 Having introduced abstract simplicial complexes and simplicial maps between them, we assigned a geometric realization, a topology corresponding to a simplicial complex. The next task is to convert the simplicial maps between two simplicial complexes into corresponding continuous functions from the geometric realization. So, today's topic is simplicial maps. Start with two simplicial complexes, K i with V i as vertex sets, a simplicial map between K 1 to, from K 1 to K 2 is nothing but a vector, a vertex map which takes a simplicis of K 1 into a simplicis of K 2, ok. So, once you have such a thing, you want to define a mod phi from mod K 1 to mod K 2, the geometric realization of K 1 and K 2. So, you can call this mod phi also as a geometric realization, it will be a continuous function, but that is the aim, but how to assign, we take the canonical definition namely if phi maps several of the vertices to the same vertex, then at that vertex, you take the sum of all the values of alpha, which are mapped onto that vertex, namely I have defined the mod phi of alpha, alpha is a function remember on the vertex sets. So, mod phi of alpha will be a function on the vertex set of V of K 2, so let V 2 be a vertex of K 2, that means it is inside V 2, mod phi of alpha will be defined as sum of all the alpha V 1 where V 1 is mapped onto V 2, there may be several points which are mapped onto V 2 by this mod phi, so take all of them and evaluate them at alpha and take the sum, ok. So, the right hand side here is a real number, a non-negative real number, this sum may be empty also, there may not be any V 1 mapped onto V 2, this sum may be non-empty, but the sum itself may be 0, for each alpha V 1's may be 0, so that is also possible, the sum total we have to see that it is less than or equal to 1, that is important, it is always non-negative that is fine, if it is bigger than 1 it does not make sense, but this cannot be bigger than 1 because all the time it is sum of, sum of the vertices evaluated at alpha, by alpha, alpha evaluated at only sum of the vertices, even if you evaluate alpha at all the vertices, namely finitely many, the sum total is equal to 1, so this should be always less than or equal to 1, ok. So, as a function it is well defined, it makes sense, one more thing we have to verify, namely when you take mod phi alpha must be a point of mod k2, then its values at all the various points, sum total must be equal to again 1, so if I vary V 2 here, then I am going to take all the points inside V 1, ok, which are mapped on to various points inside V 2, but this sum will be taken over all the V 2's, therefore all the V 1's will be taken care of, therefore this summation will become 1, ok, therefore this summation, this, this left hand side is actually well defined, defines a function from mod k1 to mod k2, ok, this summation is never infinite by the very argument I have given because they are only values of alpha taken at various points, various vertices, more over, one more thing we have to verify, look at all the points wherein mod phi alpha is not zero, that is called support of mod phi alpha, this support of mod phi alpha must be a vertex or must be a simplex of k2, ok, so why that is true, so we need to have support of mod phi alpha as a member of S2, but this is nothing but phi of support of alpha, look at all the points wherein alpha is not zero, only those things will contribute and nothing else, all of this will contribute of course because alpha of V of one of them will be equal to one of alpha phi of that, so all of them will be taken care, so support of mod phi alpha is actually phi of support of alpha, this is a simplex and phi is a simplex map, therefore phi of support of alpha is a simplex, ok, with all these things mod phi is very defined, ok, essentially you have to think of this as summing up the coordinates, so if we are working with product topology automatically this will be a continuous function, right, but we are not working product topology, what we are working with is this weak topology we have to verify that it is continuous, ok, but verifying continuity is easier in the case of weak topology, ok, restricted to any closed simplex F, ok, mod phi is actually a linear map, affine linear map, when I say linear here is affine linear namely mod phi operating upon t times alpha plus 1 minus t times beta, where alpha and beta are elements of this closed simplex, therefore this convex combination makes sense that will be also an element of closed simplex, mod phi of that is t times mod phi of alpha plus 1 minus t times mod phi of beta, can you see why, because mod phi of whatever left hand side operating upon a vertex v, ok, is taken by definition it is this function operating upon, you can see or go back to this definition, operating