 So, in the previous lecture we proved that 3 important maps or 3 maps which we see very often are continuous. So, in this lecture let us continue with some general properties of continuous functions. So, we begin with lemma let x, y and z be topological spaces and suppose we are given maps from x to y is f and g from y to z such that f and g are continuous. Then the assertion of the lemma is g compose f is continuous and the proof is obvious and is left as an exercise. So, this is one very basic lemma and another very basic lemma is the following. Let f from x to z be a continuous map. Let y contain x be the inclusion of a subspace be a subset with the subspace topology. Then the restriction of f to y that is the map f compose i this from y to z is continuous. So, proof. So, f is already given to be a continuous map and in we had proved earlier we had seen earlier that with the subspace topology the inclusion i from y to x becomes continuous. Thus, so f is continuous and i is continuous applying the previous lemma we get that f compose i is continuous. So, this proves this lemma in order to prove the next lemma which is also very important and very basic we need some notation. So, let us introduce that. So, here f from x to z is a continuous map. So, let us assume that the image f of x is a subset of y which is a subset of x. So, let i from y to z denote the inclusion. So, what this means is that this means that the map f factors as. So, we have x x to z we have map f and it turns out that the image of f is actually contained in y which means that there is this map f naught from x to y and what is f naught f naught of x is simply f of x. So, with this notation with notation as above the map f naught is continuous ok. So, let me emphasize that we emphasize that y has the subspace topology from z. So, let us prove this lemma. So, let u contained in z be open then i inverse u is simply u intersection y is open ok and every open subset of y has this description ok. So, to show that so thus to show that f naught is continuous enough to show that f naught inverse of u intersection y is open in x for every u open in z, but f naught inverse of u intersection y can be written as f naught inverse of i inverse of u which can be written as i compose f naught inverse of u and i compose f naught is f. So, this is precisely equal to f inverse of u and as f is continuous f inverse of u is open in x thus f naught inverse of u intersection y is open in x thus f naught is continuous. So, this proves three basic lemmas which are related to subspace topology and now we have a similar result very basic result about continuous maps related to the product topology. So, let us quickly come to that. So, proposition or before we prove this proposition let us introduce some notation. So, let x be a topological space and so let i be a set and suppose. So, i the size of i could be infinite and suppose for each i in i we are given a topological space y i. So, in addition to this suppose we are also given maps of sets f i from x to y i. So, given this data we can construct then we can consider the map. So, we can define a map f from x to this product of y i. So, what is f 2 f of x it just puts together all these maps f i. So, this is equal to f i of x this is an element of this product. So, with this notation we have the following proposition the following very useful proposition the map f is continuous if and only if each or f i is continuous for all i. So, let us prove this. So, obviously to make sense of a continuous map we need two topological spaces and x is has a topology and this product is being given the product topology. So, let us prove this proposition. So, since so first let us assume that f is continuous. Now, since this product has the product topology the projection maps recall we had seen the projection maps. So, what are these? So, we can look at the projection to the jth coordinate. So, what this does is given a tuple y i for this i in i it sends it to y j and we had denoted this projection map by p j. So, these projection maps are continuous right and since composition of continuous maps is continuous. So, we are given f and here we are given p j. So, the composite is continuous yeah, but the composite is precisely f j therefore, here x is going to f i x and then we are projecting on to the jth coordinate. So, we just get f j x as the composite is f j x it follows that f j x f j from x to y j is continuous for all j. So, this proves one part of the proposition. Conversely let us assume that each f i from x to y i is continuous. So, recall that the product topology has as basis open sets product of u i's where each u i contained in y i is open and the set j the set of those indices j in i says that u j is not equal to y j this is finite. So, therefore, so and as we have seen many times now to show that f is continuous it is enough to show that the inverse image of basic open sets is continuous if the topology has a basis and this the product topology is given by basis. So, thus it is enough to show f inverse of any basic open set is open. So, but notice that f inverse of this an easy set theoretic check which is left as an exercise u i is equal to intersection of i in y f inverse f i inverse of u i. So, yeah this is an easy check and so let us consider this right hand side. So, if, but this is equal to so we can divide the intersection into two parts right. So, intersection over j this is so let us call this set j i in j f i inverse of u i intersected with i not in j f i inverse of u i, but now note that if i is not in j then this u i is equal to y i and therefore, f i inverse is all of x i. So, this is equal to intersection i in j f i inverse