 I would like to present a few more examples of ocean constructions which are useful in algebraic topology. Before that let me make one remark. Suppose you have a map from Sardvan to Sattor which is a homeomorphism, similarly another homeomorphism from Y1 to Y2. X1 is a subspace of Z1, X2 is a subset of Z2 and under phi X1 goes to X2. So I am doing a junction space constructions over Z1, Y1 and over X1, similarly Z2, Y2 and X2 through maps F1 and F2. I assume that F2 is exactly equal to, you start with X2, come to X1 via phi inverse and then take F1 and then go back Y upside. If you have this data, then what you have is a disjoint union Z1 and Y1, two disjoint unions are two Y2, you have homeomorphism, phi disjoint union of psi, I mean the patch up of these two homomorphism because they are disjoint union. But now because of this relation that F1 is, F2 is precisely F composite psi composite F inverse, this compatibility, whatever you do identifications here, corresponding identifications will take place here. Therefore, under this, the homeomorphism from phi disjoint union of psi, you get a homeomorphism, for example, you get a map which is a bijective from the adjunction space F1 to adjunction space F2, ok. Once this is told, then the verification is totally obvious. First of all, you should see that there is a satiristic map which is bijection. Continuity of this one will be totally obvious because the corresponding functions are continuous on the mother space, ok. So, similarly, if you take inverse is, that map will be from Z2 disjoint in Y2 to Z1 disjoint in Y1, same thing will apply, so inverse will be also continuous. So, this is the topological invariance of adjunction space. Later on, we will see similar version for homotopy invariance of adjunction space which will require less hypothesis than for homeomorphism. In particular, we may say that homeomorphism type of the adjunction spaces depends on the ambient homeomorphism class of the adjoining maps. If the maps F from X2Y or and X1 to Y1 to X2 to Y2, they are just related by homeomorphism, that is not enough. The homeomorphism should take place on the domain at the top level named as Adjunct Z2 and corresponding homeomorphism Y1 to Y2 so that the two maps are correspond under that under those homeomorphism, that is the ambient homeomorphism class. So, here is another construction now which is a slight modification of the mapping, mapping cylinder. Remember that mapping cylinder of a function F from X2Y was defined as quotient space of X cross 0, 1 interval disjoint union Y by the relation that X, 1 is identified with Fx. So, you can think of this as adjunction space of X cross I disjoint union Y via the map which is defined on X cross 1 to Y, so this is a adjunction space, this we have seen. Instead of that, instead of X cross I, you take the cone namely X cross 0 is collapsed to a single point, then X cross X itself is identified as X cross 1 as a subspace. So, you can think of Y as defined on a subspace of this Z Cx and then take the adjunction space that is called the mapping cone of F which is nothing but another way of looking at it is it is the quotient of the mapping cylinder in which the first coordinate X cross 0 is also identified to a single point. So, recall that X here is usually identified with X cross 1 in the mapping cone but in the cone itself, but when you go to mapping cone, this X cross 1 also is not a subspace, you will have to take some X cross half or some other T, any X cross T not equal to 0 or 1, those will be the copies of that, whereas Y is always a subspace of this mapping cone. The image of X cross 1 in CF itself may not be homomorphic text but there are other copies of X namely X cross T where T is not equal to 0 or 1. So, this is the modification, so do not make the confusion that X cross 1 is still a subspace of this mapping cone of the function F. So, there are various kind of constructions, the next construction is also very useful which is just doubling the cone. Cone is defined as the quotient of X cross i, wherein you have identified X cross 0 to a single point. This time what we will do, we will take two copies, so what we will do, we will take instead of X cross i, you take X cross minus 1 to plus 1, to begin with you have a double space here. Now, you identify both the end axis, X cross minus 1 as well as X cross 1 to two different points, single points. So, this is the picture, we have X here and X cross minus 1 to plus 1 like this, the cylinder, the X cross 1 goes to a single point here, X cross minus 1 comes to a single point. So, this is called the suspension of X. There is another terminology here, what are called as reduces surface suspensions which I am not going to define right now. That will be needed when you are deep into the homotopy theory right now, you do not need it. So, in the light of that, these are called un-reduced suspensions. For example, suppose X is a single point, then what is the suspension of X? It is no identification at all, you have to take X cross minus 1 plus 1 which is just minus 1 plus 1, homomorphic to minus 1 plus 1 right. So, the cone over single point was the interval, similarly the suspension over a single point is also an interval. More interesting things happen, just take two points, a space with two points and discreet space, discreet poly, not indiscreet poly, then what will be the suspension? Now, X cross I will consist of disjoint union of two copies of the interval right. One end point consists of minus 1 cross two points, the other one will be another plus 1 cross two points. These two things are identified to single points. So, what you get is like a diamond shape now, which is nothing but up to homomorphic which is circle right. So, the two point space can be thought of as S naught okay minus 1 plus 1 cross minus 1 plus 1, then identify the end points of these, then you get a circle. Actually, you better identify it as a diamond shape, but diamond shape is also homomorphic to circle okay. So, the suspension of S naught is S1. What happens to suspension of S1? Here is a picture, forget about the shaded part, this picture if you take just the