15
$\begingroup$

Is every injective function invertible? How could I prove such thing? (Or is it just a necessary but not sufficient condition?)

If $f:A\rightarrow B$ is injective then $f(a) = f(b) \Rightarrow a = b$ for all $a,b\in A$.

If $f(x)=y$ is invertible, there is some function $g(y)=x$.

I don't see how to put these things together, so help would be appreciated.

$\endgroup$
20
$\begingroup$

You have to be precise: $f:A \to B$ is invertible if and only if it is bijective. If it is injective still you can invert $f$ but viewed as a mapping $f:A \to f(A)$. If you choose some $y \in B \setminus f(A)$ there is no $x \in A$ such that $f(x)=y$ therefore $f^{-1}(y)$ does not make sense

$\endgroup$
  • 4
    $\begingroup$ So in other words there are elements in $B$ that are impossible to reach from $A$ but every "reachable element" is still possible to invert uniquely? Or something like that? $\endgroup$ – password Sep 25 '15 at 14:00
  • 1
    $\begingroup$ Yes, that's true. For example in dealing with cyclometric function arcus sinus is defined as the inverse to sinus but viewed as a function $\sin:[-\frac{\pi}{2},\frac{\pi}{2}] \to [-1,1]$. Sinus is not invertible viewed as a function $\mathbb{R} \to \mathbb{R}$ and even as $[\pi/2,\pi/2] \to \mathbb{R}$ $\endgroup$ – truebaran Sep 25 '15 at 14:05
  • 2
    $\begingroup$ So the answer is "Yes but the inverse is not necessarily a total function." en.wikipedia.org/wiki/Partial_function $\endgroup$ – Syd Kerckhove Sep 25 '15 at 16:11
  • 2
    $\begingroup$ which is to say, it is invertible on its image. $f^{-1}: f(A) \longrightarrow A$. $\endgroup$ – hayd Sep 25 '15 at 22:52
7
$\begingroup$

A function is invertible if and only if it is bijective (i.e. both injective and surjective). Injectivity is a necessary condition for invertibility but not sufficient.

Example:

Define $f: [1,2] \to [2,5]$ as $f(x) = 2x$. Clearly this function is injective.

Now if you try to find the inverse it would be $f^{-1}(y) = \frac{y}{2}$. But notice that for $y \in (4,5]$, $f^{-1}(y)$ does not exists as $f^{-1}(y): [2,5] \to [1,2]$. So the inverse is not a function.

Additional: This is not the case in maps. Injectivity is sufficient for invertibility (as the inverse need not be a function, just a map)

$\endgroup$
  • 3
    $\begingroup$ Could you explain about the "map" bit? I'm only aware of "map" as a synonym for "function". $\endgroup$ – G. Bach Sep 25 '15 at 14:13
  • $\begingroup$ Sorry, I should not use the term "map" here. I was thinking of the relations which are either one-to-one or many-to-one (not one-to-many). An element in the domain might not be mapped to any element in the range. In this case, injectivity is also sufficient for invertibility. $\endgroup$ – user265328 Sep 25 '15 at 16:31
5
$\begingroup$

You should distinguish a left invertible, a right invertible and an invertible function.

Function $f:A\to B$ has a left invertible if:
$\exists g:B\to A: g(f(x))=x \forall x\in A$

$f:A\to B$ has a right invertible if:
$\exists g:B\to A: f(g(y))=y \forall y\in B$

(Sometimes left/right invertibles are called in the reversed order depending on the style the composition of two functions is written)

If $f(x)$ is an injection iff it has a left invertible.
If $f(x)$ is a surjection, iff it has a right invertible.

It's easy to prove that a function has a true invertible iff it has a left and a right invertible (you may easily check that they are equal in this case).

$\endgroup$
3
$\begingroup$

If $f:A\to B$ is injective, $f$ is invertible if and only if $f(A)=B$. But $f:A\to f(A)$ is always invertible if $f:A\to B$ is injective.

$\endgroup$
3
$\begingroup$

For completeness...

There is one variant of the idea of function that doesn't include a notion of codomain. e.g. with this notion, the function defined by $f(x) = x^2$ (where $x$ is a real variable) "remembers" that its domain is all reals and that its image (sometimes called its range) is all nonnegative reals, but doesn't care whether it's being construed as a mapping $\mathbb{R} \to \mathbb{R}_{\geq 0}$, a mapping $\mathbb{R} \to \mathbb{R}$, a mapping $\mathbb{R} \to \mathbb{C}$, or even something more exotic.

For this specific variation on the notion of function, it is true that every injective function is invertible.

I am under the impression that this notion of function was popular once but is no longer popular. However, it still sticks around somewhat due to inertia: e.g. people learned it that way so they taught it to others, related concepts have stuck around, that sort of thing.

For the more modern notion of function, it does "remember" its codomain, and we require the domain of its inverse to be the whole of the codomain, so an injective function is only invertible if it is also bijective.

$\endgroup$
  • $\begingroup$ When I was younger, a function had a domain definition, and was not necessarily defined on the full set. A function defined on all element was called an application. But maybe usage changed... $\endgroup$ – Serge Ballesta Sep 26 '15 at 7:05
  • $\begingroup$ @Serge: Ah, that's interesting too. I know the modern word is "partial function", and "total function" is used for everywhere defined partial functions. But IMO, introductory courses in the US seem to be in a weird place where they want to talk about partial functions, but only ever define the notion of a total function. $\endgroup$ – Hurkyl Sep 26 '15 at 7:12
  • $\begingroup$ My reference were from french courses in the 70's :-) $\endgroup$ – Serge Ballesta Sep 26 '15 at 7:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.