 Last time we proved one of the hardest results about topological manifolds, namely if any subspace of Rn is locally contractible then it is a local retract. This was the, this was the theorem. So we shall use that one now for a manifold which is inside Rn. By the very nature manifold is locally Euclidean that means it is locally contractible and by embedding theorems partly which we are assuming that every manifold is a subspace of Rn, some n. Therefore, combining these two we can do some homotopical aspects of the topology of the, of an arbitrary manifold itself. So let us take care of that one now. So this is the first thing that we derive, X be any topological manifold then it is retract of a locally finite countable Cw complex of some finite pure dimension n. Moreover, if X is compact then W can be chosen to be finite. So this is the basic theorem that we have, we can derive it from whatever we have done previously. Start with an embedding of X in R capital N for some n. Take a neighborhood U of X and a retraction R from U to X. So this is what the earlier theorem allows us. In fact, that theorem was more elaborate, only weaker version of that we are using. There is a neighborhood that neighborhood and the retraction is there that is all I am using here. Now go back to the Cw complex structure of those given by those lattice structures that will be used again. Let A prime be the collection of all n cells in P which intersect X and which are contained in U. If an n cell is too big, we do not want them. They must be contained inside the open subset U and they must intersect X, they are far away, we do not want them. So that is the collection A prime. From this A prime, delete all members which are contained in some other member already. Once a bq is taken, no smaller q, no smaller cell will be taken inside that one. It is only for capital N cells. The surface etc will be taken of course. So delete all those members of P which are contained in some other members of that one. Then this will qualify for being a locally finite Cw complex. So let W equal to W of A basically union of all these. So this W is different from the W of the previous theorem by the way. Do not confuse that with W. That W was Rn minus X. Here it is, it is a neighborhood of X now. Then X is contained inside this WA which is W and it is contained inside U because all constitutions, all n cells which make W are contained inside U. So this is by definition we have this one. And what is more is there is a retraction from U to X. It is already there. You can restrict that one to WA to get a retraction from W to X. And our previous lemma tells you that all the required properties of W namely it is a countable right. So whatever you have locally finite countable Cw complex of pure dimension N means what? All the cells are contained inside an N cell. So that is obvious because we have started with all these N cubes and then taken their subspace, phases also that is what it is. So all these things are there for this WA. From the local finiteness it follows that W is finite if X is compact. Once X is compact for each point there is a neighborhood at which only finitely many cells will intersect but we cover X with the finitely many of these only finitely many N cells will intersect it. Anything we do not intersect automatically this W will be finite by the very choice if X is compact. So this big theorem now comes very easily because of the previous theorem that hard work we have done. Now an interesting corollary is that the fundamental group, homology groups of any manifold they are all countable. If X is compact they will be finitely generated also. So what is the proof? Proof is very straight forward. From the theorem we have a retraction R from W to X. If I from X to W is inclusion map what we have retraction means what? R composite I is identity of X. Passing to the fundamental group level R check composite I check could be identity check at the 100 group level. What is that? Pi 1 of X X naught goes to pi 1 of W X naught under the inclusion induced map and there is R. R check comes back to pi 1 of X X naught. In particular this R check will be surjective. This whole thing is identity mapping this must be surjective. But W is what? W is a countable CW complex. A countable CW complex has countably many generators for pi 1. Only you have to look at the one skeleton of that. Any group which generated over a countable set is itself countable. Therefore pi 1 of W is countable is a surjective homomorphism so pi 1 of X is also countable. The next part suppose X is compact then we have seen that W can be chosen to be a finite CW complex. Therefore it is one skeleton is finite therefore there are only finitely many generators. Since this is surjective map again pi 1 of X should be also finitely generated. So this is about the fundamental group. The argument is Ditto for homology group as well. Ditto if you instead of pi 1 if you put H1 here H star for all H star to H star of X to H star of W to H star the same argument will work. So the homology group of manifolds are finitely are countable and finitely generated if it will compact. It finitely generated as a module. Therefore it cannot go on in infinitely many dimensions also. But that was there we had got if X is a CW complex structure. But here we do not have X itself will have a CW complex structure. What we had here is there is a