 So, we have been looking at this quotient map towards the end of the previous lecture and now we shall see another interesting result which again you will be able to relate with a known result, but let us just go ahead and see what this interesting result is. So, let us consider a linear transformation from a vector space V to a vector space Q right and consider a related not the same, but a related map only the difference being that this one is a mapping from this vector space which obviously looks nothing like this vector space right ok. Such that when t tilde acts on objects inside this vector space. So, what are objects inside this vector space? They are affine sets themselves. So, the way to represent affine sets as we know by now is this right this is given by whatever would have been the result if t had acted on V. So, this is in some sense what we will call an induced map right. So, this is the original map which we are interested in studying and in our attempt to do. So, we have now cooked up this new map whose domain of definition is completely different, but it maps to the same vector space U as the original map. So, because it maps to the same vector space U as the original map therefore, there is absolutely no problem in defining it in like the in this manner. Look at this t tilde it is acting on objects inside this and it is splitting out objects inside wherever TV comes from where does TV come from naturally TV comes from U. So, this also obviously then belongs to U. So, there is no confusion about where this object is taking fellows from and where it is getting mapped where those fellows are getting mapped under the influence of this induced map right. No doubts about this this is clear what this new map is doing in terms of the old map ok. But now look at something interesting we have to again be sure when we are dealing with objects inside this quotient space by the way this is also known as we said yesterday V quotiented by kernel of t we can also call it V mod kernel t in general V mod U that is also sometimes the convention ok V quotiented by U is also V mod U that is another name for that vector space. Anyway, so the point is we have to be sure that when we are taking in different different V's multiple of those V's mapped to eventually the same affine set like we have seen that is the problem. So, once again we have to be sure that for those multiple different ways in which we are referring to the same affine set this mapping must make sense which is to say that this must map to the same point in U. So, if you choose V 1 plus kernel t to be a description of an affine set and your friend chooses V 2 plus kernel t to be a description of the same affine set then after all that same object must get mapped to the same fellow here like we did with the quotient map like we did with everything that we have dealt with in these quotient spaces we have to ensure that that is indeed the case right. So, suppose V 1 plus kernel t is equal to V 2 plus kernel t. So, consider t till is action on V 1 plus kernel t and the difference with the action of t till lay on V 2 plus kernel t. So, because this object is after all the same this must vanish, but this by our definition is given by t of V 1 minus t of V 2 is not it that because of the linearity of t is nothing, but t of V 1 minus V 2 and what can we say about V 1 minus V 2 based on this remember if this is true then what do we know about V 1 and V 2 this is the U or rather here we are calling it U let us say this is the W. So, then if V 1 plus W is equal to V 2 plus W then V 1 minus V 2 the difference there of that is must belong to W. So, therefore, V 1 minus V 2 belongs to kernel of t invoking that result what does this lead to 0 right as expected. So, it does make sense this is 0 of U right. So, there is no ambiguity there is no problem with this sort of a definition of the induced map. If does not matter what name you are giving what you are calling this affine set by whether you are labeling it with V 1 or you are tagging it with V 2 as long as you are picking out the same affine set there is no difference in what the induced map does to that affine set they get mapped to identical points in U right. So, this is of course, the first sanity check in any of these operations pertaining to these quotient spaces right. So, at least this map makes sense that is the first thing we have assured ourselves all right. The next claim is the t till day is linear is that obvious do we really need to prove this yeah it is it is defined like this and t is known to be linear right of course, you might say you already use this the no, but that is ok let us say we do not yet know it is just a mapping from this to this, but we now are sure that this is also linear by virtue of the way in which we have defined it right. So, linearity is checked. So, the next check is what about the kernel of t till day yeah what can we say about the kernel of t till day it means we are looking for that particular affine set which maps to 0 of U that particular affine set or sets which map to 0 of U right. So, look at t acting on v plus sorry t till day acting on v plus kernel of t is going to give me this. So, essentially I am looking for these objects, but this is also by my definition equal to t of v yeah which essentially means v belongs to kernel t does it not see this is equal to this by definition let us mark that out using colors this equality holds true because of my definition of the induced map right. Now I am trying to find out those affine sets which map to 0. So, what exactly are those affine sets those affine sets of the affine sets that are tagged by v's what kind of v's the kind of v's that exactly belong to the kernel of t, but if that v belongs to the kernel of t then I wonder what indeed is the subject