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I read in my textbook that a function is a bijection if and only if it has an inverse. Is it the same thing to say a function $f: X → X$ is a bijection if $f(x) = x^{-1}$? If $a = x$ and $b = x^{-1}$, then I'd have $f(a) = b$, and in this situation $f(b) = a$. Would that make $f$ its own inverse and therefore mean it's bijective?

Chris
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  • What is $X$? If it's a group, then $x^{-1}$ makes sense. But if it's a topological space, what would $x^{-1}$ mean? – Henry Swanson May 01 '16 at 20:22
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    Presuming you're talking about $X = \mathbb R$, this function isn't defined at $x=0$. If you take that point out, by letting $X = \mathbb R-{0}$, $f(x) = x^{-1}$, then you do get a bijection, and $f$ is its own inverse. However, there are many other kinds of bijections very different from this one. – Dustan Levenstein May 01 '16 at 20:24
  • I think I was referring to $X$ being a set of real numbers. Though the example I'm working with is permutations, where $X$ is the set of all permutations and $f$ maps a permutation to its inverse. I don't think a permutation's inverse is ever undefined, so would that mean this would work in this case? – Chris May 01 '16 at 20:27

3 Answers3

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Your problem stems from the fact that the word "inverse" has (at least) two meanings. One describes the inverse of a function, another the inverse of a number (or element of some algebraic structure where that makes sense).

These meanings are essentially different unless you move up one level of abstraction, where you "multiply" functions by composing them. Then they can occur in the same problem - with their two meanings. That's exactly what you encountered in your example with permutations. As @DustanLevenstein points out, the function from permutations to permutations that assigns to each permutation its inverse (as a permutation thought of as a function) has an inverse (as a function).

(This is a good beginner's question.)

Ethan Bolker
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  • I was working with permutations, so it was tough for me to find a similar example. In my case I was referring to real numbers, but I'm not sure if this would be somewhat similar to permutations. – Chris May 01 '16 at 20:30
  • I think @DustanLevenstien 's comment addresses your comment. – Ethan Bolker May 01 '16 at 20:31
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I think your text book means $f\colon A\to B$ is bijective if there exists a function the other way, say $g\colon B\to A$, such that $g\circ f=\operatorname{id}_A$ and $f\circ g=\operatorname{id}_B$.

Bernard
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Your $x^{-1}$ is an example of a function with an inverse and therefore it can be a bijection but that isn't the meaning of having an inverse, or the notation.

Having an inverse means that there exists another function which reverses the function $f$, so the inverse of $x+1$ is $x-1$. The inverse of $2x$ is $x/2$. The correct notation to say there exists an inverse of $f$ is this: $\exists f^{-1}(x)$

By chance, your function $x^{-1}$ is unusual in that it is its own inverse, i.e. apply it twice and you get back where you started. This is written $f^{-1}(x)=f(x)=x^{-1}$. That type of function is called an involution, but this is a distraction, it's not fundamental to the understanding of bijections.

A bijective function means you must define the domain and range, and both the function and its inverse map every element of one to exactly one element of the other, with no members of either space not mapped. So for example the function $x+1$ is bijective from the integers to the integers because every integer is mapped to another without exception. Its inverse is $x-1$.

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