Solution
= Solution
{c}
If $x\ne y$, one of $x<y$ and $y<x$ holds. Since $f$ is <strictly increasing>, the corresponding values are strictly ordered, so $f(x)\ne f(y)$. Thus $f$ is an <injection>. Every element of the <image of a function> $f([a,b])$ equals $f(x)$ for some $x\in[a,b]$, which is precisely <surjectivity> onto that <image of a function>. Hence $f:[a,b]\to f([a,b])$ is a \b[<bijection>].