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While solving inverse function problems, I got confused in a part, like for any Inverse function to be defined, it must be one-one and onto, then in many questions why the codomain is given more than the Range as if we know that the codomain must needs to be equivalent to the Range for the Inverse function to be valid or defined, then why does they gave us the codomain different from the range?

Kindly help me solving this doubt.

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    $\begingroup$ The codomain defines the set of possible values a function may take while the range of a function defines the actual values a function takes. The range of a function is too restrictive. For example, we often want to speak of "domain shaped" objects within a space represented by a codomain of a function. This gives us greater insights into the structure of the codomain by viewing its components which may be equivalent to simpler structures. $\endgroup$ Commented Jul 9, 2023 at 21:17

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It is a matter of definition only. In some books invertible functions have to be one to one and onto, so if $f\colon A\to B$, then $f^{-1}\colon B\to A$. But in some references the function only has to be one to one. and the inverse function is defined on the range of the original function.

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It is important to note that the word "range" of a function f : X → Y used to always mean what is nowadays called the codomain. But since then its meaning has been distorted to sometimes mean codomain (i.e., Y) and sometimes mean the image im(f) of f.

Where im(f) = f(X) = {y ∊ Y | ∃ x ∊ X such that f(x) = y}.

Because of this unfortunate ambiguity, it is much better to always avoid using the word "range" unless its meaning is crystal-clear. (And even then.)

It's best to stick with the words "codomain" and "image".

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    $\begingroup$ I should add that there are some branches of mathematics where the usage of the word "range" to mean "image" is so very traditional that it is not likely to change anytime soon. One of these is when the function in question is an operator on an infinite-dimensional Hilbert or Banach space. $\endgroup$ Commented Jul 10, 2023 at 14:19

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