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During the lecture we learned this phrase:

"If $f$ is a function then $f^{-1}$ is a function iff $f$ is injective (one-to-one)."

But why? What with onto? $f$ doesn't need to be Surjective (onto)?

For example: $$f:A \rightarrow B$$ $$A=\{1,2,3\} $$ $$B=\{1,2,3,4\} $$ $$f=\{(1,1)(2,2)(3,3)\}$$

$f$ is an injective function but $ f^{-1} $ can't be a function because for 4 we can`t find a place to go.

pjs36
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BAM
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    $f^{-1}$ is supposed to be defined on the direct image of $f$, not on the whole codomain $B$ – Federica Maggioni Apr 29 '16 at 17:24
  • I would say your observation is technically correct. Functions are invertible iff they're bijective. – MathematicsStudent1122 Apr 29 '16 at 17:27
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    I'm not sure how others handle this, but I was always taught that $f$ has a true inverse iff it's bijective. We can construct $g: B\to A$ such that $g={(1,1), (2,2), (3,3), (4,1)}$, so $g$ acts like an inverse to $f$, but it's not a real inverse, because $g(4)$ exists, and $f^{-1}(4)$ is undefined. – Kevin Long Apr 29 '16 at 17:28
  • @MathematicsStudent1122 The claim doesn't say anything about invertibility. It simply says $f^{-1}$ is a function. I would be curious to know how the lecturer defined $f^{-1}$. – John Douma Apr 29 '16 at 17:29
  • If $f:A\to B$ is an injective function and $C\subseteq B$ is its image then there is a function $g:C\to A$ that serves as inverse of the function $A\to C$ that is prescribed by $a\mapsto f(a)$. – drhab Apr 29 '16 at 17:32
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    @MathematicsStudent1122 Yes, but notice how $f$ is defined as a set of ordered pairs. If the instructor defined the inverse as the reversal of ordered pairs then it is implied that the inverse has domain $Im f$ and not $B$. In that case the notion that $f$ must be bijective is false. – John Douma Apr 29 '16 at 17:33
  • @JohnDouma Yes, I see your point, thanks – MathematicsStudent1122 Apr 29 '16 at 17:34
  • I understand, thanks you! So why to choose the name $f^{-1}$ and not $g$ or something less confusing? What is the idea actually of that $f^{-1}$? – BAM Apr 29 '16 at 17:46
  • You should add the formal definitions of "function" and of $f^{-1}$ given in your lecture. Otherwise it will be difficult to write an adequate answer. – Paul Frost Jul 29 '21 at 07:41

3 Answers3

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In this context a function $f$ is actually a set of ordered pairs having the special property that for every $a\in\{x\mid\exists y\langle x,y\rangle\in f\}$ (its domain) there is a unique $b$ such that $\langle a,b\rangle\in f$.

The function is injective if moreover for every $b\in\{y\mid\exists x\langle x,y\rangle\in f\}$ (its range or image) there is a unique $a$ such that $\langle a,b\rangle\in f$.

Defining $f^{-1}:=\{\langle y,x\rangle\mid \langle x,y\rangle\in f\}$ we can reformulate this by the statement that $f^{-1}$ is a function.

In your example $f^{-1}=f$.

edit:

Notation $f:A\to B$ suggests that $A$ is the domain of the function and that the range of $f$ is a subset of $B$ (its codomain). Then injectivity of $f$ does not guarantee the existence of an inverse $f^{-1}:B\to A$ as you noticed correctly. However it does guarantee the existence of an inverse $f^{-1}:R\to A$ where $R\subseteq B$ denotes the range of $f$.

drhab
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You are completely right: A function $f:\>A\to B$ has an inverse $f^{-1}:\>B\to A$ iff $f$ is bijective.

But mathematicians are sometimes sloppy: If $f:\>A\to B$ is injective you can define an "inverse" $$g:\quad f(A)\to A,\qquad y\mapsto{\rm the}\bigl(f^{-1}(\{y\})\bigr)$$ without specifying supplementary data or making any choices.

If $f$ is not injective, but an "important" function, so that an inverse is extremely desirable, you can enforce a "restricted" inverse by restricting $f$ to a suitable chosen subset $A'\subset A$, on which $f$ is injective. Such is the case, e.g., for $f:=\sin$, whereby the latter is defined on all of ${\mathbb R}$, and is injective on the interval $\bigl[-{\pi\over2},{\pi\over2}\bigr]$. The inverse of the restriction $\sin\restriction\bigl[-{\pi\over2},{\pi\over2}\bigr]$ is then called $\arcsin$, and is a map $[{-1},1]\to\bigl[-{\pi\over2},{\pi\over2}\bigr]$.

Christian Blatter
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  • The question is tagged with elementary-set-theory. On that area functions are identified with their graphs. I think the OP had a lecture in settheory when the phrase was posed. – drhab Apr 30 '16 at 06:51
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We define $f^{-1}=\{(x,y): (y,x)\in f\}$ Then $f^{-1}$ is a relation with domain equal to the image of $f$ and codomain equal to $A$.

Suppose $f$ is injective. To show $f^{-1}$ is a function we must show that if $(x,y)\in f^{-1}$ and $(x,z)\in f^{-1}$ then $y=z$. But if $(x,y)\in f^{-1}$ then $(y,x)\in f$ and likewise, $(z,x)\in f$. This means that $f(y)=f(z)=x$ and since $f$ is injective, $y=z$, therefore $f^{-1}$ is a function.

Conversely, suppose $f^{-1}$ is a function and $f(x)=f(y)=z$. Then $(z,y), (z,x)\in f^{-1}$ and since $f^{-1}$ is a function, $y=x$. Therefore, $f$ is injective.

John Douma
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