 I said it was an exercise, but I didn't put it as an exercise here. But still it's a good exercise to think of a couple of systems like this, where here you have a periodic point. But H, the pre-image of this periodic point, does not contain any periodic point. Also, you can think of a dense orbit such that the pre-image does not contain any dense orbit. Or the system above. So that is why we call it a factor. This dynamics is contained into this dynamic. So we are going to see, we have also defined the shift map. And we are going to see that this map is a factor of this map. OK? This is a summary of what we are going to do today. We have to work in order to prove that. So let me remind you what the coding was. We have the 2x mod 1 map. And we have any point, like the red dot there. And if it is in the left interval, we are going to put the code 0 for it. And we iterate this point. And we see again in which interval it is. It is again in the left interval. So we are going to put 0 and do it again. Again 0, again 0, again. Oops. Now it's in the right side. So we put a 1, a 1, 0, 0, 1. And now we can see that this is periodic. So the code will be eventually periodic too in this case. OK? This coding is not unique as we have seen. We have problems when the point is here, or here, or here, which is the same point. We have two symbols. We can choose two symbols, 0 or 1. There will be two possible sequences that define the same point. OK. We are going to see now, on the other hand, how we obtain a unique point from a sequence. But here in the slides, it appears what I have just told you. So be patient. We have this. We are going to say that H is a semi-conjugacy. If it is, it makes this diagram commute like we have already seen. And I have not told it in the slides, but we have to require that H be continuous and surjective. OK? And I prefer the notation that F is a factor of G because it is more clear which is to coordinate to which. OK. So F is a factor of G that is the diagram. And we are going to see that the, let's call it E2 because we are going to do it later for any expanding map. So let's call it E2. We are going to see that E2 is a factor of the shift map. This is what we are going to do today. OK. That is, whoa. So we have to look for an H. An H that will go on sigma 2 to S1. OK? And makes this diagram commute. So today we are going to work in producing a continuous map from E2. We're not going only to do that. We are going to do it for any expanding map. But first let me do it. First let me do it for E2 because E2 is linear and everything is simple and nice. And later we will see for any expanding map of degree 2. That means any expanding map with two branches. But first let me do it in the simplest case. We are not working out the details. The details are, you will work out the details in the exercise session. OK? I have already told you how we code this. Now let's see the contrary. Let's assume we have a sequence and let's choose a point. OK? In the first slides we had a point and we built up a sequence. Now let's assume that we have a sequence and we want to produce a point. So this X bar is a sequence. The sequence will appear in a minute. But we are going to produce a point. So let's suppose that the sequence begins with a zero. So this image we need to define H. So we have this. We have this. And we need to produce a point. So we have a sequence and we need a point in the circle. How do we produce a point? OK? Let's suppose that the first element of the sequence is a zero. So H of X, we are going to define it to be in delta zero, the left interval. OK? We have all these options for H zero, for H X. Now let's suppose that the second is again zero. We want this. We want that E2 of H of X be again in delta zero. So it will be in the left interval of the left interval. Now the third is again a zero. So it will be in the left interval of the left interval again. Now E3 is again in delta zero. So in order to do that, we need that four iterates are in delta zero. So it will be in the left interval of the left interval. Now we have E4 is in delta one. Sorry, I have... Now this means it is in the right interval of this left interval. So the point will be here. OK? So in each iterate, this is what we want to happen. OK? We will tell you in a minute why we want this. Why we want this? Because we want that we want this be a semi-conjugacy. So we want that H composed with sigma be E2 composed with H. So we want that H composed with sigma be E2 composed with H. But if this happens, you will have that if we have H composed with sigma squared, it's going to be H composed with sigma composed with sigma, which is going to be E2 composed with H composed with sigma, which is going to be E2 squared composed with H. If you do it, you can do it n times and you will get that this happens. So we want this to happen. OK? So that is why we want E4 composed with H be H of sigma 4 of X. But sigma 4 is sigma 4 of H. OK? You understand what I'm doing? So I will go, I, depending on which digit appears here, is where I will place my H of X. OK? And I will do it one step at a time. But I had the advantage that since this map is expanding, each time I do this procedure, the interval of options is reduced by one half. OK? So in each step I have less and less possibilities and in the limit I will have just one point. This is the intuitive idea. We will work it out. You will work it out. But this is the idea of how we will get a point. OK? Have more and more steps and in each step you will be getting a smaller and smaller interval. This is more or less clear. How this is done? OK. Well, this is periodic so we will have in the limit a unique point. So how are we going to define this H? OK. This H of X going to be precisely the limit of all this. So H of X will be such that E, sorry, yes, E n of H of X will belong, this where X, OK? So for each digit we want this. So we want this to happen for all n greater or equal than zero. So this implies that H of X will belong to E to the minus one two. And so this implies that H of X, this is how we want to define this. But we have to prove that this is non-empty and that this is only a point. OK? If this is non-empty and it is only a point, then we can define this like this. And this is part of what you are going to prove today. But let me give you