of u i intersected with intersection i not in j x yeah, but this simply equal to intersection i in j f i inverse of u i. Now as each f i is continuous this implies f i inverse of u i is open in x right and as cardinality of j is finite and finite intersections of open sets are open this implies that intersection i in j f i inverse of u i is open yeah which implies that f inverse of product i in i u i is open which implies that f is continuous. So, let us just make a remark and that is one of the reasons this remark is one of the reason why we reject the box topology. So, the above proposition fails if this product is given the box topology and recall the diagonal exercise given before. So, let me say ok. So, having made this remark. So, this proposition has some nice consequences and what are these. So, let us see some consequences. So, proposition this proposition and the earlier theorems we had proved about addition etcetera addition and multiplication. Let f comma g be two maps from x to r continuous which are continuous yeah. Then we have maps f plus g. So, how do we define this map this is a map from x to r and it is defined as follows f plus g of x is defined as f of x plus g of x. Note that f of x is a real number g of x is a real number and therefore, we can add them and similarly we have another map f g of x f g from x to r. So, this is defined as f g of x is equal to f of x times g of x ok. So, the content of this proposition is these maps are continuous and the proof is easy. So, first using the previous proposition we get a continuous map from x to r 2 given by x goes to f of x comma g of x right and that this map is continuous as both f and g are continuous ok. So, then. So, if r cross r to r we have the addition map and we have the multiplication map. So, since the standard topology this is this was an exercise since the standard topology on r 2 is the same as the product topology and we check we proved that a and m both these maps are continuous in the standard topology this implies that they are also continuous continuous in the product topology right. So, therefore, what we can do is we can look at this is f and we have the addition map and we have x 2 we have the multiplication. So, thus the composite of continuous functions being continuous this implies a compose f which is equal to f plus g and m compose f which is equal to f g both these are continuous. So, this completes the proof of the proposition. So, before we end today's lecture we will prove one final proposition which is similar to the above. So, this one says the following let f comma g from x to r be continuous functions such that g of x is non-zero recall what r star was this is notation for r minus 0 for all x and x ok. So, then we can define a function h of x it is defined as f of x by g of x this h is a function from x to r this function is well defined because g of x is not 0 and h is continuous. So, the proof is very similar to the proof of the early propositions. So, proof. So, first note that as the image of g of x is contained in r star using our earlier result it follows that. So, let me just mention what that earlier result is I am talking about this lemma. So, using our earlier result it follows that the function. So, g we can view as a function g naught from x to r star and is continuous when r star has the subspace topology. So, in all these propositions and theorems r has the standard topology ok. So, we will abuse notation and continue to denote g naught by g ok. So, g is actually a map from x to r star and r star beginning with the subspace topology g is continuous. So, therefore, now can now define a function 1 upon g from x to r r star. So, this is defined as. So, first x to r star we have g and on r star we have this continuous map which takes x to x inverse or let me write y to y inverse ok. So, g is continuous this map y to y inverse that is continuous. So, therefore, the composite which is which sends x to first it sends it to g of x and then this goes to 1 upon g of x yeah. So, the composite is continuous ok. So, this composite function shall be denoted. So, this is from x to r star, but now we will just take the inclusion of r star into r and this composite is also continuous because r star has the subspace topology. So, finally, we get that once again we will abuse notation and we get that the function 1 upon g from x to r this is given by x goes to 1 upon g of x is continuous. So, with this now we define a map from x to r cross r which is given by x goes to f of x comma 1 upon g of x. So, both the coordinate functions are continuous and this implies that this function f is continuous and then we have the multiplication map ok. So, the multiplication map is also continuous. So, this is sent to. So, thus this shows that x goes to f of x by g of x is continuous. So, let me fin end by making a remark in the above we only showed that or in the proposition we proved about or in the theorem we proved about m being continuous we proved m is continuous when r cross r which is r 2 has a standard topology. However, note that the standard topology r cross r is equal to the product topology thus m is also continuous when r cross r has a product topology ok. And here over here note that r cross r has a product topology and it is the property of the product topology that we have used that this function capital F is continuous because both the coordinate functions are continuous. So, finally, remark the above results show that the set of real valued functions on a topological space x has nice algebraic properties. So, we can add them we can multiply them and if g is non-zero we can form the function f by g. So, we will end this lecture here.