 As a student I was wondering why the several books only give co-fibration for an inclusion map whereas, the vibration is for arbitrary map not necessarily quotient map. The answer was in this result that we are going to do, but there is even more satisfactory answer much much later only some 10 years back I read it from HHS book that every co-fibration is actually a inclusion map. So what are we going to do because we are studying things up to homotopy in homotopy this result says that any map can be replaced by inclusion map and that inclusion map is a co-fibration. So this was so powerful that now you can think of up to homotopy every map is a co-fibration. So this completely justifies the somewhat artificially introduced notion of co-fibration. We are not claiming that every map is a co-fibration but up to homotopy it is. So let us go through this one. So later on we will see a lot of applications of this one and so this itself is somewhat a hard work you have to do now. So here is a statement, even any map X to Y let us construct the mappings mapping cylinder M F. Remember what was M F? M F was obtained as X cross I disjoint union Y by the identification namely each X is identified with its image F X. So I am making this diagram of various functions here X is included in the M F, there is a map X to Y and M F to Y there is an F act and then Y is also included in M F by an inclusion map J. So this F act is actually an extension of F because X is t thought of as a subspace of M F. So all these things I am writing down carefully namely F act composite J is F, I think the first triangle is commutative top triangle. J composite F act is a map from M to M, M F to M F. This map is homotopic to identity of M F. So the second diagram, second triangle here, the bottom triangle is not a commutative triangle, it is homotopic commutative, it is homotopic to identity. J composite F act is homotopic to identity and this homotopic is relative to Y that is F act is a strong deformation retraction of M F to Y in particular Y is a SDR of M F. M F union X cross I is a strong deformation retraction of M F cross I. So I is a co-fibration inclusion map from C to D you know how to work because we have seen this one. J composite F finally, it is J composite F coming from F to M F, J composite F it is homotopic to the inclusion map itself a topic. So when you put this arrow identity, if you do not put this one the diagram is commutative. This put, if you put this one whatever resulting come, these things become, they will be homotopic commutative. If you come from here to here and come from here to here, it is homotopic because this is homotopic to the identity precisely the last thing he says. J composite F is homotopic to the inclusion map I. It is simply the I composite identity. So let us prove one by one more, some of them are completely obvious namely the first thing F add composite I is F we have already seen. What is, what is F add? What is F add? Each point of M F is a class X comma T right or bracket Y, X comma T or bracket Y remember that we see equivalence classes of elements from X cross I union Y right. If it is a Y part then Y goes to Y itself, if it is X comma T take it to F X ok. So it is well defined because whenever it is identified namely X cross 1 at X cross 1 it things are identified but they are identified with F X only. So it makes sense. So it is an extension of F which is defined on X cross 0 X cross 1 whatever ok. So first part we have already seen second one so we have to define homotopy here. So take G1 X cross I to M F, G2 from Y cross I to M F defined by G1 of X, T, S X is as it is you see it is everything is happening in I cross I. So that is the 1 minus S times T plus S you join T and S 1 minus S times T plus constant 1 you join by S ok 1 minus T times S plus S times 1 constant 1 that is the function you are joining them. The first part is as it is the second function GUIS is just Y ignores the S coordinate these maps fit together and induce a homotopy from M F cross I to M F on X cross I when X equal to 1 T equal to 1 it was identified with F X right. If T is 1 this is just X comma S right and it is just Y ok. So these maps fit together define M F cross I to M F ok go down to the quotient space is from identity of M F to J composite R ok relative to Y if Y is there always it is just constant function if the coordinate is Y then it is just constant function ok. What is J composite R just see that what is this R I have got it is not correct now J composite F hat it should be instead of that I put J composite R it is not correct ok identity of M F to J composite F hat ok relative to Y. For the proof of C at M F union X cross I is a strong distributed threat of M F cross I ok. So what you have to do you have to do just similar to what you have done ok I have to recall that the construction that we have done you know in module previous module this one 20 or 2019 that whatever we have done. So let me go back to this 8 here all the way this one this map will be again used ok exactly similar to that. So H 1 is X cross I cross I cross I to M F cross I H 2 is Y cross I cross I to M F cross I H 1 of X t double prime t double prime X comma S similar to what we have done in the case of arbitrary X that is precisely what we are doing here exactly same way S part is taken here X part is undisturbed H 2 of Y t prime t double prime is Y remains as it is and we are ignoring t double prime here 1 minus t prime taking the reverse of that. If you do this just like the same middle notation as in that one all that you have to do is verify that they fit together fitting together means what Y is the subspace of M F right. So on that part you have to verify that ok wherever this Y is instead of M F here. So X