What is the preimage of a function
Sophia Terry 
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.
$\endgroup$3 Answers
$\begingroup$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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