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The preimage of an output value is the set of input values that produce . More generally, evaluating f {\displaystyle f} at each element of a given subset A {\displaystyle A} of its domain X {\displaystyle X} produces a set, called the " image of A {\displaystyle A} under (or through) f {\displaystyle f} ".
This function maps each image to its unique preimage. The composition of two bijections is again a bijection, but if g ∘ f {\displaystyle g\circ f} is a bijection, then it can only be concluded that f {\displaystyle f} is injective and g {\displaystyle g} is surjective (see the figure at right and the remarks above regarding injections and ...
If f : X → Y is surjective and B is a subset of Y, then f(f −1 (B)) = B. Thus, B can be recovered from its preimage f −1 (B). For example, in the first illustration in the gallery, there is some function g such that g(C) = 4. There is also some function f such that f(4) = C.
The fibers of are that line and all the straight lines parallel to it, which form a partition of the plane . More generally, if f {\displaystyle f} is a linear map from some linear vector space X {\displaystyle X} to some other linear space Y {\displaystyle Y} , the fibers of f {\displaystyle f} are affine subspaces of X {\displaystyle X ...
Then a pullback of f and g (in Set) is given by the preimage f −1 [B 0] together with the inclusion of the preimage in A. f −1 [B 0] ↪ A. and the restriction of f to f −1 [B 0] f −1 [B 0] → B 0. Because of this example, in a general category the pullback of a morphism f and a monomorphism g can be thought of as the "preimage" under ...
If f were not injective, then the non-injective elements can form a distinct element of its kernel: there would exist a, b ∈ G such that a ≠ b and f(a) = f(b). Thus f(a)f(b) −1 = e H. f is a group homomorphism, so inverses and group operations are preserved, giving f(ab −1) = e H; in other words, ab −1 ∈ ker f, and ker f would not ...
The preimage by f of an element y of the codomain is sometimes called, in some contexts, the fiber of y under f. If a function f has an inverse (see below), this inverse is denoted . In this case () may denote either the image by or the preimage by f of C. This is not a problem, as these sets are equal.
If is any set then its preimage := under is necessarily an -saturated set. In particular, every fiber of a map f {\displaystyle f} is an f {\displaystyle f} -saturated set. The empty set ∅ = f − 1 ( ∅ ) {\displaystyle \varnothing =f^{-1}(\varnothing )} and the domain X = f − 1 ( Y ) {\displaystyle X=f^{-1}(Y)} are always saturated.