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I got into an argument with my professor today in my beginner foundations in computer science course. We are using the textbook Discrete Structures, Logic, and Computability. At the bottom of the third page in Chapter 2, the book states:

$f^{-1}(\{1,3\})\ = \{a,b\}$

Given that $$f : \{a,b,c\} \rightarrow \{1,2,3\}$$ and $$f(a) = 1, f(b) = 1, f(c) = 2$$

She insisted that $f^{-1}(\{1,3\})$ is undefined, whereas I agree with the book. My thought process is that the pre-image finds the union of the sets in the domain that are associated with each element in the codomain. Who is correct?

Emre Tekmen
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  • Your thinking is right, but the words you use aren't quite right: the pre-image of a set $A$ comprises the elements in the domain that are mapped to some element of $A$. See my answer for the formal expression of this. – Rob Arthan Feb 06 '24 at 21:40

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Your professor is wrong: if $f$ is a function $X \to Y$, and $A$ is any subset of $Y$, then $f^{-1}(A)$ is standard notation for the set $\{x \in X : \exists a \in A \cdot f(x) = a\}$. There is no requirement that $A$ is contained in the image of $f$.

Rob Arthan
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