Check the injectivity and surjectivity of the following functions

A function f : A --> B is said to be a one - one function or injective mapping, if different elements of A have different f images in B.

(surjective mapping) If the function f :A → B is such that each element in B (codomain) is the f image of atleast one element in A. then we say that f is a function of A onto B.

now, Let's start to check whether function is injective or surjective mapping.

(i) f : N → N given by f (x) = x²
We can see that for x, y ϵ N,
f(x) = f(y)
⇒ x² = y²
⇒ x = y
⇒ f is injective.
Now, let 2 ϵ N. But, we can see that there does not exists any x in N such that
f(x) = x² = 2, therefore,  f is not surjective.
Therefore, function f is injective but not surjective.

(ii) f : Z → Z given by f (x) = x²
We can see that f(-1) = f(1) = 1, but -1 ≠ 1
⇒ f is not injective.
Now, let -2 ϵ Z. But, we can see that there does not exists any x in Z such that
f(x) = x² = -2
⇒ f is not surjective.
Therefore, function f is neither injective nor surjective.


(iii)f : R → R given by f (x) = x²
We can see that f(-1) = f(1) = 1, but -1 ≠ 1
⇒ f is not injective.
Now, let -2 ϵ R. But, we can see that there does not exists any x in R such that
f(x) = x² = -2
therefore,  f is not surjective.
Therefore, function f is neither injective nor surjective.

(iv) f : N → N given by f (x) = x³
We can see that for x, y ϵ N,
f(x) = f(y)
⇒ x³ = y³
⇒ x = y
⇒ f is injective.
Now, let 2 ϵ N. But, we can see that there does not exists any x in N such that
f(x) = x³ = 2
⇒ f is not surjective.
Therefore, function f is injective but not surjective.

(v) f : Z → Z given by f (x) = x³
We can see that for x, y ϵ N,
f(x) = f(y)
⇒ x³ = y³
⇒ x = y
⇒ f is injective.
Now, let 2 ϵ Z. But, we can see that there does not exists any x in Z such that
f(x) = x³ = 2
⇒ f is not surjective.
Therefore, function f is injective but not surjective.