retraction. It is like W is dominating X. So you should be happy that we have proved some very, very handful I mean a result which on which we can lay our hands on. So here are few examples for you I mean exercises for you which you can try out. Of course always there will be a team of members which you who will help you out. Now let us go to the next topic. Let us begin that topic of classification of one dimension manifold. It will be again some strictly lengthy topic. So this is again another very useful and on which we can put our hands and you know that we have learned something. So more generally let us look at any classification problem in science. First of all what do you do? We have to collect samples plenty of them and when we are sure oh we have got all sort of varieties we feel that oh this must be covering all possible varieties all possible objects that is the meaning of that one. Only after that we go for a systematic you know classification as such namely what do you mean by this belongs to this class? What is the meaning of something is connected? What is the meaning of something is compact and so on just like that. What is the meaning of you know what is the meaning of and so on you know you had this biological biology classes you had all these phylums that is why that is the similar one. Even biology was studied in this way of classification. So the final step will be that you should determine all possible mutually different types mutual different classes all possible. In the in the run for that you may find that you have not corrected some kind of thing there must be something like this. So this is what you search for such a such a search what you mean by search such a scientific study always leads to the search just like how the classification of elements was completed. So when you systematically do that you will see that there are certain missing items here there must be something missing here. So you search for it and then you find that it is there that kind of thing is also possible. So for any now coming back to any manifold is locally connected and hence its connected components are all open as well as closed. Therefore every manifold is actually as a topological space the disjoint union of its connected components. Therefore studying connected manifolds is enough why because then you can put disjoint the asteriologically the other things are just disjoint union of these things. So each one you have to study connectivity you can put the connectivity here okay. Quite often we put the compactness and that is not for completeness there is no way of actually reducing the general classification to compact phases this is only for technical reasons. So that things become somewhat easier for us okay. So right now as far as one dimension manifold is a control we are only putting connectivity condition that is all okay. So what are examples of one-dimensional connected manifolds we have okay you have to collect them first samples. So you can take any open subset of R the R manifolds R itself is one any open subset again I want to take connected so they are only intervals but now there are different kinds of intervals open interval half closed intervals and closed intervals and that is the end of all types of connected one-manifolds inside R right. So you would like to go by the way what you mean by all of them now within this class you take any two open intervals we know that they are homeomorphic to each other whether finite or infinite if it is one half closed then it will be always homeomorphic to any other half closed interval if it is a closed interval any other closed interval will be homeomorphic to this closed interval as much we know already actually we have all these things linear isomorphism from finite to infinite we have even the map x equal to tan x also x equal to tan x gives you the homeomorphism from minus pi by 2 comma pi by 2 to more of R. So with this these kind of things equipped ourselves we are sure that there are only three types of manifolds one-dimensional manifold connected one-dimensional manifold inside R open interval half closed intervals and closed intervals okay right once you go out of R just go to R2 there are many other kinds of things namely all conic sections will come circle ellipse hyperbola parabola and so on right pair of lines of course if they if they are parallel then they are manifold but that is just the case which you have already considered any line is homeomorphically the real line there is no problem the circle is quite different circle is a different type okay you can rigorously prove that if there are many other right now you feel that it is true then we we pass on we have to just collect them right now we are collecting ellipse is not a different object geometrically yes but topologically ellipse and circle are homeomorphic curvature triangle no problem it is again homeomorphic to circle square any polygon convex polygon boundary of a convex polygon is homeomorphic to a circle okay hyperbola is not connected you have to take one lap and then it is like a line okay it may be curved but it is homeomorphic to a line okay so this way we can keep on collecting but we don't know anything more even if you go to R3 R4 you know you may have very twisted things and so on but you don't know whether there are any different types of of manifolds so at this day you would like to ask whether we