10
$\begingroup$

I believe a function has to be bijective to be invertible, in fact Wikipedia says that an invertible function is another name for a bijection. https://en.wikipedia.org/wiki/Bijection

But in answers to the question Is a bijective function always invertible?, it is stated that a function only has to be injective to be invertible.

$\endgroup$
2
  • $\begingroup$ For the record, I fail to see why this was closed as "unclear what you're asking". I found it perfectly clear what the OP's question was about. I would understand it being closed due to the OP putting in too little effort - the answer I gave he could probably have figured out himself - but closing it because it's "unclear what you're asking" still seems wrong... $\endgroup$
    – fgp
    Feb 13, 2015 at 1:15
  • $\begingroup$ @fgp I can't also understand why it's 'Unclear'. Must I bring an equation to you in order to it become "Clear!" $\endgroup$ Feb 13, 2015 at 6:08

6 Answers 6

15
$\begingroup$

That depends on subtleties of your definition of inverse function. If you expect a function $f\,:\, A \to B$ to have an inverse $f^{-1} \,:\, B \to A$, then $f$ needs to be surjective. If it is not, there are some $b \in B$ that aren't reached by $f$ at all, so how would you define $f^{-1}$ for such elements of $B$?

However, any injective $f\,:\, A \to B$ can be made surjective by simply restricting $B$, i.e. by setting $\widetilde{B} = \{f(a)\,:\, a\in A\}$ and redefining $f$ as $f \,:\, A \to \widetilde{B}$. That redefined $f$ then has an inverse $f^{-1} \,:\, \widetilde{B} \to A$.

Note, however, that while this makes the concatenation $f^{-1} \circ f$ the identity funtion on $A$, it does not make $f \circ f^{-1}$ the identity function on $B$, only on $\widetilde{B} \subsetneq B$.

Thus, if $f \,:\, A \to B $ is injective but not surjective, it has a left-inverse, because we can find an $f^{-1}$ as above with the property that $f^{-1} \circ f$ is the identity on $A$. But it does not have a right-inverse, because we won't find a $f^{-1}$ such that $f \circ f^{-1}$ is the identity on $B$ - the best we can achieve is an identity on $\widetilde{B}$.

Similarly, if $f \,:\, A \to B$ is surjective but not injective, it has a right-inverse, but no left-inverse.

$\endgroup$
7
  • $\begingroup$ But redefined $f$ is not $f$. Because the function is defined by a single domain and codomain. Inverse function have to be bijective. $\endgroup$
    – Mihail
    Feb 12, 2015 at 18:39
  • 1
    $\begingroup$ @fgp do you mean that if $f$ is surjective but not injective then it has no left-inverse but a right-inverse? Wikipedia explains how left-inverse implies injectivity. $\endgroup$
    – rb612
    Oct 17, 2017 at 6:46
  • 1
    $\begingroup$ @rb612 Yes! If $f$ not injective, then applying the function loses information, because different things are mapped to the same thing. You can't undo that loss of information afterwards, meaning there can't be a $g$ such that $g\circ f$ is the identity. But if $f$ is surjective, you can avoid it before it happens, i.e. when forming $f\circ g$ by picking a suitable $g$ -- one that maps $x$ to one of the $y$'s that $f$ then maps back to $x$. Note how you need $f$'s surjectivity to be certain that you can find a suitable $y$ for every $x$. $\endgroup$
    – fgp
    Oct 17, 2017 at 12:35
  • 1
    $\begingroup$ @fgp thank you for clarifying! That completely makes sense. Should I edit the answer provided to correct the mistype in the last sentence of your answer? $\endgroup$
    – rb612
    Oct 17, 2017 at 20:52
  • 1
    $\begingroup$ @rb612 Oh, I hadn't noticed how confused my answer was regarding left- and right-inverses -- I had mixed up the two pretty consistently. Sorry for that, and thanks for pointing it out! Should be fixed now, but if you could check once more, I wouldn't hurt. $\endgroup$
    – fgp
    Oct 19, 2017 at 10:29
5
$\begingroup$

A function has an inverse if and only if it is both surjective and injective. (You can say "bijective" to mean "surjective and injective".)

Khan Academy has a nice video proving this.

edit: originally linked the wrong video.

$\endgroup$
2
$\begingroup$

Hint: if function $ f : A \rightarrow B $ was not surjective, how would we define $ f^{-1} : B \rightarrow A $ for an element that was not in the image of $ f $?

$\endgroup$
1
$\begingroup$

An invertible function shall be both injective and surjective, i.e Bijective! where every elemenet in the final set shall have one and only one anticident in the initial set so that the inverse function can exist!

$\endgroup$
1
$\begingroup$

This depends on the definition of inverse, which varies in the literature across different areas of mathematics. In probability theory, the inverse of $g:\mathcal{X} \to \mathcal{Y}$ is often defined even though $g$ is not bijective. (The argument below requires no formal knowledge of probability).

Suppose we have a random variable $X$ (a function) defined on on some sample space $\Omega$ such that $X : \Omega \to \mathcal{X}$. Furthermore, define the function $g:\mathcal{X} \to \mathcal{Y}$ for some set $\mathcal{Y}$, not necessarily bijective.

If we define the function $Y := g \circ X$, then it is clear that $Y: \Omega \to \mathcal{Y}$. The inverse function $g^{-1}: \mathcal{Y} \to 2^{\mathcal{X}}$ is defined such that $g^{-1}(y) = \{x\in\mathcal{X} : g(x) = y \}$, which can be the empty set if $y \in \mathcal{Y}$ is not mapped under $g$.

This definition is extremely useful for finding distributions of transformations of random variables.

$\endgroup$
0
0
$\begingroup$

If your function $f: X \to Y$ is injective but not necessarily surjective, you can say it has an inverse function defined on the image $f(X)$, but not on all of $Y$. By assigning arbitrary values on $Y \backslash f(X)$, you get a left inverse for your function.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .