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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 and are the domain and image of , respectively, then the fibers of are the sets in {():} = {{: =}:}which is a partition of the domain set .Note that must be restricted to the image set of , since otherwise () would be the empty set which is not allowed in a partition.
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 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.
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.
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 ...
In mathematics, particularly in the field of differential topology, the preimage theorem is a variation of the implicit function theorem concerning the preimage of particular points in a manifold under the action of a smooth map. [1] [2]