 So, with a lot of background from topology, now we return back to our homotopy theory. Earlier, we had defined notion of homotopy and then while dealing with loops and fundamental rope, we modified it. Namely, path homotopy was defined as homotopy which keeps the endpoints fixed. This idea will now be generalized completely. In order to study homotopy properties of spaces, we need to work out from smaller pieces of the space to larger chunks of the space. This demands that whatever good work we have done, whatever information we have collected on smaller spaces is not lost when we move to larger spaces. So, the notion of homotopy needs to be strengthened by allowing us to exercise control over these smaller pieces. And that is precisely what leads us to what we call as relative homotopy, namely the homotopy which don't change on whatever smaller piece is nice already. Today, we have to keep the functions as they are, don't change them. So, this section will begin only a small beginning actually, only a few definitions and observations. Okay, even this D has to be developed step by step. R to the subset A of X where X is at a value space and two functions X to Y which are continuous functions such that on A they agree F A put G A for every A in A. Okay, so on a smaller subset the functions are the same. Then we say F is homotopic to G relative to A. If there is a homotopy H from F to G such that H of 80 is F A for every A in A. We write this as F is homotopic to G without any just simply without any equality there relative to A. The simplest notation for the most useful concept. If A is empty then it is the standard original homotopy that we have no controls. If X is the interval and A is 0 and 1 this is precisely the path homotopy that we have. Okay, so the relative homotopy that we have taken is a perfect generalization of the older concepts. Okay, so all the properties that we have all the deductions that we have done for older things they are not lost by this new definition. You do not have to recheck them again. Okay, this is one important thing you have to do. Now once you have defined what is the homotopy between X to Y relative to some subspace we can talk about homotopies of pair. Here you take a pair X A means A is the subspace of X and Y should be another topological space and B should be here instead of A take B to be a subspace of Y. Okay, nothing wrong but it should be too restrictive that the same A should not need not be subspace of both X and Y. In general by a pair of topological spaces we mean as a topological space the second entry there must be subspace that is all. Okay, so such a pair has same homotopy type if there exist functions from F from X A to Y B G from Y B to X A such that F A is equal to identity of A equal to G A. Okay, on A F A is identity of A is G restricted to A and G composite F is homotopy to identity relative to A and F composite G is homotopy to identity relative to A. So we keep writing the relative to A specifically. Okay, if I just write equivalence it may be confused with the old moreover relative to what we have to keep telling. So that is so this G composite F is from relative to B should be. Okay, so it is easily checked that above relation is an equivalence relation among all topological pairs X A where A is fixed. Okay, if you take all topological pairs X A here you take Y also A only topological pairs A has to be fixed. In other words A to B you can take a homeomorphism. Okay, that is that is a consequence of this one right. If F A is identity and G A F composite G is identity of a relative to A means F composite G restricted to A is just identity map right. Therefore G and F are inverses of each other when you restricted to A. So they will be homeomorphic A and B are homeomorphic and that is one reason why I have written both Y and Y A here also instead of B. That is why there is nothing wrong but in the definition you would take A to B arbitrary B. For topological pairs when they are equivalence and so on then you fix A and look at all topological spaces which contain A as a subspace then you have this relation equivalence relation whether they are homotopy equivalent to each other with respect to A with relative to A. So that will be a homotopy that will be a equivalence relation and such types are called homotopy types the classes are called homotopy types. I have already told you what is a retraction but let us recall a retraction is a continuous function from X to a subspace such that on the subspace it is identity r A equal to A for every A. If such a retraction exists then we call A is a retract of X. Retraction is an action it is a function. A retract is the result A is a retract of X. A function F from X to X which is homotopic to identity is called a deformation of X. Any function which is homotopic to identity will be called deformation of X. Often some people call the homotopy itself as a deformation because that is the action of the deforming. From identity map you have taken some other function. If F is a homeomorphism we are not bothered so much about homeomorphisms you know. So maybe F of X is smaller than X but it is homotopic to identity map then it is called a deformation. So everything is taking place inside X that is important. So anything is homotopic to identity is called a deformation of X. If r is a retraction and it is homotopic to identity. You see retraction is a map from X to A but A is a subspace of X. Therefore we can view r as a map from X to X itself. Then we can talk about whether r is homotopic to identity. If that is the case r is retraction and it is homotopic to identity then it is called a deformation retraction. Retraction which is homotopic to identity homotopy is taking place inside X. r itself is an image of r is A but identity map is there from X to X. So it comes slowly and sits inside A. So that is the meaning of a deformation retraction. Again if such a map exists then A will be called as deformation retract of X. So you have deformation retraction and deformation retract. Retraction and retraction. Okay retract subspaces, deformation retract is a subspace. Well deformation retraction is the action the homotopy the map itself. If the homotopy from identity to r in the previous definition namely when you will define deformation retraction if that homotopy is