boundary, there is a circle. When you take the suspension, what is this space? It is nothing but union of two copies of discs because the cone over circle is a disc identified along the common circle, the base. So, it will be nothing but the two sphere, suspension of a one sphere is a two sphere. Exactly same argument will tell you, suspension of any n sphere is n plus 1 sphere. When I say is n plus 1 sphere, I just mean homomorphic, that is all you can directly write down a neat homomorphism. If you use cost data science data and so on, it will become neat instead of the corners here, it will become round okay. There is a nice description of this suspension just like the cone at the function level. Namely suppose you start with a function from x to y and S of x, S of y are the suspensions, then there will be SF from Sx to Sy just defined by SF of a class Xt close to the class of Fx and t. To begin with, if you do not take class here, this is a map at the level of x cross minus 1 plus 1 to y cross minus 1 plus 1, which is just F cross identity. Whatever is identified in the domain, okay, all of them go to same points which get identified. Therefore, this map will factor down to give you a continuous function at the suspension level. The argument similar you have seen, cx to cy, cf, whenever x is a continuous function from x to y, it is similar to this one. So, you have suspension level and this suspension has all those typical properties. If you take x to x identity, S of identity is identity function of the suspension Sx and if you have x to y and y to z, another function g, then if you take g composite F and then S of that, namely suspension of that, that will be suspension of g, composite suspension of F. Because all those things are happening in the first coordinate here, g composite F is, okay, g composite F, that is all, okay. The t coordinates do not disturb the functoriality, the functional compositions on the first coordinate here, okay. Now, here are a few observations which I will leave to you to think about it. I will go through them slowly, to think about it. Finally, you have to come up with a proof, okay. Sx can be obtained by gluing two copies of cx, the course which you can write it as c plus x and c minus x along x, which is x cross y, identified. And a homeomorphism type is, the homeomorphism type is independent of gluing homeomorphism F from x, x. You can take any homeomorphism and identify c plus with c minus of x. The resulting space always homeomorphic presence. Construction of Sx is functorial property, which I already explained what is the meaning of this sentence. If x is a homeomorphism, then Sf will be also a homeomorphism. If g is the inverse, then Sf inverse will be Sg, okay. The last thing is, which also I have explained. Now, you have to write down some rigorous proof of this, that is all. The suspension of SN is one-dimension higher sphere, namely SN plus 1. To help this one, I am giving you the topological result here, namely, if we glue two copies of n-dimension discs along their boundary, okay. Via a homeomorphism, the resulting space is homeomorphic to n-dimensional sphere, okay. It is like gluing, simplest case n equal to 1, you take two intervals, close intervals, identify the boundaries by a homeomorphism from one to the other, namely a bijection, a goes to a prime, b goes to b prime, identify what you get to circle. So this generalizes to all the dimension dn, okay. So that is how S of SN becomes SN plus 1. If you do this one, then this is no, this is, there is no exercise here, this is not exercise. So this is like that, because you would have done this exercise already. And you would have also learned that the cone over the sphere, okay, is a disc of one-dimension higher. If you put them together, you will get this one, okay. So no exercise is to write down the details for each of them. Next, I will take one more example. This is called topological joint. This is actually a far more generalized than those suspension and so on. You can get suspension as a special case of this and you can see this, okay. So let X and Y be two non-empty topological spaces. If empty space in the product is empty and so on, so you do not want to get into that. We define the topological joint X star Y of X and Y to be the quotient space of X cross I cross Y, okay. So it is not just X cross I, it is not I cross Y, it is X cross I cross Y. If both X and Y are I, then this would be a cube, I cross I cross I, okay. That is the simplest picture you will have, all right. Now what I am going to identify each X1, 1 and Y, I mean when this T parameter is 1, it will be identified with X2, 1, Y. The third coordinate must be the same. Then first coordinate any arbitrary X, they are all identified as single point. For each Y, this is called. So at one end, I am killing the identifying the X to a single point for each Y, okay. You can write X cross Y cross 1, right. Instead of X cross 1 cross Y. In that each for each Y, the X is identified single point. So that is one thing. Same thing I want to do the other way around. At 0, for each fixed X, I identify Y1 to Y2 for all Y1 and Y2. X is the same. So for every X in X and Y1, Y2 and Y, okay. I started non-empty spaces. If Y is empty, then you have to define what is X cross Y. So if you take product, then it will be empty. So you do not do that. You would like to have X star Y as X itself, okay. Okay. Maybe you would like to define it as empty. That may be better. So I am not very sure of this one. Okay. If you need some change, you can define it like this, okay. Inputively, what is this topological joint? X cross I cross Y. It has union of lines parameterized by X comma Y. This is just line segment. Union of lines parameterized by X comma Y. Okay. But when you identify all the X to a single point, what you have is from a single Y to that space, you have a line segment. Only for Y2 do we have a line segment and so on. So I will let you know X1 with a better example. Okay. Here this is not an example. We are trying to explain this in the abstract sense only. So what it is, is look at the class some X, some Y. Okay. So T is never identified to anything. So