upon a vertex is the left hand side this alpha operating upon vertices which are mapped onto that, right, so this is operating upon some v2 and then you are taking the sum, but before taking the sum you can pull out this t, t, it is t times alpha v2 plus 1 minus t times beta v2, then you are taking the sum, so t and 1 minus t will come out of respective things leaving mod phi of alpha and mod phi of beta in the integral, so this linearity is very easy to verify, ok, so if you restrict mod phi to closed simplex f, it is continuous because it is affine linear, ok, since this is true for every mod f, ok, where f range is over all the simplex is of k that is even for continuity from mod k1 to mod k2, this is one of the criterion for continuity, that is the definition of the weak topology on mod k1, ok, so we are successful defining mod of phi for each simplexial map phi from k1 to k2, ok, this is in each chunk of simplex on a triangle on a line on tetrahedron so on, it will be linear, only affine linear, ok, there is no origin for all of them, so this is what is going to be what we are going to say, you know, unifying the concept of linear approximation now, so we have generalized just the linear map which is very rigid, ok, so now we can cut down one linear map in a small portion, another linear, another linear and so on, combination of linear, so this is what is going to happen with this one, so in some sense you can think of this mod phi as a linear map, ok, all simplexial maps give back on mod k on the geometric derivation something like a linear map, especially there will be continuous also, another interesting thing here is that under composition this mod is compatible namely first you take phi and then follow it by psi, these two are simplexial maps, you get psi-composified, if you take the mod of that it is modulus of psi-compositive model of phi and modulus of identity is identity that is very clear because look at this definition here, if phi is identity map then alpha v1 will be v1, ok, there is no summation nothing, so phi of alpha will be just alpha, right, similarly if you take another psi on this one, ok, phi mod phi, ok, if you take another component psi of that by variation of definition it is psi of this one you have to take, psi of alpha mod psi of alpha is again you have to take similar definition, so it will be the compositor, so composition is also not difficult to verify, there are finite summation, finite summation you can interchange the summation also if you like there is no need to do that but you can say that the total summation is same thing as once you take this summation and other summation, so composition is also not difficult to verify, ok. Hello sir, like for phi map where simplex is mapped to simplex, is it same happening with mod phi too? Mod phi is called geometric simplex, it is going to be inside geometric simplex, mod f you have to take bracket f closure of f, it will go to phi f closure, it will go inside that that is all, that is what we have to verify, support of phi alpha, this will tell you, this precisely tells you that, no, restricted reach implies it is inside, it is taking inside that one, support does not go out of that, phi of support, this tells you the story, support of mod phi alpha is phi of support of alpha, phi of support of alpha if support of alpha is f and phi of support of alpha is phi f, it is all, that is simplex, what you may have is here you may have an edge, there you may have just a vertex because both the vertices of the edge might have mapped to the same point, here you may have tetrahedron, there you may have a triangle only, if you have an edge here it will not, the image will not be tetrahedron because there are only two elements, no, function, saturated function will take only smaller number of points, if at all, at most that many, these are all finite sets, phi of any of those that will have at most that many elements, but it is simplex in the other simplexial complex, K1 and K2, of course K1 and K2 may be the same also, that is why you can talk about the identity map, identity map is a automatically a simplexial map, now we go ahead what we do with simplexial maps and so on, by a triangulation of a tupholoacal space, we mean a pair, a simplexial complex, K is a simplexial complex and F is a homeomorphism from mod K to X, you start with a tupholoacal space X, it has no structure of a simplexial complex or anything, alright, but K is a simplexial complex and mod K is a tupholoacal space, F to mod K is a, F from mod K to X is a homeomorphism, which is the definition of a triangulation, the word triangulation is borrowed from when you take a surface, you cut it into triangles, but now we can use it for 1 dimension, 0 dimension, n dimension, 50 dimension, all of them, the word