# Inverse of sets

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.

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? • 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. Apr 16, 2019 at 17:05 • @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. Apr 16, 2019 at 18:03