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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.

Asaf Karagila
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xeonphi
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3 Answers3

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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 $f$..

Brian M. Scott
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  • Is your point that the phrase 'preimage of a function' directly ignores and implies the opposite of the fact that the preimage requires a set for the argument of a function? – 000 Dec 09 '12 at 00:03
  • I don't quite understand. In layman's terms, how would you define it? – xeonphi Dec 09 '12 at 00:04
  • @Limitless: Not really. My point is that it isn’t the function that has a preimage; the function is the mechanism by which preimages of subsets of the codomain are constructed. – Brian M. Scott Dec 09 '12 at 00:04
  • @BrianM.Scott I misphrased what I was trying to say, sorry. In my comment, I meant $f^{-1}$ by 'function'. That is, $A$ is the argument in $f^{-1}[A]$. – 000 Dec 09 '12 at 00:06
  • @xeonphi: As I said, I wouldn’t: given a function $f:X\to Y$, I don’t consider the expression the preimage of $f$ meaningful. Subsets of $Y$ have preimages under $f$. – Brian M. Scott Dec 09 '12 at 00:06
  • @Limitless: You can always view $f^{-1}$ as a function from $\wp(Y)$ to $\wp(X)$; from that point of view it’s the $f^{\leftarrow}$ of my answer. For each $A\subseteq Y$ it gives you the preimage of $A$ under $f$. None of that is problematic. What I’m saying is that at least as I (was taught to) use the words, ‘the preimage of $f$’ is meaningless: $f$ doesn’t have a preimage. Subsets of $Y$ have preimages, and specifically preimages under $f$. – Brian M. Scott Dec 09 '12 at 00:10
  • @BrianM.Scott Yes, I understand perfectly now. You've made a very good point. – 000 Dec 09 '12 at 00:18
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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 $f$.

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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.

Asaf Karagila
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