is only triangulation, there you do not see triangles, if you go to 3 dimension, there will be triadrons and so on, okay, if X has similar in measure theory, you call the time call volume, 2 dimension volume is area, 1 dimension volume is length, but we do not have many words, so 4 dimension, 5 dimension, we are just taking volume, it is just like that, so if we select a specific triangulation on it, then we call it a simplexial polyhedron, okay, there may be several triangulations for a given tupholoacal space or there may not be any, if there is one, it is a triangulable, if you fix one, then it is called a triangulated manifold or triangulated space or a simplexial polyhedron, sometimes just polyhedron, okay, so do not confuse this polyhedron, it is a convex polyhedron inside RA, convex polyhedron is a special case of a polyhedron, that is all, now we will have some examples here, one by one, okay, recall that the joint of two tupholoacal spaces were defined as a quotient of X cross I cross Y, wherein X cross 0 cross arbitrary Y1 was identified with X cross 0 cross Y2, Y1 Y2 varying over Y, similarly on the other end X cross 1, X1 cross 1 cross Y was identified to X2 cross 1 cross Y, X1 and X2 varying, Y is kept fixed, so this was the definition of what, of the topological joint of X and Y, right, and we had also defined the joint of two simplexial complexes, so these two are not extental, there is a close relation between them, so if K1 and K2 are any two simplexial complexes, then the mod of K1 star K2 is the joint of K1 and K2 is homeomorphic to mod K1 star mod K2, okay, K1 is a polyhedron, K2 is a polyhedron, I mean mod K1, K1 star K2 you do not know that it is a polyhedron, not only it is a polyhedron, it is triangulated by just K1 star K2, mod of K1 star K2 is homeomorphic to this, is the conclusion here, indeed this homeomorphism is such that remember both K1 and K2, mod K1 and K2 are subspaces of this joint and here K1 and K2 are sub complexes of K1 star K2, therefore here also mod K1 and mod K2 will be sub complexes of mod K1 star K2, under this homeomorphism these subspaces are mapped to the corresponding spaces, so that will be easy, that will be, that will follow by just looking at the homeomorphism that you are constructing, okay, so what do we do? We construct the map at the mother level, so this is in the left hand side remember what we have is K1 star K2, I am defining map from mod K1 star K2 to mod K1 star K2, so this is a quotient space of mod K1 cross i cross mod K2, so instead of defining it on the joint I am defining a map on the mother itself, then I am verifying that it factors down to the quotient, this is the way continuous functions are defined on quotient spaces, okay, so how do we define this one? An element of this remember can be thought of as a line from a point here and a point there, line segment, right, so a point alpha and a point beta and a point this some value t, so this is a product, right, there is some identifications later on, so I am not going to use identification, I am directly taking the product, assign it to 1 minus t times alpha plus t times beta which is finally supposed to be the line segment, that is why I am defining it like that. Now alpha and beta, what are they, the 1 minus t times alpha plus t times beta should make sense here, right, right, so how does this make sense here? Alpha and beta are functions from vertex set V1, alpha vertex set V1 to i, this vertex set V2 to i, remember K1 star K2 was defined as vertex set of K1 star K2 is a disjoint union of vertices of K1 and K2, V1 and V2 for example, therefore alpha which is function on, function on V1 and beta function of V2 makes sense provided you think of this as 0 on the rest of the chord, that is the meaning of this, on V2 definitely it is 0, this will be 0 on V1, inside V1 whatever definition is there you take that, here definition you take that and now 1 minus t times alpha plus t times beta make sense is whatever, okay. So first of all you have to see that if I put t equal to 0, okay, then beta is not at all coming into picture and this t is 1, this will be alpha. Similarly if t is just 1, then alpha does not come into picture, this will be just beta. Therefore the very definition of this identification makes that mod phi, this phi defines a factors through a continuous function from phi, namely phi twill for mod K1 star K2 to mod K1 K2, okay. So definition is fine, now we have to verify that it is a bijection, continuous the universe is a bijection, okay. The bijection is already taken care by the very fact, by the expression here, what is the meaning of expression on this side, okay. Here you must remember that simplices here are just disjoint union of simplices here and simplices there, right. So modulus of those simplices are