 So, in the first part of this course, we saw several examples of topological spaces. Now we will see the notion of continuous maps between these, which will help us study their topological spaces, their topological properties later on. So, this lecture is about continuous maps. So, definition continuous map, so let X and Y. So, the definition is a very simple one, bit topological spaces, let F from X to Y be a map of sets. So, we are just given a map of sets between X and Y and X and Y happen to be topological spaces, they have topologies. So, we say that F is continuous if for every open subset V in Y. So, recall what this means, so Y is a topological space and therefore, it has a topology tau. So, we say that V is open in Y, if V belongs to that topology tau, the set F inverse U V, which is contained in X is open, is open in X or in short given an open set in Y, its inverse image should be open in X. So, recall the meaning of the definition of F inverse V, this is those X in X, so that F of X belongs to V. So, let us see some very simple examples of continuous maps. So, let X be a topological space and consider the identity map. So, this F from X to X and it is given by, it is the identity map, so it is F of X is equal to X. So, we claim that F is continuous. So, to see this, we need to let U be an open subset, so what we need is F inverse U should be open. So, then F inverse U, but F is the identity map, so this simply equal to U, F inverse U is just equal to U. So, therefore, so as U is open in X, this implies F inverse U is open, this implies F is continuous. So, this is a really silly example. So, let us see a slightly more complicated example. So, let X be a topological space, you do not have to write this always. So, let X let Y contained in X be a subset and we give Y with the subspace topology. So, let us denote this inclusion map by I. So, we claim that then, so Y now has a topology which is a subspace topology coming from X and of course X is has a topology. So, we can ask if this inclusion map is continuous. So, then we claim that I is continuous. So, what do we need to do? So, let U contained in X be an open subset, then we need to show that this I inverse U is open in Y in the subspace topology, but I inverse U is simply U intersected with Y by definition. So, I mean what is I inverse U? I inverse U is those Y in Y says that I of Y belongs to U, but I simply the inclusion. So, this same as saying that Y belongs to U. Right. So, therefore, I inverse U is equal to U intersection Y. So, this shows that I is continuous. This inclusion map is continuous. Let us take a third example. So, this is the same example as above. So, let X be a topological space and let I from Y to X be inclusion of a subset. So, let tau be a topology on Y. So, here we are saying tau is any topology on Y. It need not be the subspace topology. For instance, tau could be the discrete topology on Y. For example, or it could be any other topology on Y such that, but we have this condition such that the map I is continuous. We are given that this map I is continuous. So, then we claim that tau contains the subspace topology tau sub Y. So, we are given that tau is a topology on Y such that the inclusion map is continuous and we need to show that tau sub Y is contained in tau. So, how do we see this? So, this is easy. So, since so, let U be an element of tau sub Y and recall what the subspace topology was. This is let us call this V. This is the intersection of all U in X where U is open in X. U intersection Y sorry not X where U is open in X. So, we have to take we are given that in the inclusion is continuous when Y is given the topology tau. So, what this means is and we have to show that ok. So, tau sub Y is contained in tau. So, we just we start with any V in tau sub Y then there is U in X open such that U intersection Y which is precisely equal to I inverse of U is equal to V. This is by the definition of the subspace topology and now as I is continuous this implies that I inverse of U is open in tau. That is I inverse of U is an element of tau right, but I inverse of U as we have seen is just equal to V. So, this implies that V belongs to tau. So, thus we have proved tau sub Y is contained in tau. So, the point of this example is that. So, this example. So, let us make a remark this example shows that the subspace topology on a subset Y is the smallest topology which makes the inclusion map continuous right. We have proved that if we put any topology on Y for which the inclusion map is continuous then that topology contains a subspace topology. So, in other words the subspace topology is the smallest topology which makes the inclusion continuous. Let us see a similar example. So, this is 4 let X i for i in i be a collection of topological spaces and consider the product with the product topology. So, recall what the product topology was recall that this is the topology generated by the basis B equal to such that A U i is open in X i and B when we look at the cardinality of those i's for which U i is not equal to X i this cardinality is finite. So, now from this product so we have natural maps projection maps. So, this is sorry X j right what is do. So, we have an element over here that looks like X i for i in i and we just send this element to the jth coordinate X j and so these maps we should denote by P j. So, we claim that these projection maps P j are continuous right. So, what we have to do to prove continuity we have to take an open subset in X j and show that the inverse image of that open subset is open in the product topology right. So, for this let U contained in X j be an open subset. So, then P inverse U. So, what is P inverse U? P inverse U is P inverse U is a subset of the product of X i's is a subset of product. So, this is P j inverse U sorry and it is exactly equal to is exactly equal to. So, those