Check the injectivity and surjectivity of the following functions :

(i) $f:N \to N\;\;given\;\;by\;\;f(x) = {x^2}$

(ii) $f:Z \to Z\;\;given\;\,by\;\,f(x) = {x^2}$

(iii) $f:R \to R\;\,given\;\,by\;\,f(x) = {x^2}$

(iv) $f:N \to N\;\,given\;\,by\;\,f(x) = {x^3}$

(v) $f:Z \to Z\;\,given\;\,by\;\,f(x) = {x^3}$

(i) $f:N \to N$ given by $f(x) = {x^2}$

Injectivity
$f({x_1}) = f({x_2}) \Rightarrow x_1^2 = x_2^2 \Rightarrow {x_1} = {x_2}$
Therefore, $f$ is one-one ie.e, f is injective

Surjectivity
There are many such numbersof co-domain which have no image in domain N.
e.g., 3$\in$co-domain N, but there is no pre-image in domain of f.
Thus, f is not onto i.e., f is not surjective.
Hence, f is injective but not surjective.

(ii) $f:Z \to Z$ given by $f(x) = {x^2} \Rightarrow Z = \{ (0,\; \pm 1,\; \pm 2,\; \pm 3,\;.....\}$

Injectivity
Let $- 1,\;1 \in Z,\;\;\;f( - 1) = f(1) \Rightarrow 1 = 1$
But, $- 1 \ne 1.\;\;\;\;\;\therefore \;\;f$ is not one-one i.e., f is not injective.

Surjectivity
There are many such elements belonging to codomain which have no preimage in its domain Z.
$2 \in Z\;\;(co - domain).\;\;But\;\,{2^{1/2}}\not \in Z\;(co - domain)$
Therefore, Element 2 has no pre-image in its co-domain Z

Therefore, f is not onto i.e., f is not surjective.

Hence, f is neither injective nor surjective.

(iii) $f:R \to \;\;given\;\;by\;\,f(x) = {x^2}$

Injectivity
Let $- 1,\;\,1 \in R,\;\;\;f( - 1) = f(1) \Rightarrow 1 = 0$
But, $- 1 \ne 1 \Rightarrow f$ is not injective.

Surjectivity
$- 2$ belong to co-domain R of f but $\sqrt { - 2}$ doesnot belong to domain
R of f. Therefore, f is not surjective.
Hence, f is neither injective nor surjective.

(iv) $f:N \to N$ given by $f(x) = {x^3}$

Injectivity
Let ${x_1},\;{x_2} \in N,\;\;f({x_1}) = f({x_2}) \Rightarrow x_1^3 = x_2^3 \Rightarrow {x_1} = {x_2}$
i.e., for every $x \in N,$ f has a unique image in its co-domain.
Therefore, f is one-one and thus f is injective.

Surjectivity
There are many such members of co-domain of f which do not have pre-image in its domain e.g., 2, 3 etc.

Therefore, f is not onto and thus f is not surjective.
Hence, f is injective but not surjective.

(v) $f:Z \to Z,$ given by $f(x) = {x^3}.$ Here Z is the set of integers.

Injectivity
$f({x_1}) = f({x_2}) \Rightarrow x_1^3 = x_2^3 \Rightarrow {x_1} = {x_2}$
Therefore, f is one-one and thus it is injective.

Surjectivity
Many members ofco-domain of f have no pre-image in its domain.
e.g., 2 belonging to its co-domain has no pre-image in its domain of Z. Therefore, f is not surjective.
Hence, f is injective but not surjective.