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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} ".
Proof: if is a closed subset of , then () and () are both closed since each is the preimage of when restricted to and respectively, which by assumption are continuous. Then their union , f − 1 ( U ) {\displaystyle f^{-1}(U)} is also closed, being a finite union of closed sets.
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 ...
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.
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 ...
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 ...
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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.