 Hello and welcome to the session that is because the following problem today give examples of two function f such that from n to z and g from z to z such that g of f is injective but g is not injective. Now let us write the solution. Let us consider x is equal to x and g of x is equal to mod of x. Now let us take g of x. g of x is equal to mod of x which is equal to x if x is greater than equal to 0 and minus x if x is greater than 0. Therefore, g of x is not 1 1 since g of 1 is equal to g of minus 1 is equal to 1 but 1 is not equal to minus 1. Now let us find g of f of x which is equal to g of f of x which is equal to g of x because f of x is equal to x and g of x is equal to mod of x. Hence g of f of x is equal to mod of x. Now domain of g of f is n that is set of natural numbers which is given to us for every natural number a mod of a is equal to a that is g of f has a unique image. Therefore, g of f is 1 1 and hence g of f is injective. Hence g of f is injective and since g of x is not 1 1 therefore, g is not injective. Thank you for the problem. Bye and have a nice day.