have really reached the end dead end okay so do we get any other types of one-dimensional manifolds if we look for them inside R3 R4 Rn and so on where to look for them we don't have to look outside because any topological manifold is a subspace of Rn that result we have already proved in fact everything will be inside some R3 beyond R3 you don't have to look because one-dimensional manifold by the theorem of which we haven't proved of course yet is embeddable inside R3 okay so here is a surprisingly surprising and present answer that we have exhausted the list okay the answer is no there are no more no more of them so here is a statement of the theorem which we would like to prove while proving we may get some other insight and then we may find a new element so that is possible of course so this is a way conjecture so this is what that is a way conjecture statement this is what we want to prove what is that that xbx connected one-dimensional manifold half-door and second countability is in in our back of our mind okay remember that those two conditions are there okay so here I am stating even for smooth case also but you can ignore that okay right now so take topological manifold then axis homeomorphic to one of the following what are these open interval 01 half closed interval 01 closed interval 01 or the circle s1 that's it there are only four types one can see that each of this type is a different type they are themselves not homeomorphic to each other if they are then I won't put them in the list I will cut cut down my list the problem is that this list is complete why there is not a fifth element here that is the hardest thing to prove okay so in particular what does this mean boundary of axis empty suppose boundary of axis empty along with it is a connected one-dimensional manifold boundary of axis empty here and empty here but for the most two and three it is not right so boundary of xmt it means axis homeomorphic to open interval 01 or the circle if this term is true suppose we just prove this part not using this this list okay not using this suppose we just prove this part namely that any manifold one-dimensional connected manifold without boundary is homeomorphic to open interval or the circle then we can complete the proof of what happens with the boundary axis non-empty so other two will come so what I am telling tell you is that proving this list will be you know the task of proving this one will be halved by this one instead of four of them I have just look at the one and four I have to only classify them okay so how let me see that suppose we assume this part that one and four are the only elements if axis boundary less then take the boundary take a manifold one-dimensional connected manifold with boundary its interior is again a one-dimensional connected manifold without boundary this time therefore the interior must be either 01 or the circle s1 to agree so can the circle s1 be the interior of a manifold with the boundary so that is what you have to see okay and what happens to that if at all so it follows the interior of x cannot be homeomorphic to s1 why because then it will be a compact subset of x okay and hence s1 will be closed inside x but boundary points are closed in the closure so all the boundary points will be inside s1 but then s1 will not be the interior so that is a contradiction okay so s1 if it is a if it is contained inside the arm then it will be disconnected so s1 itself is closed as well as open that is our idea okay so the other one 01 01 can be the interior of some closed manifold yes of course you just look at this list this list itself tells you close 01 0 close take the open interval that is the interior here also 01 open interval is the interior of 01 1 1 and 0 are the boundary points what we want to say that is these are the only two cases now can there be another point there is one point in the boundary this is the case it could be 0 and 1 closed but 01 closed is homeomorphic to 0 closed 1 by reflection around the half okay so the second case is there are two points 0 as well as 1 and the boundary that is this case can there be a third point that is a question okay so it can be very easily seen that you cannot have a third point as an interior point as a boundary point of this one why because once 01 is there similar to s1 that itself will be compact and it is a closed subset of x the manifold x that we are looking for okay a third point which is outside 01 close 01 if it is a closure if it is the boundary then it must be in the closure of this but it must be 01 itself the same argument which we use for s1 once 0 comma 1 the closure is there it cannot have any more boundary points so that completes the picture that these must be all the four of them so we have to only classify manifolds connected to one dimension manifolds which are boundary less so that is the gist of I have given slightly different argument here using you know house dwarfness but you do not need to even say compact house dwarf if you say if you have a compact subset of a house dwarf space then it is a closed subset that is what I have used so there is like different way I have put it without compact so this kind of argument also we will use later so I think we will stop here now and proceed with the classification next time okay we have just made a proposition and what we have to do next time we will continue with this one thank you