cross I affects when X cross 1 is identified to M X so that is like Y. So there also you have to verify what happens to the map. So then they agree as a function so then continuity follows because all these whatever they agree all those spaces are close of spaces. So they fit together to define M F F H from M F cross I cross I M F will be you know this part X cross I disjoint union Y cross I one more I I have taken then identification you have to verify. So it gives you map from M F cross I cross I to M F cross I which is a long different retract of M F cross I into M F union X cross I ok. So M F union F cross I is a different retract of M F cross I. Once you have this the proposition will tell you that the inclusion map of M F into M F cross I is a what sorry inclusion of X into M F is a co-fibration. So that is D. D tells you that I is a co-fibration. What is I? I is the inclusion map of X into M F ok. Finally what is E? E is j composite F is homotopic to I. J composite F will be equal to J composite F add composite I right. J composite which J composite F composite J composite F add composite I. Now you have already shown that this composite this is homotopic to identity. Therefore the whole thing will be homotopic to identity composite I which is same thing as homotopic to identity. So this follows easily from our these two if you put these two things together ok A and B together ok. So essence of this one you know this is all this may be difficult to remember. What you have done? Start with F any map it is replaced by I is inclusion map ok. This F is getting replaced go back to this one. This I is replaced by this one ok the inclusion map which is homotopic to this one. As if it is homotopic to F it is not exactly homotopic to F because things are not taking place here but inside here ok. So J composite F is homotopic to I. So that is the meaning of that F is replaced by I inclusion map of homotopic ok. Instead of Y we have M F. But what is the relation between M F and Y? They are the same homotopic type they are homotopic to I and F F is homotopic to I. In fact this is not an orderly homotopic to I. This is a strong deformation attract. Why is a strong deformation attract? Up to strong deformation attract this F becomes I. We cannot define a inverse here but the J plays the role of homotopic inverse. So it is also same. So F composite J sits inside M F, map into M F and it is homotopic to I ok. So that in that sense that a arbitrary continuous function has been replaced by a inclusion map into some other space. And what is this space? This is this is homotopic type of the codomain. It is strong deformation attract. Codomain is strong deformation attract of the J. It contains a codomain also. It is enlarged codomain of this map. F has some codomain Y. Now that Y we have enlarged. And the enlargement is does not lose the homotopy invariant. I mean homotopy information is not lost ok. So this is the gist of theorem and essentially we use the construction of this number 8 namely going back all the way of taking I cross I and constructing a strong deformation attract. That was the fundamental result in all these things. Because the mapping cylinder is a device that enables us to replace an arbitrary map F from X to Y by an inclusion map which is a co-fibration of the homotopy. Observe that M F contains lots of copies of X and a copy of Y. X cross I, so for each T X cross T is there except at X cross 1 there are identifications ok. From B of the above theorem we have that Y is a strong deformation attract of M F. The mapping cone there is a mapping cone constructions here is called the co-fiber of the vibration ok. In the next section we shall give a number of applications of a mapping cylinder. So to sum up the hard work is somewhat over now. So next module we will reap the harvest. Lot of interesting applications will be there now for this ok. So that is all we will not continue now we will stop here. So let us take the next thing later ok. So there are some exercises here as usual I will just go through a few of them but I am not giving you you are giving you any solution as such. The first thing is you should know all these things these are all more or less point sectorological things. Even any two points in the interior of the disk there is a homeomorphism F from D N to D N such that F P goes to Q. P and Q have both interior points. On the boundary F is identity. The boundary is undisturbed any point goes to any point. First prove this one for a closed interval A and B fixed give me a homeomorphism which takes any point strictly inside A B to another point and it must be homeomorphism ok. I think you will be able to do this. So for all disks you must able to do this. So for the if you once you do it for a unit disk it will be done for all unit disks in all R. Ok next thing is how far are these convex subsets related to the standard convex namely above. Let us look at that one. Let X be a convex polygon inside R N inside R 2 with N vertices N greater than 4 3. A convex polygon like a triangle or a quadrilateral and so on ok. I am not assuming any regularity here any convex polygon ok. Show that X is homeomorphic to the cone over the boundary of X. Boundary of X is a you know when you say triangle there is some confusion a full triangle or the boundary triangle ok. So here I mean convex polygon I mean the entire convex hull of the whole thing and the boundary it consists of only sides of the polygon right. So on the boundary that is a topological space you take the cone over that then you have to show that that cone is homeomorphic to the convex polygon itself ok. The full triangle is homeomorphic the cone over the triangle ok. So that is the first exercise here. Now choose any N distinct points