elements X i's says that the jth coordinate is in U right. So, in other words you know. So, note that if ok. So, we can write. So, note that we can write P j inverse of U to be equal to it is actually equal to a product U i where U j is exactly equal to U and U i is equal to X i for i not equal to j ok. So, if i is not equal to j then X i can be anything and if i is equal to j then we need that X j should belong to U. So, that is that is a description of this set P j inverse U. So, you can convince yourself that this is correct that is an easy exercise and clearly. So, clearly this is in B right what is B? B is this correction because each of the in P j inverse U each U i is either X or it is U U is open in X j and each U i is either X i or U U is open in X j and X i is obviously open in X i and this collection i and i says that U i is not equal to X i this contains only one element which is j. So, in our case this collection X i is equal to just this one j and which has finite cardinality ok. So, therefore, this P j inverse U it is in B right. So, and which is contained inside tau sub B right the product apology generated by B ok. So, therefore, so this implies that that is the projection maps ok. Now, similar to the earlier example let us show the following. So, let tau be a topology in this product of X i's such that all the projection maps. So, then we claim that tau contains the product apology. So, let us denote the product apology by tau sub B ok or prod ok. So, how do we see this? So, first note that so let. So, if B is the basis that we define for the product apology this product apology and product U i is an element of B right. So, for this element so let j be equal to the subset of those i's for which U i is not equal to X i yeah. So, then cardinality of j is finite by our definition of this set being in B. So, notice that what we can do is so this product. So, note that this product i in i U i it can be written as the intersection of j in j P j inverse of U j ok. So, yeah so I will leave this as an exercise this is left as an exercise. So, for instance if we have U 1 cross X 2 cross X 3 and so on let us say our i is z ok and we have topological space X i not z let us say natural numbers for each i in n. So, then this intersected with X 1 cross U 2 cross X 3 cross and so on this is equal to U 1 cross U 2 cross X 3 cross X 4 and so on yeah. So, this is rough the basic idea. So, using this we can easily prove this assertion in the square. So, we will use this right. So, now our aim is to show that. So, we need to show that the product apology is contained in our topology tau right. So, using the earlier result we have proved result or let me just say using the earlier result we had proved it suffices to show that this basis B for the product apology is contained in tau yeah, but the basis B every element. So, now let us pick up an element if this product U i is an element of B then using this identity we can write it as U i is equal to intersection a finite intersection because the cardinality of the subset j is finite U j. Now each p j inverse of U j is open in tau as the projection maps are given to be continuous right. So, each of these sets is open and we know that in our topology finite intersection of open sets is open since a finite intersection of open sets is open this implies that this intersection j capital J is a finite set. So, this is open. So, this is in tau which implies that this product is in tau. So, this implies that B is contained in tau this implies that tau product is contained in tau. So, therefore, so in view of the above we see that this product apology is the smallest apology we can put on this set product x i for which the projection maps for which all the projection maps are continuous. So, this product apology has this nice property ok. So, let us say make some more remarks now we will explain why the box topology is not such a great topology. So, let us see a natural map which is not continuous when the product is given the box topology. So, let me explain. So, we take x equal to r and r with a standard topology and for each natural number. So, by natural numbers I mean the set of integers positive integers let x n be the topological space r with again with a standard topology right. So, consider the diagonal map. So, this is the map from x to product n in n x n. So, what is this map do it sends x to x and so on ok. So, this is the diagonal map let us give it a name let us call it delta ok. So, we claim that if this thing this product is given the box topology then delta is not a continuous map ok. So, this is just one example. So, for instance. So, in the. So, how do we see this we have to construct a set which is open in the box topology, but whose inverse image is not open in x yeah. So, for this consider the set. So, we can look at this set ok. So, let me just write it as u n contained in the product where each u n is a set minus 1 upon n comma 1 upon n ok. So, then one checks easily. So, this is open in the box topology. In fact, it is in the basis b 1 which we defined and check that delta inverse of this set is just equal to singled in 0 which is not open in x. So, that is the map delta is not I am sorry is not continuous if this product is given the box topology ok and I will end this with an exercise. So, let x be a topological space and let x i be equal to i for i in i, i is any set and. So, then consider the diagonal map this diagonal map and give this the product topology now ok. So, then show that delta is continuous in order to show prove this exercise we will need the following lemma which we will prove in the next class, but you can try it if you want to. So, let f from x to y be a map of sets between topological spaces x and y right. Let b be a basis for the topology on y if f inverse v is open for every v in v then f is continuous ok we will stop here.