just nothing but points of this way and then you have to take 1 minus t times this plus 1 minus t times that, some total has to be 1, that is why you have to take, cut it down by say 1 minus t plus t, then you take the sum, okay. So there is a map, let us see why this is, why this is a continuous bijection. Suppose Fj, F1 is K1, inside K1, F2 inside K2 are some faces, then restrict the entire thing to just these faces, okay. The image never goes out of that in the definition. So let us look at phi hat, phi twiddle from mod F1 star mod F2, this will be subspace of mod K1 star mod K2, right, to F1 star F2 mod, which is nothing but disjoint union of F1 and F2 mod. Okay, now remember this is a full simplex, okay, this and all subsets of this one will be there inside K1 star K2, alright. How many elements are there here? Cardinality of F1 plus cardinality of F2. How many are there here? They are also exactly same and this map is nothing but, you know, this is automatically an isomorphism. In this picture, which is nothing but, you know, it is 1 minus t times alpha plus t times beta dilute all the elements here. So every element here can be broken up into an element here plus element here. The element terms, they may be 0 also, it does not matter. 0 times 1 minus t times 0 plus t times this one, when you take, it is nothing but this, all the functions here, that is what it gives you. So this is a homomorphism, okay. So here you can directly verify that this is actually linear, affine linear isomorphism, okay. From this one, automatically everything follows, okay. Why? Because if you look at the interior of these things, they are not going to, they are not going to intersect each other. So the bijection on the boundary will be bijection on the boundary from boundary to boundary, interior into here. So on the union also it will be a bijection. The second point is easier, namely continuity of the entire thing is verified by continuity restricted into each, each simplex and simplex is just like this. So that is continuous. Therefore it is continuous on K1's mod K1 star mod K2. The converse also for continuity on from this way biject, sorry the inverse, again the same thing on F1 distribution F2, where F1 and F2 we range your all of them. If you verify the continuity, restricted to that, then it will be continuous from K1 star K2 modulus to the K1 star mod K2, the inverse map, okay. So inverse continuity is also verified by this, this homomorphism. Therefore the simplicial joint corresponds to topological joint when you take the geometric realization. So that is the meaning of this. And this feed today is very canonical way we have obtained. You see there is no other structure here, it is just the structure of these intrinsic structure of these two things are used. So this will automatically give you K1 star K2 star K3 all of them. So SOJ2D or K1 star K2 K2 star K1, okay. So this is what you call as canonical bijection, homomorphism. Under a simplicial map, if K1 to K2 there is a simplicial map, K2 to K3, then you go there, correspond a simplicial map, there will be commutative diagram under these homomorphisms. This is the so generic thing but when you specialize, you can take, pay anything but K2 second one will take a single point. Then by definition K star V is nothing but a cone, the cone over K. First take the cone over K and then take the geometric realization, the same thing as take the geometric realization and then take the cone over K, okay. So this one is a cone over a simplicial complex K is in fact the topological cone over mod K, alright. Because it will be K star, mod K star V star with a single point space in the case of topological zone is just a cone, that is what we have seen. In particular K star V mod is contractible, okay. Similarly, if you take X star two points, S naught, S naught has two points, right. You see the suspension we have defined, right. So you can take K star two points here and then take the modulus that will correspond to suspension of K, okay. So this also this part we have already seen, repeated application of S naught star, S naught star, S naught gives you N plus, S naught star, S naught, yeah, S1. So N plus 1 copies will give you S n. This is the old thing. In particular, if you start with some simplicial complex K such that geometric realization of that is a sphere of some dimension, let us say J, then if you take the joint of these two, then take the geometric realization between the same thing as joint of the two spheres, corresponding sphere P1 and P2, which will be equal to S P1 plus P2 plus 1. So this is what we have seen, joint of these two. So this is a special case of this one. So you can use, you can use many special cases to derive various results from this joint. So let us say some more examples, simpler