A1, A2, AN on the standard circle ok. Now construct a homeomorphism F from the boundary of this polygon to S1 so that the vertices of X XSN vertices are mapped into A1, A2, AN. Here you will have to you will have to be forced to take the A1, A2, AN in a particular order namely whichever way suppose V1, V2, AN are the vertices of the polygon return in a particular order consecutively then A1, A2, AN must be also consecutive ok. You can shuffle them you cannot shuffle them. So that is understood here. So V1 should go to A1, V2 should go to A2, VN should go to AN and the entire thing must be a homeomorphism of the boundary of X to S1 ok. Next construct a homeomorphism G from X to D2 which is the interior of S1 together which extends the given F in B ok. That must be extension on the boundary it must be your F and the whole thing must be a homeomorphism G. Do the same thing as in B and C with the right half disc G with three of the points on the boundary being 0 1, 0 0 and 0 minus 1. So I am taking half disc here inside D2 inside R2 instead of taking a convex polygon I am taking another convex set. The half disc is also convex set ok. Its boundary consists of a line segment on the y axis namely minus 1 to 1 and half circle on the left ok on the right half disc here. So take 0 1, 0 0 and 0 minus 1 these three points ok. Map them to three distinct points on the circle construct a homeomorphism on the boundary and then construct a homeomorphism of the half disc to the full disc. So that is what you have to do in D ok. Similarly now this exercise assume that n is greater than or equal to 4 now ok at least four points. Let a1, a2, a3 be three consecutive sides consecutive vertices of X sorry and vertices of X. Why be the quotient space of X obtained by identifying the points of the edge a1 to a2 then a3 to a2 reverse all. So identify them by the row t times a2 plus 1 minus t times a1 that will lie lie on the edge a1 a2 right should be identified to t times a2 plus 1 minus t times a3. In other words a2 will remain as is a1 and a3 are get identified. So this is the only identification the two edges are identified ok. So for all t less than equal to a1 you we make this identification. Whatever you get the entire convex polygon ok but I would have to write after a defiantification again is homeomorphic to d2 ok. For this you have to assume that there are more vertices. So like this there are other exercise also here then over the cones ok. So later on these are on loops and homotopes of the maps and so on. So this is on our first part of the thing. So this one we have already used if X is host of space the inclusion map is a co-fibration then a is a close of space. I will indicate the proof all today. So you can just write down my memory if you have understood it or you have to work it out yourself ok. So this also I have indicated namely pi1 of xa can be identified with homotopy classes of loops from s1 to I mean continuous functions from s1 to x where 1 goes to a. And the homotopy classes here are keeping one fixed throughout the homotopy. So that is that meaning of pi1 of xa this has been already used ok. So you write down a detailed proof of that one. Now suppose you have a map which is null homotopy defined on the circle then pi1 of xx0 is trivial for each point x0 belong to s1. What I am saying take any loop in x suppose it is null homotopy. If any loop in x is null homotopy then pi1 of xx0 is trivial for each point of x ok no matter where it is this should be true for all loops. So this is a straightforward application of this exercise. So if you have a path connected space ok the just homotopy classes without any base points ok free homotopy classes is equal to the set of conjugacy classes of elements in pi1 of xx0. Pi1 of xx0 is a group in a group you know what is the meaning of conjugacy classes that is what you have to show that conjugacy classes are in one known correspondence with homotopy classes of maps from s1 to x. Base points are not fixed here so that is the difference ok. From this you can deduce this theory this exercise also. This this exercise actually I have solved it in the in the theory part itself roughly. I have given you sufficient hints so this is easy to work out for it. As a path connected space if pi1 of xx is abelian if and only if what happens for each b in zx for all path is from tau from a to b the h tau which is obtained by uh congealing by the path tau pi1 of xa to pi1 of xb the all this map issues are the same that from all maps whatever tau for all all path is tau h tau is the same map same homomorphism same bijection whatever you are. This is an exercise which will help you to prove that the comb space in the comb space 1 comma 0 is not a you know strong deformation retract first prove this exercise 4 point level. What does it say? Suppose some point is a strong deformation retract then for every open subset u of x containing x0 that means every neighborhood of x0 there is another neighborhood v inside x0 inside v inside u the neighborhood of x0 inside u such that inclusion map u to v is null homotopy in v u to v you can write a homotopy to a constant function that is the meaning of that starting with the inclusion map to a constant function that is homotopy if you use this cleverly then you can show that 0 and is not a str of the comb space no hand waving okay for that I have already told you what you have to use is the comb space is not locally path connected or locally connected at each point on the y axis it fails to be locally path connected use that use this exercise you can complete the proof of this sdr okay so let us stop here the next time we will see many more results so as I told you hard work is more or less over for a while for a while huh okay thank you