Is it the domain of the inverse or the defined domain of the inverse (in the case the function is not invertible)? The concept is very confusing for some reason I cannot grasp.
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$\begingroup$Let $f:X\to Y$ be a function. Let $A\subset X$ and $B\subset Y$. Then $f(A)=\{f(x)\in Y:x\in A\}$ is called the image of $A$ under $f$ and $f^{-1}(B)=\{x\in X:f(x)\in B\}$ is called the preimage of $B$ under $f$.
$\endgroup$ $\begingroup$I would not use the terminology preimage of a function at all. If $f:X\to Y$ is a function, and $A\subseteq Y$, the preimage of $A$ under $f$ is $$f^{-1}[A]=\big\{x\in X:f(x)\in A\big\}\;.$$ That is, in my terminology subsets of the codomain of $f$ have preimages under $f$, and these preimages are subsets of the domain of $f$.
You can, if you wish, define a function $$f^{\leftarrow}:\wp(Y)\to\wp(X):A\mapsto f^{-1}[A]$$ that takes each subset of $Y$ to its preimage in $X$ under the function $f$, but I would never call this function the preimage of $f$..
$\endgroup$ 7 $\begingroup$If $f\colon A\to B$ is a function (any function) then the preimage is actually a function $\tilde f\colon P(B)\to P(A)$ (where $P$ indicates the power set) defined as follows: $$\tilde f(X)=\{a\in A\mid f(a)\in X\}$$
Note that the domain of $f$ is $A$ so $A$ itself is $\tilde f(B)$, however it is possible that $\tilde f$ is neither surjective nor injective in cases where $f$ fails to satisfy some of these properties.
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