 Hello and welcome to the session. Let us understand the following question. Check the injectivity and subjectivity of the function. Given to us f is a function from n to n given by fx is equal to x cube. Now let us write the solution. Let us check for 1 1. Let x y belongs to n such that f of x is equal to f of y. Then f of x is equal to f of y which implies x cube is equal to y cube. Now on taking cube root implies x is equal to y. Therefore f is 1 1. Now let us check for on 2. We know set of natural numbers does not contain cube roots. Let y belongs to n in co-domain. Then f of x is equal to y which implies x cube is equal to y which implies x is equal to y to the power 1 by 3. But for all y belongs to n in co-domain y to the power 1 by 3 does not belongs to n in domain. Therefore f is not on 2. f is 1 1 and f is not on 2. Now since f is 1 1 and not on 2. Therefore function is injective but not surjective. You understood the problem. Bye and have a nice day.