relative to A the points of A are not moved. Then this is called a strong deformation retraction. Okay and A is called then strong deformation retract. All right. Finally one more definition. If the inclusion map A to X is a homotopy equivalence. See treat A as some other space Y then you know we have defined our homotopy equivalence. Right. But if it is an inclusion map itself is a homotopy equivalence then we say A is a weak deformation retract. Okay inclusion map is a homotopy is a weak deformation retract means there is a map X to A it may not be a retract but it is homotopy to identity. That is why it is called weak deformation retract. Okay. One thing is the name should tell you that strong deformation retract implies deformation retract which is obvious because we put extra condition in strong deformation retract namely during the homotopy the entire homotopy should be identity on the subspace A. Deformation retract is what the retraction map is homotopy identity. Okay automatically tells you that inclusion map is homotopy equivalence. Okay it is homotopy inverse is R. All right. So weak deformation retract does not have an inverse which is a retraction that is all. So it is some homotopy equivalence its inverse homotopy inverse from X to A may not be a retraction. Therefore SDR implies DR implies WDR strong deformation retract implies deformation retract implies weak deformation retract. The caution here is if you are referring to various books they may differ in the definitions. So take care. Okay. So what are what are the definition first you check. There are different the same words may mean different things for different authors. Let us work on some examples. Simple examples familiar to us. Okay first and then later on we will try to do more and more complicated ones. Take any convex subspace of Rn. Then every point X is a strong deformation retract of X while proving that a convex set is what contractible is precisely what we had seen. What was the homotopy? X comma t goes to t times X plus 1 minus t times whatever the point X naught every point. So let us say X naught then t times X plus 1 minus t times X naught. Okay that is the strong homotopy it will not move the point X naught at all. Okay everything is joined to this one. So this follows from the observation that parameterization of the line segment is continuous in terms of certain points. So this we have used already anyway X t go into 1 minus t times X plus t times X naught. It shows that when t equal to 0 it is identity X goes to X and t equal to 1 it is X naught. Okay if X naught comma t if you take it is always X naught no matter what t is that is the meaning of X naught never moves. Therefore this homotopy is already relative to X naught. Right therefore what is the constant map X naught all the from X to go into X naught okay that is a retraction obviously and it is homotopy identity therefore X naught is a strong deformation that track. Okay identity is homotopy the constant map X naught c X naught and this homotopy is relative to X naught okay. So pictoriously suppose you have a pond like set like this at any point so all the points are coming here this point is not moving okay at the end result will be constant map. So this is the picture of a strong deformation you have to take any point and then join. So this is actually true of a star shaped set also with an apex X naught at X naught will be what a strong deformation type of the whole space whenever a space is star shaped at a point apex point is a strong differential track okay another remark here for any space X the subspace X cross 0 is a strong differential track of X cross i along the these intervals okay for each point I can just push the whole X cross i to X cross 0 okay there is nothing special about X cross 0 either you can take X cross T naught then push the entire of interval at each point around each point X above each point two single point X naught T naught okay so it is easy to see that X cross any T the inclusion of X inside X cross i is a strong differential track of X cross okay so more generally take any contractable space okay then X cross Y where Y is a point of Y is a subspace of X cross capital I capital Y okay and Y is contractable so X cross little Y will be a strong differential track of X cross Y capital Y the deformation retract Y the homotopy can be used to prove this one without moving the points of X X comma whatever point okay maybe you can generalize this one further now let us work out another familiar example to begin with I can start at S1 inside R2 okay so I say that the circle the unit circle is a strong deformation delta of R2 minus 0 the origin you have to throw away okay how do you do that once again you know the convexity kind of convexity whatever which is not a convexity convexity about S1 is what it is namely joining joining it works namely 1 minus T times X plus T times X divided by norm X take any X inside R2 minus 0 okay take a real number T be in inside I H of XT I am defining it 1 minus T times X X is a nonzero vector so I can divide by norm even norm 1 to get a unit vector so that is the map from C minus 0 or R2 minus 0 to circle X by norm X I am joining that function X by norm X to the identity map X 1 minus T times X plus T times X by norm X putting T equal to 0 this identity putting T equal to 1 it is the function which takes X to X by norm X which is a retraction if it is a unit vector already X by norm X is equal to X which is a retraction and it is homotopic to identity therefore this is a deformation retraction during the homotopy if X is always norm 1 then dividing by X has no effect 1 minus T times X plus T times X is X so no point of S1 moves during the homotopy so this is a relative homotopy therefore it is a strong deformation retraction okay now the same argument will work even if you take okay not the whole of C cross I not the whole of C minus 0 but some kind of annulus annular regions and so on like you can take just all the vectors between of length half bigger than half and less than 3 by 2 so it is an annulus you should not include 0 that is all okay not only that now instead of S1 and R2 this whole thing generalizes exactly ditto Rn and Sn minus 1 is that clear not only that what you can do is why 0 itself you can take a disk you can take a sphere centered