Q.Show that the function $f : \mathbb{R} \to \mathbb{R}$ defined by $f(x) = x^2$ is neither one-one nor onto. Suggest a restriction of the domain and codomain that makes it a bijection.

Not one-one: $f(2)=4$ and $f(-2)=4$, so $f(2)=f(-2)$ but $2 \neq -2$. This single counterexample shows $f$ is not one-one.

Not onto: since $x^2 \geq 0$ for every real $x$, the range of $f$ is $[0,\infty)$, which is a proper subset of the codomain $\mathbb{R}$. For example, $y=-1$ has no real $x$ with $x^2=-1$. So $f$ is not onto. …

Give one explicit counterexample pair for injectivity and one explicit unattained codomain value for surjectivity, then restrict bot …

• Restricting only the domain and forgetting the codomain must also be narrowed to $[0,\infty)$ for onto to hold.
• Using $x=1,x=-1$ but forgetting to state clearly that $1 \neq -1$ is what breaks injectivity. …