 So, last time we had introduced the definition of mod k where k is a differential complex. So, today we will study the topology on mod k. Before that let me just recall what was the definition and a few notations. So, we took mod k as subset of the power set, the product set I cross I cross I taken V number of times, number of vertices in V whatever that many copies of I is taken which same thing as taking functions from V to I, additional condition that is that the support of alpha many points wherein alpha is not 0, they must be the numbers of S and the sum total is equal to 1. After that we introduced two notations for each simplex, closed simplex F bracket close the bracket this is all V for which its support is contained inside F, the same thing as for every V it should k inside k minus F it is 0. The second one on this open simplex is all alpha in case as that alpha V is 0 if and only if V is in k minus F that means if it is in F it is not 0. So, it is same thing as support of alpha is equal to F, then boundary of F was defined as closed simplex minus open simplex. So, let us come to the topology now. I will first treat the finite case because it is simple that is all and it gives you motivation, k is finite is same thing as V is finite. Then mod k is a subset of I cross I cross V number of times as many as elements of V, so it is a finite Q I cross I cross V the finite Q. So, we can just simply take the subspace topology from this I power V which is I power n where n is the cardinality of V. It is a familiar object for us. It is inside R power n. So, take the induced topology namely a subspace topology then you call it as geometric realization of k. The geometric realization of k and denoted by mod k is not just a set it has a topology. What is this topology? At least in the case of case finite it is the subspace of I power V defined in a particular way. It is a subset of I power V already it takes a subspace topology. You already notice that this given by a closed condition. So, mod k is a closed subset of this one in particular it is compact. Closed subset of a compact that is compact. And each closed f is again a closed subset of mod k that is also given by closed condition. Its conditions are all given on the coordinate functions equal to something intersection several of them. So, it is it becomes a closed set. Note that its closed simplex f as a natural convex structure coming from that of I raise to V. I raise to V is a cube right I cross I cross Y that is a convex set. So, that convex set the same structure is there inside mod f the closure of closed simplex f and closed simplex f itself is a convex subset for if alpha and beta are inside this closed simplex and t is between 0 and 1 and t times alpha plus 1 minus t times beta is again is an element of f. What is the meaning of this? Take this one say this is a gamma. Gamma of V is not 0 implies that V must be inside f. If gamma of V is not 0 if gamma of V is 0 for alpha it will be 0 outside but beta also it will be 0. So, the combination also it will be 0 right. If V is a point outside f then this will be 0 that is why it is inside. So, moment gamma of V is 0 for every V outside k minus f or inside k minus f outside f then it is an element of closed simplex. We may identify mod f with a convex subset of I raise to f you see I raise to f is a subset of I raise to is actually a quotient of I raise to V because f is subset of V you can take only those those things but this I raise to f is another copy of I cross I cross I and so on. So, there is an identification of this one also and of course, inside that we have to again take the condition 2 that subtotal is equal to 1. So, what is the thing I have I have just excluded all those coordinates for which this point is 0 anyway. So, that is why I have taken only this part that is all. Now comes the more interesting topology the relative interior of closed f, a relative interior of a set inside a topology topological space is not exactly the interior of f as a subspace of the topological space x. The relative interior has to be defined inside the space it is a tricky thing. So, this can be done for geometric things. So, that is what we are going to do. For example, suppose you have open interval your closed interval contained in R2 as subspace of R2, but it is inside R. In R the closed interval has its interior which is open interval, but as a subspace of R2 it has no interior. So, this open interval is the relative interior of this mod the simplest irrespective of where it is contained in. So, we are defining it irrespective of where. So, this is what interior of f is precisely those alpha for which alpha v is not very final. This is the another notation now which is we have already defined it as open simplex f. Now, you know why it is called open simplex f it is interior of mod f. So, inside this bracket f, round bracket f, open f is an open subset. The boundary of f by the very definition will be a closed subset and that will be the union of property at least one of them one of the vertices is 0. So, it is inside that proper subset union of subsets. Second point again I am telling you is closed simplex f is homeomorphic to the geometric realization of mod f where f is thought as a simplicial complex. Given a simplex you take all its subsets that is a simplicial complex. Then what is the geometric realization of this? You have to take all functions