as T runs from 0 to 1, you will get a line. This line you can think of as starting at X and ending at Y. In other words, you see in abstract sense it is creating lines from X to Y. When the cone what it has created, an extra point was taken and there was lines from every point of X to that extra point. Right. That is the cone. Here what I have done, here we do, instead of just one extra point, a space is already given. X is given, another space is given. Now for each point of Y, you draw a line from X to Y. Where do you draw, that is the point. So it is not happening in any RN or Euclidean space or any vector space. It is abstractly happening. Okay. That is why you take X cross I cross Y and do the identification. Okay. So to justify this explanation, let me give you a simple example. There is a typo here. Start with X to be a line segment PQ and Y to be another line segment UV. Okay. Denote two line segments in R3. X equal to PQ this is. Okay. Which are skewed. The two segments must be not intersecting non-parallel. That is called a skew lines. So how to ensure that? There are many ways of ensuring it. Namely, you can look at fix one point as origin. Then look at Q minus P, U minus P and V minus P. So you get three vectors. These three vectors must be independent. So that will ensure that P minus Q and U minus V are not parallel vectors. So line segments are not parallel. Neither they intersect because if they intersect, there will be only two vectors here. Okay. The third vector will be already one of the independent either 0 or one of the difference of these two. Some of these two. That is all. Okay. So the topological join of this X and Y, the two line segments which are skewed is nothing but the convex hull of this. Take the convex hull of all these four points. Namely, what is called a tetrahedral. The convex hull of three independent points would have been a triangle. Okay. So there is a fourth point which is not in that plane. So what you get is a tetrahedral. Which is same thing as take arbitrary point in PQ from P to Q. Join it to some arbitrary point in U and V. Okay. Between U and V. Take the collection of all these. So that will be the join of X and Y. It is also equal to the just the convex hull of PQ, U, V, which is tetrahedral. So in some sense the join is generalizing even the construction of convex hulls. Okay. Now we can do some elementary algebra of sets on this star. The star is joint X star Y. X star Y contains both X and Y as subspaces. Only at one point the points of X are identified at namely when the T coordinate is one. Similarly, when T coordinate is 0, points of Y are identified. If you take T coordinate not equal to 0, the whole X cross Y will be there. So both X and Y are subspaces. In fact, to take X going to X, 0, Y, there is no identification of X here. The Y coordinates gets identified. So this is a nice copy of X on the 0th phase. Similarly, Y going to any X comma 1 comma Y would be a inclusion of Y. Okay. X comma 1 comma Y. These are subspaces of X star Y. Okay. Just similar to the phase of the cone or suspension and so on. The topological joint is both commutative and associative. What do you mean of this? Up to homeomorphism, X star Y is same thing as Y star X. So this follows by the commutativity of the product. X cross Y is homeomorphism to Y cross X. Okay. Similarly, it is associative. X star Y bracket star Z is same thing as X star Y star Z. All that you have to do is first have this law in the product space. X cross I cross Y cross I cross Z. You put the bracket wherever you like. So when you take quotient space, that is the X star Y star Z or X star Y star Z. But the identifications are exactly same. Exactly same whether you do it this way or this way. Therefore these two spaces are homeomorph. Okay. Similarly, this is easier. This is commutative. Okay. Suppose X is a singleton space, then what I have? I have X cross I cross a single point. This is X cross I. And then I make identification. Okay. So what is this? At 0, I am defining all the X to be single point. Right? So that will become like a cone. When I come to 1, there is already only one point there. Right? Why? Why there is no identification. All the points of Y should be identified but there is no identification. Therefore, this is just the cone. Okay. So this is one of the reasons why we, even if a single point, it is not X. The operation is not X, it is CX. So if you want an operation to be, to have identity, then you have to take Y as empty set. That is one of the reasons why I have taken. If Y is empty, then you define X star empty space as X. This is just definition. It does not follow logically because if you take the product space with an empty set, it is empty already. Okay. So what we know is if you take one of the things as single point, then this is the cone. If you take one of the two points, then what you get is S naught of S naught star X by definition. These are two points. This will be exactly same as the suspension. Okay. So with the kind of explanation I have given, it should be clear that why this is true. S naught cross I cross X will consist of two copies of X cross I. When you perform identifications around X, we will get two cones and they will get identified because of identifications for S naught. Okay. So you get S naught cross star X is S X. The special cases of this I have already told you. Once it is S X, S of S naught. So S naught star S naught. We discussed this. What is this? It is S of S naught which S 1 and S naught of S 1 which is nothing but S of S 1 which is S 2 and so on. Right? So S naught of S naught, S naught star of S naught is homomorph of S 1. Inductively, we can show that S n star S m is homomorphic to S of n plus m plus 1. All right? You can write S n star S m as inductively S naught star S n minus 1 here star with S m. Okay. By associativity, you take the first this star that will be equal to by by induction star S of n plus m. But now I am taking S naught of that. By this S naught of that will be one more dimension higher namely S n plus n plus 1. The induction star said n equal to m equal to 0. Okay. So that is a proof. Okay. So we can stop here.