Let f:X-->Y be a function.

If f(x_1)=f(x_2) then, x_1=x_2 the function is said to beinjective(Bourbaki's terminology). Alternate form isone-to-one

If for any y in Y we can find an x in X such that f(x)=y the function is said to besurjective(Bourbaki's terminology). Alternate form isonto

If f is both injective and surjective it isbijective.

It is not difficult to show a function has an inverse function (invertible) if and only if it is bijective.

There is a useful theorem (Cantor–Bernstein–Schroeder theorem)* that

if an injective funtion exists between X to Y and Y to X. Then there exists a bijection between the sets.

*)It was originally proposed by Cantor and proved using Axiom of Choice. Later Bernstein, Schroeder proved it was not necessary.

Right because for aas i was asked if monotone implies that the function is invertible, said no because it needs to be continuous toorealfunction to be invertible it needs to be surjective, in a sense it needs to satisfy the Intermediate Value Theorem. Now if a function is continous then it does satisfy IVT and you have an invertible function when it is striclty monotone (because it gives a unique value for each distinct x). That is a sufficient condition but not necessary (not able to come up with an example though).