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First, I wan't to make clear that we use the same terms:

$X, Y$ are sets with $X \neq \emptyset$ and $Y \neq \emptyset$. $f:X \rightarrow Y$ is a funtion.

$f^{-1}(A)$ is the preimage of $A \subseteq Y$ (and NOT the inverse function).

Image

$f$ is a function from domain $X$ to codomain $Y$. The smaller oval inside $Y$ is the image of $f$.

Source: commons.wikimedia.org

Question:

What is the preimage of the codomain of a function if the function is not surjective?

Example:

$f: \mathbb{R} \rightarrow \mathbb{R}$

$f(x) := x^2$

a) $f^{-1}(\{-1,1,2\}) = \{1, \sqrt{2}, -1, \sqrt{2}\}$ and especially $f^{-1}(\{-1\}) = \emptyset$ or

b) $f^{-1}(\{-1,1,2\})$ is undefined, as $-1$ has no preimage?

Which one is correct? Does it depend on the author?

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Your example has nothing to do with your question. The codomain in them is $\mathbb{R}$ and its preimage is $\mathbb{R}$ too. Generally, the preimage of the codomain is the domain. –  Michael Greinecker Oct 20 '12 at 6:35

3 Answers 3

up vote 2 down vote accepted

Definitely (a). The definition of preimage is $f^{-1}(A) = \{x \in X : f(x) \in A\}$ so it's defined for all subsets $A \subset Y$, regardless of how $A$ relates to the image of $f$.

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Tim Gowers had a lot more to say about the subject of the notation "$f^{-1}(A)$" on his blog: gowers.wordpress.com/2011/10/13/… (note that he uses the term "inverse image" for what here is called preimage). –  kahen Oct 20 '12 at 6:35
    
@Ted: Thanks for your clear, short answer. I've now realized, that my problem was that I didn't have the definition of the preimage in mind. –  moose Oct 20 '12 at 6:52
    
+1 for the suggestion of taking a look at Tim Gowers's blog post (and his neighbouring ones). Terrific stuff. –  Peter Smith Oct 20 '12 at 6:59

This emended picture from http://yorkporc.wordpress.com/2011/05/28/transforms-pre-images-and-kernels-and-null-spaces/ should avail.

enter image description here

The pink is the preimage/inverse image of $S \subseteq Y$ (though this picture shows $S \subsetneq Y$) under $T$ $= T^{-1}[S] = \{x \in X : T(x) \in S\}$.

Moreover, it evinces Arthur Fischer's statement: "There is no requirement that everything in $S$ is in the range/image of $f$." In $Y$, the pink is strictly smaller than $S$ thus $T(T^{-1}[S]) \subsetneq S$ here.

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I definitely go with (a).

Note that Wikipedia says the following:

The inverse image or preimage of a particular subset $S$ of the codomain of a function is the set of all elements of the domain that map to the members of $S$.

Put into set-theoretic notation $$f^{-1} ( S ) = \{ x \in X : f(x) \in S \}.$$ There is no requirement that everything in $S$ is in the range/image of $f$. (Actually, the attentive reader should note that the same definition could be used even if the set $S$ is not a subset of the codomain. And this makes sense since "codomain" is an indefinite concept (set-theoretically speaking). What is the codomain of $x \mapsto x^2$? Is it $\mathbb{R}$? $\mathbb{R}^{\geq 0}$? $\mathbb{C}$?)

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My prof. definied in lecture about analysis: Let $f:X \rightarrow Y$ be a function. Then $Y$ is the codomain of $f$. So it is obviously defined defined. The "$f:X \rightarrow Y$" is part of the definition of the function. –  moose Oct 20 '12 at 6:58
    
@moose: Note that according to this somehow the functions $f:\mathbb{R}\to\mathbb{R}$ and $g:\mathbb{R}\to\mathbb{C}$ defined by $f(x)=x^2=g(x)$ are different. But yet they agree everywhere! Notationally you can tell them apart, but if I just tell you that I've got this function $h$, and $h(x)=x^2$ for every real number $x$, you cannot tell me what the codomain of $h$ is. (In fact, the codomain of $h$ is $\mathbb{R} \cup$ $\mathrm{GL}_7 ( \mathbb{R} )$.) (cont...) –  Arthur Fischer Oct 20 '12 at 7:24
    
(...inued) In this way codomain becomes a concept related to intent, and intent is something that is hard to put into concise mathematical terms. (One possibility is to use $f : X \to Y$ to denote a triple $(f,X,Y)$ where $f$ is a function with domain $X$ and range/image a subset of $Y$. But now we've just added a lot of extra baggage to what is a relatively simple concept: $f$ associates to each element of $X$ a unique element of $Y$; leaving the possibility that some elements of $Y$ are not associated to an element of $X$.) This is likely a distinction that can be left until later. –  Arthur Fischer Oct 20 '12 at 7:24

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