I recently came across this question on MathOverflow asking if there are any polynomials of two variables with rational coefficients, such that the map is a bijection. The answer to this question is almost surely “no”, but it is remarkable how hard this problem resists any attempt at rigorou...
In the bijection above, FF appears on the left and is called the left adjoint of GG, while GG appears on the right and is called the right adjoint of FF. Moreover, the isomorphism homD(FX,Y)≅homC(X,GY)homD(FX,Y)≅homC(X,GY) looks almost identical to the property that a ...
I recently came across this question on MathOverflow asking if there are any polynomials of two variables with rational coefficients, such that the map is a bijection. The answer to this question is almost surely “no”, but it is remarkable how hard this problem resists any attempt at rigorou...
Bijection- A 1-1 function that satisfies that "image of f is Y". Inverses Function- The function g of a bijection f that satisfies that "if y = f(x), then x = g(y)". One-way Function- A bijection that satisfies that "the computation of its inverse function is very hard, and ...
(L) Under one estimation noise level, \(\exists\) a bijection between \(X_n\) and \(Y_{(r)}\). (R) With two noise levels, \(Y_{(r)}\) and \(Z_{(s)}\) may be generated by the same \(X_n\) for some r and s Full size image As we are interested in the overall...
What is the cardinality of rational numbers? Cardinality of the Rational Numbers: For the cardinality of rational numbers we can write: Rational numbers can be written in bijection with natural numbers, in this sense we will say that their cardinality is the same. ...
Isn't a surjective solution valid though? I guess bijections are the "best" but is there a reason a surjective solution gets that warning? Any relation to group theory would help my brain understand the problem. Sign in to comment.
So, if we take the set, I, of all functions in S zero at the origin, and we mod out by I^2, the set of everything with a double zero at the origin, then d is a bijection between I/I^2 with R*^n. As an example, let's take S to be the set of all polynomials in x ...
So it seems the first function is properly defined since for any value x,y the function always returns a distinct pair of integers. I'm confused as to how i would prove it is a bijection. Can i just compute the inverse to prove it (since any function with an inver...
It is evident that, the concepts quasi sg-openness and sg-continuity coincide if the function is a bijection. On quasi sg-open and quasi sg-closed functions Arriving home from work, Quasi saw Esmerelda getting the wok out and said: "Great - Chinese food!" Esmerelda replied: "Nope - I'...