defined on f the f is a vertex right now. 2 i that is why I have defined mod f as can be identified with i raised to f with condition summation of equal to 1. And the first condition is automatic because every subset of f is already inside the simplex complex. So, this mod f this bracket f is a subset of some simplex complex mod k. But its geometric realization you know is it can be identified with geometric realization of the simplex the simplex complex f. Each map f from f to i can be identified with the map f with the alpha from v to i which is 0 outside f and you are doing the same condition. Now inside k inside mod k what happens to the intersection of f 1 and f 2. Look at f 1 and f 2 are two simplexes ok. I can talk about mod f 1 and mod f 2 they will be subspaces of mod k right. I want to say that it is precisely equal to the mod of f 1 intersection f 2. f 1 intersection f 2 if it is empty it should be empty. Those which have supports inside f 1 those which have support inside f 2 if f 1 and f 2 are disjoint ok. There cannot be any such element right. They will be totally distinct thing. Therefore, if this is f 1 intersection empty this is empty. But suppose this is non-empty then this is a simplex. This is a phase of both of them. This is contained in the f 1 and f 2 contained both of them. And I want to say this is equal and this is very straightforward. Namely this is point set topology. You see just you look at what is the meaning of an element here. The support of that element is inside f 1 intersection f 2. So, it is inside f 1 as well as f 2 and gone or something. So, the two subspace topologies on the intersection coming from f 1 or coming from f 2 they coincide ok. So, this is very important thing we have to observe. Why because modules f 1 and f 2 subspace of mod k itself directly. So, whether you come from f 1 or come from f 2 it must be instant. So, the family of closed subsets mod f or you can write it as now bracket f closed bracket f what mod f they are the same ok. So, whatever tentatively I have put at bracket f is now the same thing as mod f. So, f is a phase of k as soon as it is a phase of k it is a closed subset. Look at all these that is a family of closed subsets ok. Everything in k must belong to one of them that is by definition. Every element of mod k has support one of the phases. So, this is a cover for mod k, but it is not an open cover it is a closed covering, but it is a finite because k itself is finite I have taken ok. Clearly g contained inside mod k take any set it will be closed if we find only g intersection mod f is closed in mod f for every f in a k ok. This is the property of any finite closed covering for a topological space. It is an elementary property how you define a continuous function on two closed sets continuous on each of them and on the intersection they agree automatically it is continuous on the whole space. So, this is true for finite covers which are closed on each of them if you verify continuity and on the intersections the functions agree that is function is defined on the whole of it then it will be automatically continuous right. So, it is same thing as saying that g intersection mod f is closed for every f if and only if g is closed. So, this has the standard wording here namely it is a weak topology with respect to the above covering if you have not if you have forgotten your concept topology. This property gives us an idea how to get a good topology on mod k when k is infinite. This property is going to be our guiding principle for defining a topology when k is infinite ok. So, that is not finite. So, let me complete this one. Let V be an infinite set and k is always a subset of i raised to v it is a closed subset of i raised to v ok by definition. You can put the following metric now first I will put a metric on on this mod k of course it will be from i raised to v for i raised to v v is too big. So, I do not want to put a metric on that all right. What is it take any two elements alpha and beta alpha will look like a finite some alpha v i right because alpha is not this alpha is nothing but alpha of v i right. We have seen that it is a i th coordinate v i th coordinate function that is the meaning of this one and that is not 0 only finitely many similarly beta is a finite combination like this ok. So, this is just a tentative notation if you identify v i with an element of mod k how do you do that? It is v i operating upon v i is identity is 1 and v i operating upon v j is 0 right. So, that is the element these are the standard basis elements for i raised to v ok. So, in that sense this summation makes sense ok summation beta v i belonging to mod k we know that alpha i and beta i are 0 except for finitely many i which is same thing as that these two are finite sums ok. Then I can define distance between alpha and beta to be just the square root of the sum of the squares of the differences just like in the Euclidean squares. Take the coordinate functions take the differences take the square of that take the sum and then take the square root this makes sense because only for finitely many i these are non-zero no i raised to v I cannot do that because this is too few then infinitely many coordinates may be non-zero here ok. So, for mod k I can do that because mod k is contained inside i raised to v for which so this makes sense direct verification just like in a direct sum of vector spaces this