Continuous injections fixing two points

Detailed Solution

A continuous injective function on an interval must be strictly monotone (a standard theorem). Since $f(0)=0<1=f(1)$, $f$ cannot be strictly decreasing, so it must be strictly increasing on all of $\mathbb R$ — this is (B). (This in turn forces $f(x)>0$ for $x>0$ and $f(x)<0$ for $x<0$, since an increasing function satisfies $f(x)>f(0)=0$ for $x>0$ and $f(x)<f(0)=0$ for $x<0$; so (C) is also a true consequence, but (B) is the more fundamental fact from which it follows.) Option (A) is too strong: $f(x)=x^3$ is continuous, strictly increasing, satisfies $f(0)=0$ and $f(1)=1$, yet $f(2)=8\neq2$. Option (D) is false: $f(x)=\frac{4}{\pi}\arctan(x)$ is continuous, strictly increasing, satisfies $f(0)=0$ and $f(1)=1$, but its range is the bounded interval $(-2,2)$, not all of $\mathbb R$. Thus the correct option is $\boxed{\mathrm{B}}$.