Indian Statistical Institute B.Math & B.Stat : Continuity and Bijections

$Problem-1.$ Let $f: \mathbb{R} \to \mathbb{R}$ be given by $f(x) = x(x-1)(x+1)$. Then show that $f$ is onto but not 1-1. The injectivity part is very much obvious. See that $f(0)=f(1)=f(-1) = 0$. Now for the surjectivity, observe that $f$ is an odd degree polynomial. Hence it must be surjective! (Think if not, then $\exists$ a real number which has no pre-image (say $r$ ). Now consider the polynomial $g(x) = f(x)-r$!! what else g(x) is also an odd degree polynomial with no real roots!!! ) $Problem-2.$ Another Problem with a kick of continuity. Let $f: \mathbb{R} \to \mathbb{R}$ be given by $f(x) = x^3-3x^2+6x-5$. Then show that $f$ is both onto and 1-1. Again note that $f$ is an odd degree polynomial so it is surjective. Since $f$ is a polynomial it is continuous on $\mathbb{R}$. Also $f'(x)=3x^2-6x+6 = 3(x^2-2x+2) = 3\{(x-1)^2 + 1 \} > 0$ $\forall x \in \mathbb{R}$. This shows that $f$ is strictly increasing. Thus $f$ is continuous and strictly increasing hence must be 1-1. $Problem-3.$ Let $f: \mathbb{R} \to \mathbb{R}$ be given by $f(x) = x^2 - \frac{x^2}{1+x^2}$. Then show that $f$ is neither onto nor 1-1. Clearly $f$ is not injective since $f(x)=f(-x)$. Also $f(x) = \frac{x^4}{1+x^2}$ which shows that $f$ assume non-negative values. Thus the $Range_f \subset \mathbb{R} = Co-domain_f$. Thus $f$ is neither surjective. $Problem-4.$ Let $\phi :[0,1] \to [0,1]$ be a continuous and 1-1 function. Let $\phi(0) = 0, \phi(1) = 1, \phi \big(\frac{1}{2}\big) = c, \phi \big(\frac{1}{4}\big) = d.$ Then show that $c > d$. This follows form the property of continuous functions. On an interval I ( not necessarily compact ), if a function is continuous and 1-1 it is strictly increasing. To prove this you can use the Intermediate Value Theorem.