will become a metric ok and we have a metric topology now if its k is finite this is just the Euclidean metric in r raised to n because all those things I have to involve now ok. So, whatever topology we have done we have analyzed for k finite that is metric topology only from Euclidean space alright, but we are using a different property of that space and then we will have to do this one this metric topology on infinite thing is not a good thing for us so we will modify it ok. Let us denote this mod k with the metric topology kd suffix d because we are not finally interested in this one which is only a intermediate step clearly it coincides with the Euclidean topology when v is finite. Therefore, topology induced on each mod f from mod kd is the same as the intermediate topology that we have been taking namely of the simple existence on that is what we have done so far ok. For each mod k for each mod f the topology that we have defined using an isomorphism using a homomorphism of this one with a standard simple act that topology is the same as the subspace topology from kd. Now I want to redefine the topology and the whole of k by declaring a subset of a subset a of k is closed different only a intersection f is closed for every f is closed. So, this property which we have verified for when k is finite this becomes the definite defining property for the topology. A subset is closed we find only intersection with each f is closed. So, this is the weak topology with respect to the family of closed subsets. Clearly then this topology is finer than kd if a is already closed inside kd then a intersection f will be closed inside kd as well as mod f because mod f has the same subspace topology from kd. So, all closed subsets of kd are inside this topology weak topology. This is a funny name this is called weak topology for some different reason, but with respect to this kd it is stronger than the topology of kd because everything closed subset inside kd is also closed here it is a finer topology than kd. The metric topology has certain properties the mod k is even better it has more open sets. So, one good reason for taking this weak topology I have put it in a bracket because this is not weaker than mod kd, but it is actually finer. Instead of the metric topology is that constructing continuous functions on k becomes simpler and is coherent with our theme that the simplices are the building blocks of simplicitial complex. So, this is the motivation I will restate it as a theorem take any topological space x okay or a again this x should be y they take any topological space y a function from mod k to y is continuous if and only restriction map of f to mod f to y is continuous for every f. What is f? f is a simplex in k. We write this one all the time instead of writing k is equal to a comma s it is actually for every f inside s the simplex okay. Likewise a function from mod k cross i to y is continuous if and only h restricted to each mod f cross i is continuous for every phase f in k okay the first one is actually same as definition of the topology and mod k right. Take a subset here which is closed its inverse image is closed here is what I want to show the inverse image here is nothing but f inverse of f restricted to f inverse of that set okay by definition if this is closed but if this is closed for every f then that is closed because this is the victor power and gone earthly okay this is fine but now same thing is claimed for the product for this careful proof of this one has to be written down and that uses functions spaced topology okay to prove the second part for first part I have done we note that i is locally compact in fact i is compact right whenever locally compact host of space I can use exponential correspondence theorem right. So I am just recalling the exponential correspondence theorem if x is locally compact y and z are any topological spaces exponential map e from y power x cross x to y even by e of fx is continuous and any function from any topological space z to y power x is continuous if and only if e composite g cross i g cross identity is continuous so these two theorem this was part a part b and so on of that exponential correspondence theorem. So I use now x to be the closed interval 0 1 because there is a locally compact right to put x si and z dash k then you can define g from this k mod k to this y raise to i y is arbitrary space by the formula g z must be a function from i to y to the h of z t remember h is given from mod k cross i to y so first coordinate is in mod k and second coordinate is in i then to take e composite g cross identity that will precisely h that is the definition. Hence h is continuous if and only if g is continuous by the exponential correspondence. Now g is continuous from the first part applied to y power g if and only if g restricted to each mod f is continuous for every mod f inside k right which is same thing as now go back to exponential correspondence this g restricted to f is continuous if and only if e composite g cross i restricted this one which is same thing as h of f restricted to i h restricted to mod f cross i is continuous. So exponential correspondence is used first go to the power here and then come back both both sides you have to restrain on this part has to be used we will stop here next time we will study maps between temperature complexes what kind of maps we have to study there okay thank you.