Why is a bijection that preserves connectedness on $\mathbf{R}$ must be monotone?

Solution 1:

Suppose that $f$ is not monotonic. Then there are numbers $a,b,c\in\Bbb R$ such that $a<b<c$ and that $f(b)$ is greater than both $f(a)$ and $f(c)$ or that $f(b)$ is smaller than both $f(a)$ and $f(c)$. Suppose that we are in the first case. You have $f(a)>f(c)$ or $f(a)<f(c)$ or $f(a)=f(c)$. In this last case, $f$ is not injective, and we're done. If $f(a)>f(c)$, then, $f([b,c])$ is not an interval, since it contains $f(b)$ and $f(c)$, but not $f(a)$. And if $f(a)<f(c)$, then, $f([a,b])$ is not an interval, since it contains $f(a)$ and $f(b)$, but not $f(c)$.

The other case is similar.