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 \( f \) whose curve is known is 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 simpler 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.