hints about this. Let me give you some hints. Why this exists and why it is not empty and why it is a unique point. From the picture here it was more or less clear what happened. OK? But we have to prove it. And in the linear case it's not that difficult. In the expanding map case we are going to do it. The first thing is you have to prove that this consists of two to the n intervals of lengths one over two to the n plus one. And this can be done by induction. OK? It is clear for the first step and then you assume it holds for one step and you prove it for the next one. And of doing directly without passing through this proving inductively that this intersection consists of only one interval of lengths one over two to the n plus one. You can prove directly the second step if you like. OK? For each fixed step this is an interval of lengths one over two to the n plus one. Here you have it. In any case if you finish doing all this you get that h is a well-defined function. So we only have that f h is a well-defined function and it is an exercise but it is pretty easy to prove that because of its form e2 composed with h equals h composed with sigma. This follows more or less easily from the definition. OK? Because you apply e and you are considering the sequence from there on. OK? And so this implies this. Or if you like, if you apply sigma here you are just considering this. OK? And this is just you renumber this e to the minus n with one less n. So by the definition, because of the definition it's from the definition sorry it's easy to prove this it's just working out the symbol. So we are going to prove that it is continuous for a general expanding map in some minutes but I would like to give you as an exercise for the precise example of 2x mod 1 that h is continuous. And it's a good exercise just to have in mind you have to use the topology we have defined for this. We have to use this metric. The metric, remember the metric we used here we have defined that two points were 1 over 3 to the n plus 1 close if and only if their first n plus 1 entries were equal. If two points are close enough their first n entries are close enough. So then this is going to be the same for n steps. So they will belong to the same interval of lengths 1 over 2 to the n plus 1 so their images will be close enough will be 1 over 2 to the n. So work it out. I am just giving hints. And it is surjective. I want you to think about this. I will not tell you why but it has to do with what we spoke about coding. It will be surjective but in general it will be non-injective. This I have already told you hints. It is non-injective because we have for instance this case that has two symbols. One is 0 1 1 1 1 1 and the other is 1 0 0 0 0 0. No, 0 1 0 0 0 0. 0 1 1 1 Here it is 0 and then 1 1 1 1 Yes and the other is 1 Yes 1 0 0 0 0 What is wrong with it? If you put 0 here and then all 1s then you get this point. If you put 1 first and then all 0s then you get this point again. Let me tell you what we are going to do in the next class. I cannot erase this. What is the idea? The idea of this is that we are going to take any f now from s1 to s1 expanding and of degree 2 with two branches. And we are going to prove if you do all the exercises you have got already that this is a continuous surjective map and this is a factor of this. But we are going to prove that the same construction applies and in general for this case for any expanding map of degree 2 we will have there is a continuous surjective map from the shift map on to the expanding map. It will be the same construction and I don't know you this is hard to visualize because it is an abstract space it is totally disconnected space I see it in my mind as a counter space so that the two these two will go to the same point when this is separated into two and the irrational points correspond to points that are inside the intervals of the counter set but that is how I picture it myself in my mind you may choose any picture you like and we are going to see that for any expanding map we have the same the same not the same but a similar semi-conjugacy and then we are going to use these semi-conjugacies to prove that every couple of expanding maps of degree 2 are conjugated in particular every expanding map will be conjugated of degree 2 will be conjugated to this one non-frequent in dynamics that you have such a beautiful classification and you have a very beautiful model for all a family in turn you will get approximations of this this happens in very particular cases so the next thing we are going to do is to repeat the procedure we have done for the 2x but one map to any expanding map of degree 2 and then in the next class we are going to use these two these maps to construct a conjugacy we are going to take this is to remember that if we have a degree 2 expanding map there is a unique fixed point you remember why this was why if we have an expanding why if we have a degree 2 map expanding we have a unique fixed point yes because we have done it in the first class we have proven that P and F was equal to this the number of fixed points for F to the N and so in particular the number of fixed points is degree of F minus 1 which is 2 minus 1 which is 1 so for any degree 2 map there is a unique fixed point a unique fixed point now we don't have this beautiful model anymore we are going to have well let's put P here P and since we are in the circle this will be P2 so now we are going to take another point we want some other point playing the role of the one half we want another point playing the role of the one half this point will be the pre-image of P that is not P let us draw it here it's not going to be in the middle why? there is a unique Q such that F of Q equals P and why is it a 2 to 1 map yes when we lift when we lift this 2R that's 2 yes that's it it's because the difference between F of X plus 1 and F of X is 2 so we will have F of X here F of X plus 1 here and in particular this will hold for P which is P we can choose it to be P and so here if this we choose it to be P this is going to be P plus 2 in the middle we will have exactly one that will be since the derivative F prime is also greater than 1 okay so it is monotone F is monotone and so there will be a unique point here that will be F of P plus 1 okay and this point is going to be called