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Prove that the inverse of one-one onto mapping is unique.

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MCS013 - Assignment 8(d)

A function is onto if and only if for every y in the codomain, there is an x in the domain such that f(x)=y.
So in the example you give, f:RR,f(x)=5x+2, the domain and codomain are the same set: R. Since, for every real number yR, there is an xR such that f(x)=y, the function is onto. The example you include shows an explicit way to determine which x maps to a particular y, by solving for x in terms of y. That way, we can pick any y, solve for f(y)=x, and know the value of x which the original function maps to that y.
Side note:
Note that f(y)=f1(x) when we swap variables. We are guaranteed that every function f that is onto and one-to-one has an inverse f1, a function such that f(f1(x))=f1(f(x))=x.


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