In these papers the set of interpolation points is either givenДЄs the set of common zeros l of functions that are entire in C l or as points that are in some sense 'close' to points on an integer (at the interpolation nodes) is the subject of and other papers. And we proved that X has to be equal to why that is that f s 1 to 1.The structure of entire functions in C t with a given growth estimate and with given values at points of a discrete set And you have a lovely with eggs And he have said with why So we conclude X equals to what? So we had X equals F N y. And if w equals set, that means that the f W equals G f said. Then we can conclude from here that w equal set. But since I've composed with, he was a rejection and in particular it is 1 to 1. But this can also be written as if composed with he evaluated at w equals have composed with he evaluated at Set. This means that f of G f W equals F of G, F said. Let's take two elements x and Y element of B And because he is an to we know that there are elements WN set elements of a such that G f W is X and you have said, is why Now let's suppose to prove that F is 1 to 1 little booths that f of X equals f of y. Let's accept that if campus with me is our rejection, and that is until so we have to conclude that f is won't do it. That means that isn't now for the other direction. We took an arbitrary element of B, and we saw that there's an element of a message that you have said is wise. But using the assumption that F is 1 to 1, then we can conclude that G f said equals two. Now, if composed with you, can be written as f of G f set, and this is equal to F F. Until then, we can find an element set of a such that, if composed with the evaluated at set, is the God do f F Y. And I returned element of B and they noted why? And then let's consider its image and the F So F y is an element of C, and since if complexity is a bijection in particular, it is. And it goes to be and it goes from B to C Uh, yeah, so let's consider. Remember that, uh, the domain of G u S A.

So let's consider, uh, uh, element and arbitrary element of B. If f composed with the is a bijection and if it's 1 to 1, then we have to prove that he is under. And now if we part C, we haven't even relieved. And when we started assuming that G f X was equal to G F Y, and we concluded Ted X was equal to White, that means that he's one too. And because of the assumption that if composed with key is 1 to 1, we conclude from here that X is equal to y. We have this and well, we can write f of G f X, as if composed with G F X and the same on the other side. So let's consider what this If these two uh, if this equation is valid, then we can take it from both sides. And we want to prove that X is equal to y. And let's suppose that the F X is equal to U. So let's take two elements x and Y elements of a. Now, for Part B, we have to conclude that use 1 to 1. I noticed that this year why is an element of the so we have found an element of the such that its image is under F is X, and this means that epicenter except salvatori. This is because of the assumption that have composed with is under now every composed with the F Y can be within this f of g f. We know that there exists a Y in a such that, if composed with G evaluated at y equals X. We want to conclude that F is on top, so let's take an arbitrary element. Okay, so here's a diagram of the functions in part.

Onto vs one to one discrete meaning