Preimage of a Function

From the definition of the function

Finding the preimage(s) of a value $ a $ by a function $ f $ is equivalent to solving equation $ f(x) = a $.

Example: Calculating the preimage of $ 1 $ by the function affine $ f(x) = 2x + 1 $ is to solve $ 2x + 1 = 1 \iff x = 0 $. So the fiber of $ 1 $ by $ f $ is $ 0 $

Example: Find the preimage of $ 4 $ by the 2nd degree polynomial function $ g(x) = x^2 $. Solve the equation $ x^2 = 4 \ iff x \pm 2 $. So the preimages of $ 4 $ by $ g $ are $ -2 $ and $ 2 $

From the curve of the function

Finding the preimage(s) of a value $ a $ by a function $ f $, which has a known curve, is equivalent to find the abscissae of the intersection(s) of the curve with the ordinate line $ y = a $.

If a function $ f $ is such that $ f(x) = a $, the image of $ x $ by the function $ f $ is $ a $, and the preimage of $ a $ by the function $ f $ is $ x $.

Hash functions are combinations of nonlinear functions specifically created so that their antecedents are not easily computable. It is generally faster to calculate from the function f all the pairs (x, a) (or as much as possible, where the most probable) and to compare with the value of the antecedent sought.