Q F of Q there will be a unique point such that F of Q is F of P plus 1 thank you so there will be a unique point here doesn't have to be in the middle but there will be a unique point such that F of Q is F of P plus 1 so both will project onto P and this will play the role of 1 half in our 2 X mod 1 map so we will have here something like this and here something like this and we are going to call this delta 0 and this delta 1 okay so we will have the theorem I have already announced for any expanding map of Degree 2 we have that F is a factor of the shift in the theorem I already state how we are going to construct this H we are going to call it H of F H sub F in order to distinguish it from the H for the doubling map so we are going to take any expanding map and so we will want H sub F of course we will want that this happens H F composed with sigma equals composed with H F but in particular I mean this is formula is something natural because we are going we want in particular this will happen so if we have some H of X H of X will belong to some point we will want it to belong to delta X F 0 so H of sigma X we want this to be F of H F so we want this but this is this if we define this to be like this this will belong to delta X 1 we want this to happen we want we want this is how we are going to define we want that H of X belongs to delta X 0 so the dynamics and the fact that we want this to be a semi conjugacy will give us where they belong when we iterate okay so this will belong H of sigma X and in general if we put an N here we will have this and we will have this this is what we want to happen we want to happen this because we want this will be the rule to construct this will be the rule to construct these two rules we are going to use only two rules this and this with these two rules the construction is really natural so this is and we don't have to think a lot to see how we construct the map the map is already constructed it's already restricted but by our desires what we want this map to do already defines the map okay we want we want F N of H of X belongs to delta X N for all N this implies that H of X will have to belong to F to the minus 1 delta X N for all N greater or equal than 0 and this happens if and only if this belongs so in order to be all we have to do is to show that this is not empty this is unique this map is continuous this is subjective and this is it I don't want to prove anything else so I will spoil a little bit your exercises because I have to do it in general but I would like you to do it in your exercise class so how do we do to prove that this is well defined first let me I will give it as an exercise but I will give you hints at the end of the class so in order to prove that this is well defined we have to prove that for all N non-empty and it's an interval it's a close interval but we have to prove that if the length of this interval goes to 0 or another way to prove it is to prove that if two points belong to this intersection then there must be equal okay let's let's prove that so this is going to be proven by induction it's going to be a guided exercise for this afternoon let's prove that there's a unique point in this intersection so let's suppose that we have two points two points here so this will imply but this this length the length of delta xN it's bounded they are less than one okay there are two halves they are bounded let me we have not proven this yet so f is expanding okay let me f is expanding on a compact set so in particular the derivative of f of f exists for every point but it is bounded by a constant which is greater than one it's bounded from below by a constant which is greater than one this is because s1 is a compact set and f' is continuous okay so this is f of fN-1 and what happens with this what happens in general when we have a derivable map this is equal to what this is f' at some point let's put z times the length of x-y I must say that this is greater than one but this is smaller than one but it's greater than some gamma positive so here we will have that this this will be equal to the derivative intermediate point but the derivative is bounded from below by lambda okay so this is going to be greater or equal than f this is x this is y so in our example we will have this is greater or equal than lambda x-y okay where to erase so we have that this can you see why we are going to have equality now because we have that this in one step this is but these are again in the same delta xn minus one okay can you see that so we apply the same procedure and we will get that this is greater or equal than lambda two and we do it n times and we will get that this is greater or equal than lambda n but then we will have these two points will be less or equal this is less over than the n and this is less or equal than the length of delta xn which is less or equal than one but this will happen for all n greater or equal than zero these are equal so the only thing we have used is that they are expanding oh and I have to finish okay h is continuous, h is subjective we will follow in the same way as 2x mod one map so I will not do it or maybe we can do it tomorrow and so let me give you some hints to prove to prove that this is an interval so define this define this prove by induction that this is an interval I will give you hints to do that for the first step it is obvious that this is an interval because delta x0 is an interval by definition and so we want to prove that each time we have an interval we do this intersect it with f minus n of this and we get another interval okay by induction we have to prove that f to the n of each of the extremes go to p f to the n plus one of each of the extremes go to p for the first it is obvious because we chose it to be like that we have this and we have this f of q goes to p so you have to prove that fn plus one is injective in the inside so the extremes are going to go to p but in the inside it is going to be injective let me give you some hint some extra hint for the exercise of today once it is injective you will have one point inside that goes to p that goes to q because the two extremes go to p you will have one that goes to q and then you produce the next delta xn zero xn plus one okay we discuss this today in the afternoon but today we have finished with this thank you