Calculus (3rd Edition)

Example: f: {-1, 0, 1} $\rightarrow$ {0,1,2,3} defined by f(-1)=1, f(0)=0, f(1)=1. A function should only have one and only one image. But when D has 2 elements and R has 3 elements, two images should be there for a single preimage which is against the definition of a function. Therefore, such a function doesn't exist.