Let $A = \mathcal{P}(\{1, 2, 3, 4\})$. Let $f$ be the following function.

$f : A \rightarrow A$ defined by $f(X) = \{1, 2, 3, 4\} \setminus X$.

Does $f^{−1}$ exist? What is $f^{−1}$?

I'm quite puzzled by the question because I'm not quite sure what to do about the sets when trying to calculate the inverse. Please help.


2 Answers 2


When $f$ is a function from set $S$ to set $T$ the inverse of $f$ is a function $g$ from $T$ to $S$ such that each of $f$ and $g$ undoes the other: for every $s \in S$ you have $g(f(s)) = s$ and for every $t \in T$ you have $f(g(t))=t$. A function $f$ may or may not have an inverse. When it does, you can show it has only one, so you call it $f^{-1}$.

Your example is tricky in several ways. First, the domain $A$ is a set whose elements are themselves sets - it's the power set of $X = \{1,2,3,4\}$ . The function $f$ takes a subset $X$ of and produces another subset. So, for example, $f(\{1\}) = \{2,3,4\}$ .

Second, the codomain of $f$ is the same set as the domain, so the $S$ and $T$ in my first paragraph happen to be the same: #{1,2,3,4}$.

The third other potentially confusing thing in the example is that $f$ happens to be its own inverse. If you apply it twice you are back where you started.


Let $g:A \rightarrow B$ be a function. In general, you can always define, and always exists, for $b \in B $ $ g^{\leftarrow}(b) = \{a \in A | f(a)=b\} \subseteq A$.

$g^{-1}$ exists as a function $g^{-1}:B \rightarrow A$ if $g$ is a bijection.

In this case $f(X) = X^c$ is the complement. What do you think? Is $f$ a bijection? If so, $f^{-1}$ is the only function such that $f^{-1} \circ f =$ identity of $A$. What is $f \circ f$ in this case?

  • $\begingroup$ The notation $g^{-1}$ is used in two different ways here. The first paragraph makes $g^{-1}(b)$ a subset of $A$; the second paragraph makes it an element of $a$. In view of the level of the question, I would expect this discrepancy to cause a problem for Andrea, especially when, as here, the elements of $A$ are themselves sets. $\endgroup$ Apr 16, 2019 at 17:05
  • $\begingroup$ @AndreasBlass changed the notation from $f^{-1}$ to $f^{\leftarrow}$ for the inverse image; but it's a common practice, better learn it sooner than later. $\endgroup$
    – dcolazin
    Apr 16, 2019 at 18:03

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