Not the range, but the codomain, i.e. for every y in the codomain, there is an x in the domain with y = f(x).kathrynag said:onto, not entirely sure. I know it has something to do with showing entire range is used
An isomorphism is a function that has a one-to-one correspondence with another function. In the context of inverse functions, an isomorphism is a function that can be reversed to obtain the original function.
To prove that f and g are isomorphisms, you need to show that they are both one-to-one and onto functions. This means that for every element in the domain of f, there is a unique element in the domain of g that maps to it, and vice versa.
If f and g are inverses of each other, it means that applying f followed by g (or vice versa) will result in the identity function. In other words, the compositions of f and g will cancel each other out, leaving the input unchanged.
Proving that f and g are isomorphisms is important because it guarantees that they are bijective functions, which have many useful properties in mathematics. It also allows for the easy manipulation and understanding of functions by being able to reverse their effects.
No, f and g must be both one-to-one and onto to be considered isomorphisms. If they are not, then they cannot be reversed to obtain the original function and therefore do not meet the criteria for an isomorphism.