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It is a surjective relationship. It may or may not be injective, and therefore, bijective.
No. The function y = x2, where the domain is the real numbers and the codomain is the non-negative reals is onto, but it is not one to one. With the exception of x = 0, it is 2-to-1. Fact, they are completely independent of one another. A function from set X to set Y is onto (or surjective) if everything in Y can be obtained by applying the function by an element of X A function from set X to set Y is one-one (or injective) if no two elements of X are taken to the same element of Y when applied by the function. Notes: 1. A function that is both onto and one-one (injective and surjective) is called bijective. 2. An injective function can be made bijective by changing the set Y to be the image of X under the function. Using this process, any function can be made to be surjective. 3. If the inverse of a surjective function is also a function, then it is bijective.
It is a injective relationship. However, it need not be surjective and so will not be bijective. It will, therefore, not define an invertible function.
counter example: f(x)= arctan(x) , f:R ->(-pi/2 , pi/2) g(x)=tan(x) , g:(-pi/2, pi/2) -> R (g(x) isn't surjective) f(g(x))=arctan(tan(x))=x f(g(x)): R -> R Although, if two of the three are surjective, the third is surjective as well.
Domain, codomain, range, surjective, bijective, invertible, monotonic, continuous, differentiable.