2

I recently started reading a book which has challenged my understanding of functions and inverse functions quite a bit (mostly the format of the notations that are used); I know that we write a function $( f )$ that maps elements of its domain, $X$, to elements of its co-domain, $Y$, as:

$f:X\rightarrow Y \qquad x \mapsto y \qquad \text{where} \quad x\in X \quad \text{and} \quad y=f(x) \in Y $

$\\$

This is the point where my understanding starts to deviate from the book's content. The book goes on to define the inverse of the function $f$ as the map

$g:Y\rightarrow X \qquad y \mapsto x \quad \text{where} \quad y \in Y \quad x=g(y) \in X $

and then mentions that, since the map $g$ maps EVERY element of its domain ($Y$) onto its co-domain ($X$), it must be injective and since the same holds for the map $f$, we conclude that both $f$ and $g$ are bijective maps.

The problem with this statement, I feel like, is that it excludes the possibility of the co-domain being larger than the image of $f$ (it restricts the co-domain into being the image of $f$)

$\\$

$f:[-\pi/2,\pi/2] \rightarrow \mathbb{R} \quad \text{where} \quad x\mapsto f(x)=\sin x \qquad [1]$

this function is not invertible as it's not bijective (The co-domain isn't the same as the image of $f$)

whereas the function

$f:[-\pi/2,\pi/2] \rightarrow [-1,1] \quad \text{where} \quad x\mapsto f(x)=\sin x \qquad [2]$

is invertible. Is this correct? If it is, then, are we to say that the functions [1] and [2] are different functions? If it isn't, then doesn't that go against our definition and conditions for inverse functions?

Bernard
  • 175,478
Amin Parvaresh
  • 439
  • 2
    Yes, functions [1] and [2] are formally different. Recall that a function is the data $(X,Y,f)$ where $X$ and $Y$ are the domain and co-domain, and $f$ is a mapping rule. By changing $Y$ we change the function, even if the mapping rule is the same.

    About the statement: what are the assumptions on $f$? If $f$ is assumed bijective, then its image is $Y$. If it is assumed injective, then it still has an inverse, when we consider $f$ as a function onto its image (which, as you have remarked, is formally a different function).

     – Randy Marsh Dec 09 '19 at 20:05
  • So, is the following statement correct? "If a function f is injective it doesn't have an inverse; however, we can simply change its codomain into its image and the new function will be invertible" – Amin Parvaresh Dec 09 '19 at 20:13
  • Yes, it is correct if we insist that the inverse of a function $X\to Y$ must be a function $Y\to X$. Some sources, e.g. Wikipedia, use the more relaxed definition, i.e. $f\colon X\to Y$ is invertible if there is $g\colon f(X) \to X$ such that $f(x)=y\iff g(y)=x$. The strong vs. relaxed definition is not a problem, as long as one is consistent. – Randy Marsh Dec 09 '19 at 20:22
  • A binary relation from $X$ to $Y$ is any nonempty subset of the cartesian product $X\times Y$ ; i.e. any set of ordered pairs. The domain and range of a relation is the set of all first coordinates and the set of all second coordinates respectively. A function is by definition any such binary relation $f$ with the property $(x,y),(x,y')\in f\implies y=y'$. – Oliver Kayende Dec 09 '19 at 20:25
  • The inverse of a binary relation is simply that relation obtained by interchanging the coordinates in each of the ordered pairs. Every relation has an inverse. Every function has an inverse. However, the inverse of a function, although still a relation, need not be a function. If the function is an injection then it must have a function inverse. – Oliver Kayende Dec 09 '19 at 20:28
  • It is in the matter of surjectivity that a given codomain is necessary. That is, surjectivity is not an intrinsic property of any function but requires an ambient codomain. – Oliver Kayende Dec 09 '19 at 20:34
  • This question and the answers to it might be of interest: https://math.stackexchange.com/q/3209762/391136 – Randy Marsh Dec 09 '19 at 20:40

1 Answers1

2

Note.- By "relation" is meant : a set of couples, that is, a set of ordered pairs.

A relation from a set A to a set B is a subset of the cartesian product A cross B ( a subset of all the possible ordered pairs with first element in A and second element in B).

A function is a relation, and every relation has an inverse relation.

For all relation R, the inverse of R ( noted R^-1) is

{ (y, x) | (x,y) belong to R }.

(In words, the inverse relation of relation R has , as elements, the " same" ordered pairs as R, but in reversed order.)

But saying that any function has an inverse relation does not mean that this inverse relation is automatically a function.

There are functions such that their inverse relation is also a function, and in that case, one can talk about their " inverse function".

Example, function f such that f = { (1,2) , ( 3,4)}

There are functions such that their inverse relation is not a function.

Example, function g such that g = { (1,2), (3,2) }

Other example :

The set of all pairs of real numbers (x, y) such that y = x² is a function.

It's inverse relation is the set of all pairs (x,y) such that x = y².

A vertical line test will show you that this inverse relation is not a function.

enter image description here