Let f:A->B be a map between sets A and B. Let Y subset= B. Then the preimage of Y under f is denoted by f^(-1)(Y), and is the set of all elements of A that map to elements in Y under f. Thus

 f^(-1)(Y)={a in A|f(a) in Y}.

One is not to be mislead by the notation into thinking of the preimage as having to do with an inverse of f. The preimage is defined whether f has an inverse or not. Note however that if f does have an inverse, then the preimage f^(-1)(Y) is exactly the image of Y under the inverse map, thus justifying the perhaps slightly misleading notation.

For any Y subset= B, it is true that

 f(f^(-1)(Y)) subset= Y,

with equality occurring, if f is surjective, and for any subset X subset= A, it is true that

 X subset= f^(-1)(f(X)),

with equality occurring if f is injective.

Preimages occur in a variety of subjects, the most persistent of these being topology, where a map is continuous, by definition, if the preimage of every open set is open.

See also


This entry contributed by Rasmus Hedegaard

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Hedegaard, Rasmus. "Pre-Image." From MathWorld--A Wolfram Web Resource, created by Eric W. Weisstein.

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