at any other point then you have to just translate the whole thing you know you choose your coordinates as as if the center of the sphere is 0 that is all the same arguments will work okay not only that you do not have to even throw away the center suppose you throw away any point inside the sphere inside the disk okay the interior of the disk then Rn minus that point will be will be will be can be pushed back to the sphere the sphere will be a strong deformation retract of Rn minus that point so you should try to see all these variations of this theme but this is the simplest theme this can be modified to get many other easy examples okay less obvious is the fact that let us look at now minus 1 plus 1 cross minus 1 plus 1 d1 cross d1 okay the square now you throw away the origin 0 0 okay so that is my S I want to say that the boundary square is a strong deformation delta of this what I have done I have taken the square then I have thrown away the origin this space can be can be pushed or you can see the boundary of this one will be strong deformation tract of entire space there are many ways to do this okay instead of the circle I have a square now okay the only difficulty is geometrically it is very easy you have to just push the whole thing which just means that you are taking stereographic projection okay but writing down formula is not that simple as in the case of round sphere okay the geometric idea is the same so I will give you one one simple solution here you are welcome to do whatever you like different different kind of thing this is the simplest okay because the the square has four corners so I divide it into whole thing into four equal parts the four triangles then it is in each triangle I can write down a simple formula on the intersection of two trans nearby triangles namely on this line the two formulas will agree okay the two formulas will agree here they will agree here they will agree here and so on so they will give you one single continuous deformation a homotopy from this entire field square minus the center point here to the boundary okay during this homotopy the boundary will not move okay so you can write down formula so here I have written down formula for each triangle you have to have a different formula here okay for example x coordinate is one y coordinate is changing this the first first formula is for this triangle okay take any y here x coordinate is one y coordinate is changing okay send it to y by x so that is the formula send it to y by x x is not zero because it is zero here x is half to one here so I can divide y by x when x is one it is y it is identity map okay so those are the projections here so other things you can work out once you have written r which is retraction the the homotopy you can write down in a single formula namely write take the retraction and the identity and join them let us solve okay writing the retraction itself will be truncated here there is no one single formula like x by norm x so here are a few remarks every singleton subspace of a space is a retract you take the constant map at that point but deformation retract weight deformation retract they are the homotopy concept they are not always always true okay if x is a host or space and if a is a subspace which is a retract then a must be a closed subspace okay so this is one of the reasons why in a joint adjoining the you know the adjunction space etc we assumed the subspace on which adjunction is taking place to be closed yesterday one of you asked why do you want to assume it is closed okay this is one of the reasons just in a host or space subsets which are not closed they are not retracts very easy to uh get examples which are not retracts okay note that if a deformation if a is a deformation retract of it then the inclusion map is homotopy equivalent this i have already told you retraction r as above is it so much of inverse in this terminology to say that x is contractible is equivalent to say that every singleton subspace of x is a deformation retract in the convex set it is actually strong deformation retract is what we have shown but that is not necessary okay if you take an arbitrary contractible space every point is a deformation retract so that is that that is more or less black bar from the definition but if you want a strong deformation retract that may not be true okay so i will show that example and that will be the end of today's lecture so comb space is an example which is just to warn you that you have to be careful with the definitions here so what is this comb space it is a subspace of the i cross i the square consisting of either y is zero that means the x axis is there or x is zero the y axis is there okay or x is one by n y could be anything one one by two one by three one by four and so on okay so those are the teeth of the comb okay so there are infinitely many teeth for this comb and they are all clustering they are coming closer and closer at the teeth namely at x equal to zero okay so this is the picture of the comb space so the topology around this point namely around zero to zero zero to zero one along the the y axis here okay topology of this space is we are namely it is not locally connected not locally path connected and so on now that creates a problem you can push the whole thing down to the x axis once you have pushed that you can push the whole thing whichever point you want along the x axis so that shows that every point on the x axis is a strong deformation without moving that muster let us say half what I do I will push everything down to zero here and then to half everything down to the x axis and then push to half okay all these line push to half so every point on the x axis is a what it says strong deformation retract of this one but there are points on this line namely the y axis okay other than zero zero none of these points is a strong deformation retract you can you can get a homotopy to that point because every constant function is homotopic to identity here because this is a contractable space but the homotopy will not be identity on that point the point itself have to move all the way to the x axis and then come back okay so this is this is because of locally here the topology is complicated it is not locally path connected rigorous proof you have to work out that is an exercise okay thank you