examples. Now the convex hull of our E naught, E1, E n inside r infinity is easily seen to be geometric realization of standard simplest delta n. That is what we have seen actually by definition. We shall now use symbol, now mod delta n for both of them. Whether it is inside r n plus 1, r n plus 2 or r n plus 3 or r infinity, we will use the same symbol mod delta n for geometric realization, the standard n symbol. All spheres and disks are triangular. It means what? There is a polyhedron, there is a simplicial complex, mod of that simplicial complex will be isomorphic to these spaces. So what are these? Suppose n is 0. What is D naught? D naught is a single point, what is S naught? S naught is two points. So there is nothing to prove there. So singleton, double-ton, they are triangular. The simplest polyhedron is mod F for any q simplex F, which is if its dimension is q, it is homeomorphic to delta n mod delta q, that is what we have seen. So we have been telling that this mod F is homeomorphic to the unit disk D q inside r q. So just to recall all these things, I will tell you how we have done it. Namely start with a set of vertices V naught, V 1, V q for F. An element of mod F is described by describing its coordinate functions, values of the coordinate alpha naught, alpha naught, alpha q. It is a function from this set into i such that sum of alpha i is equal to 1, right, where each alpha is nothing but alpha of V i. Thus mod F is subspace of i raised to q plus 1 elements here, consists of elements alpha naught, alpha q with summation of i is equal to 1. You can see that it is the convex of if you take V naught, V 1, V q corresponding to E naught, E 1, E q. Rotate through an angle pi by 4, this will be having a pi, it will make pi by 4 angle with the x axis or x 1 axis, x naught axis whatever. Rotate it so that it will become, it will be contained inside some horizontal plane r q cross t. Next, choose the origin at the barycenter of mod F. What is barycenter of mod F? It is V naught plus V 1 plus V q divided by q plus 1. You think of that as origin. Then you take x going to x by mod x, x by norm x to get the homeomorphism of the boundary phases, union of all the boundary phases to the round sphere s q minus 1. This x by norm x is modulus 1. So, here is a picture. This is a triangle. You can say this one is E naught, E 1, E 2 and the span of that, its triangle is making 45 degrees. You rotate it, bring this point to the x y plane, the 001 x y plane by rotating it rotation because I do not want to change this geometry. Just for a while. So, then this has become horizontal in the horizontal plane. You choose the orthocenter or barycenter as the origin. Then you take x by norm, project it. All of them, this boundary of this triangle goes to the sphere homeomorphically. This is just like a projection map. Once you have a homeomorphism of the boundary to this one, you can use the so-called cone construction to say that the full disk is homeomorphic to the full simplex. So, I will describe that one. So, this is what is going to have triangulation of the pair. This is supposed to be the full disk d cube and the boundary simultaneously of the pair gets triangulated. Triangulation means now what? You have a simplexial complex in the boundary of simplexial complex giving you, when you take mod, respectively it will be d cube and s cube minus 1. So, I will describe this cone construction once for all. Again and again, this will come. Remember, the boundary of f is nothing but all the simplices except f itself. So, this, I want to say that this is going to give you s cube minus 1. So, this part we have seen. Now, you take the cone over this, that is going to be back, the same thing as f is going to give you f back when you take the modulus. If you take the cone over f, see for each subset of f, you have add the extra point. So, that will give you the full set f. The maximum set f, one subset, one point was missing, the cone will come there and it will be the full set. So, this is one way of looking at the boundary of Bf. The cone over that one is homomorphic to gain mod f. Mod level of Bf is homomorphic. So, this is what we have to use on both sides now. So, this is what I am saying. Now, epic to the following fact, simplexial complex f is isomorphic to star of, this is the star, a single point is a cone. This is the join of the single point and Bf. Therefore, mod f is homomorphic to the cone over Bf. I take Bf and take the modulus of that and then take the cone. First you take the modulus or take the first take the cone or then modulus, this gives the same thing, which is in turn homomorphic to the cone, which again is homomorphic to the full disc. This part also we have seen. So, cone over any sphere is homomorphic to the disc. Let us stop here.