Question

Check the injectivity and surjectivity of the following functions:
$\left(i\right)f:N\to N$ given by $f\left(x\right)=x^2$

$\left(ii\right)f:Z\to Z$ given by $f\left(x\right)=x^2$

$\left(ii\right)f:Z\to Z$ given by $f\left(x\right)=x^2$

$\left(iv\right)f:N\to N$ given by $f\left(x\right)=x^3$

$\left(v\right)f:Z\to Z$ given by $f\left(x\right)=x^3$

Answer 1

$f:N\to N$ is given by $f\left(x\right)=x^2$
It is seen that for $x,y\in N,f\left(x\right)=f\left(y\right)$
$\Rightarrow x^2=y^2\Rightarrow x=y(\because x$ and y are positive numbers)
Therefore, f is injective.
Now, $2\in N$ but there does not exist any x in N such that $f\left(x\right)=x^2=2.$
It means there exists at least one element in co-domain which does not have any preimage. Therefore, f is not sujective.
Hence, function f is injective but not surjective.

$f:Z\to Z$ is given by $f\left(x\right)=x^2$
It is seen that f(-1)=f(1)=1 but $-1\ne 1$.
Therefore f is not injective.
Now, $-2\in Z$. But there does not exist any elements $x\in Z$ such that $f\left(x\right)=x^2=-2$.
Therefore, f is not surjective.
Hence, function f is neither injective nor surjective.

$f:R\to R$ is given by $f\left(x\right)=x^2$. It is seen that f(-1)=f(1)=1 but $-1\ne 1,$
Therefore, f is not injective.
Now, $-2\in R$. But there does not exist any element $x\in Z$ such that $f\left(x\right)=x^2=-2$
Therefore, f is not surjective.
Hence, function f is neither injective nor surjective.

$\left(iv\right)f:N\to N$ given by $f\left(x\right)=x^3$
It is seen that for $x,y\in N,f\left(x\right)=f\left(y\right)\Rightarrow x^3=y^3\Rightarrow x=y$
Therefore, f is injective. Now, $2\in N$. But, there does not exist any element x in domain N such that $f\left(x\right)=x^3=2.$
Threfore, f is not surjective.
Hence, function f is injective but not surjective.

$f:Z\to Z$ is given by, $f\left(x\right)=x^3$
It is seen that for x, $y\in Z,f\left(x\right)=f\left(y\right)\Rightarrow x^3=y^3\Rightarrow x=y$
Therefore, f is injective.
Now, $2\in Z$. But, there does not exist any element x in domain Z such that $f\left(x\right)=x^3=2$. Therefore, f is not surjective.
Hence, function f is injective but not surjective.