right. This essentially means that v plus kernel t is equal to 0 plus kernel t is it not because v belongs to kernel t therefore, v minus 0 also belongs to kernel t therefore, v plus kernel t is the same as this, but what is this object is this not the additive identity element sitting in the quotient space right. So, this is equal to the 0 of v mod kernel t which means what have I shown I have shown that only objects that correspond to the 0 in my domain in this case the domain of t till there happens to be v mod kernel t which is my quotient space. So, only the 0 element in the quotient space when passed through this induced map can produce the 0 in the co domain of the induced map, but that is the classic definition of injectivity which means that I can now safely answer this question as the 0 of v quotiented by kernel t implying that t till day is injective right. It is linear it is injective yeah. This new sort of an induced map that I cooked up based on any arbitrary map that I had any arbitrary linear map turns out it is linear it is also injective right. What else? I am going to make a claim that ok and it might seem like something big, but it is really very trivial and straightforward. Look at the definition every object that belongs to the image of t has a representation like so. So, go ahead and define an affine set by that corresponding v call it v plus kernel t and that object by your very definition by the way you have defined this induced map it implies that there will be a pre image in v mod kernel t which maps you to any point in the image of t right. Please ask if there is a need out about this is this clear right yeah. What I am saying is that t till day the induced map is a surjection between the domain and not the entirety of the image or entirety of the co domain which might be u, but rather within the restriction of u what restriction of u the part of u that forms the image of t. So, you have v quotiented by kernel t and you have u I am not saying that this is a surjection on u all of it no I am just looking at the part of u that corresponds to image of t that is residing inside u right it is a subspace of u. Now, if that part is what I am focusing my attention on and I pick out arbitrary objects from here that means there must be some v in the vector space v for which t v is equal to this point. So, if I call this point say u then there exists v such that t v is equal to u, but if there exists v such that t v is equal to u this also means that there exists v plus kernel t the affine set which when passed on as an argument to the induced map also gives me the same u which means that any object that I pick out in the image of t has a pre image in this under the map that is t till day. So, t till day is a surjection between v quotiented by kernel t that is v mod kernel t and the image of t and already we have seen before those that v I mean t till day is an injection between this and this. So, if it is an injection between v mod kernel t and u it is also an injection between you know if you if you just think of this small aspect and its pre image whatever that is right that is also an 1 to 1 because the whole map is 1 to 1. So, of course, a restriction of that is also going to be a 1 to 1 or an injection. So, what am I going to claim with all of this if I stitch them together the final claim is going to be this t till day is a linear bijection between sorry v quotiented by kernel t and image of t what do we have the moment we have a linear bijection like. So, we have an isomorphism which means that that is just another way of saying that v quotiented by kernel t is isomorphic with the image of t what does this result remind you of this is the first isomorphism theorem yes it is the rank nullity theorem is it not basically it is saying you get rid of all the redundancies whatever t does to the kernel is uninteresting. So, if you just choose to focus on an appropriate appropriately constructed vector space you can basically capture all the non trivial actions of any linear operator t and this quotient space exactly provides you with that handle right. If you apply the dimension criteria in case these are finite dimensional then this is exactly the rank nullity because you see the difference between the dimension of v and the dimension of kernel t is equal to the dimension of the image t which is nothing but rank nullity theorem right. So, this is the first isomorphism theorem, but we have never invoked anything about dimensions here at all we have not even gone through any construction of basis or anything if you notice in this entire result it is just an isomorphism we have just got any map based on that map by an by a superior understanding if we may call it that of quotient spaces we have constructed an induced map and now by invoking the bijection property and checking that it holds for this induced map we can infer about this. So, remember this vector space looks nothing like v, but it helps in capturing the action of t completely apart from what t does to the kernel of t which is nothing interesting it just takes it to 0, but anything that it takes it to some non-zero that is this image of t is completely captured by this right. So, this is where we will bring this discussion on quotient spaces to a close ok and this is where the syllabus for mid-sem also ends whatever what we are going to do subsequently is not going to be part of your mid-sem, but it will obviously be incredibly important because we are launching into the second part of the syllabus hereafter. We have not invoked anything about dimensions so this is much deeper than just finite dimensions we are not using dimensions see isomorphism may be between vector spaces that are not necessarily finite dimensional right, but here nothing like that is being used.