I am actually proving that it is a quotient map okay so therefore what you get is that Q prime is a quotient map after taking the product therefore this part is nothing but Y union over F Z cross 0 because X cross i part is just hanging around there goes into this one so that is A F okay so this will show you that A F is a deformation this part what is this part this was deformation retract that is A F is what it is Y union H this was a deformation retract now this I showed that it is equal to A F okay so there is some it will take some time to understand why such things work okay you have to you have you will get lost in the notations here proof is this picture is tells you what is going to happen once you have proved that A F is deformation retract AG deformation retract is just changing T 2 1 minus T symmetric argument okay so you can stop here