 Hello and welcome to the session, let us understand the following problem. Check the injectivity and subjectivity of the function. Given to us f is a function from z to z given by fx is equal to x square. Now let us write the solution. Let x1, x2 belongs to z such that f of x1 is equal to f of x2 which implies x1 square is equal to x2 square which implies x1 is equal to plus minus x2 which implies x1 is equal to x2 and x1 is equal to minus x2 but x1 is equal to minus x2 is not possible. Therefore, f is not 1 1. Now let us check for on 2. We observe that the range is not equal to co-domain. We find that range is a set of all square of integers and according to the given condition, the set of all integers range is not equal to co-domain. Therefore, f is not on 2. Since f is neither on 2 nor 1 1 therefore function is neither injective nor surjective. I hope you understood the problem. Bye and have a nice day.