If $f: A\to\mathbb R$ one-to-one but not monotone, there exist $x,y,z\in A$ with $x f(z)$ (wlog) The following result is part of the folklore, but I'd like to have a standard reference for something that I am writing:
If $A \subseteq \mathbb R$ and $f: A \to \mathbb R$ is one-to-one but not monotone, then there exist $x,y,z \in A$ with $x < y < z$ such that either $f(x) < f(y)$ and $f(y) > f(z)$, or $f(x) > f(y)$ and $f(y) < f(z)$.
The proof is easy, but boring and lengthy (unless I am missing something), and on the other hand, the result is essentially trivial. That's why I hope for a reference. I've checked, among the others, Bartle and Sherbert's book, but I couldn't find anything sufficiently close to the result.

We start from the fact that, by the standing assumptions, there exist $x,y,z,w \in A$ such that $x < y$, $z < w$, and either (i) $f(x) < f(y)$ and $f(z) > f(w)$, or (ii) $f(x) > f(y)$ and $f(z) < f(w)$.
In fact, we may assume wlog that (i) holds (otherwise we replace $f$ with $-f$) and $x \le z$ (otherwise we replace $f$ with the function ${-A} \to \mathbb R: a \mapsto f(-a)$, where ${-A} = \{-a: a \in A\}$.
Now we distinguish three cases: 
Case 1: $x = z$. Then either $x = z < y < w$ or $x = z < w < y$  (in fact, $y = w$ would imply $f(w) < f(z) = f(x) < f(y) = f(w)$, absurd). 
If $x < y < w$, then $f(x) < f(y)$ and $f(x) = f(z) > f(w)$, namely $f(x) < f(y)$ and $f(y) > f(z)$, and we are done. 
If $z = x < w < y$, then $f(x) = f(z) > f(w)$ and $f(w) < f(z) = f(x) < f(y)$, i.e. $x < w < y$ implies $f(x) > f(w)$ and $f(w) < f(y)$, and again we are done.
Case 2: $y = w$. It is perfectly analogous to the case $x = z$, and we may omit the details.
Case 3: $x < z$ and $y \ne w$. We have a couple of subcases:
Case 3.1: $f(z) \ge f(y)$. Then $f(x) < f(y) \le f(z)$, i.e. $f(x) < f(z)$, and on the other hand $f(z) > f(w)$, so we have $x < z < w$, $f(x) < f(z)$, and $f(z) > f(w)$. Done.
Case 3.2: $f(z) < f(y)$. If $y \le z$, the claim is proved, since then $x < y < w$, $f(x) < f(y)$, and $f(y) \ge f(z) > f(w)$. So we may assume that either $x < z < y < w$ or $x < z < w < y$, and we have two more subcases:
Case 3.2.1: $f(x) > f(z)$. Then $x < z < y$ (which is true in any case), on the one hand,
and $f(x) > f(z)$ and $f(z) < f(y)$, on the other. Done.
Case 3.2.2: $f(x) < f(z)$. Then $x < z < w$ (which is true in any case), on the one hand,
and $f(x) < f(z)$ and $f(z) > f(w)$, on the other. Done.

It is likely that the proof can be slightly shortened here and there, but I believe that it cannot be made into a one- or two-line proof, so I still beg for a reference.

This is a theorem about maps $f:\ X\to Y$ between ordered sets. Te claim is: If there are data points $$(x_k,y_k), \quad y_k:=f(x_k),\quad x_1\leq x_2\leq x_3\leq x_4,$$ testifying nonmonotonicity of $f$ then we can select $x_{k_1}<x_{k_1}<x_{k_3}$ from the $x_k$ such that the points $(x_{k_i},f(x_{k_i})\bigr)$ form a $\wedge$ or a $\vee$. Note that injectivity of $f$ is not required.
In the testimonial the point $(x_1,y_1)$ is paired with $(x_r,y_r)$, where $2\leq r\leq4$. In any case $x_r>x_1$, and wlog we may assume $y_r>y_1$. The remaining pair is then strictly descending.
When $r=2$ and $y_3<y_2$ then  the triple $x_1<x_2<x_3$ forms a $\wedge$. When $y_3\geq y_2$ then  the triple $x_1<x_3<x_4$ forms a $\wedge$.
When $r=3$ and $y_2>y_3$ then the triple $x_1<x_2<x_3$ forms a $\wedge$. When $y_2\leq y_3$ then $x_1<x_3<x_4$ forms a $\wedge$.
When $r=4$ and  $y_2>y_1$ then the triple $x_1<x_2<x_3$ forms a $\wedge$, if $y_2\leq y_1$ then the triple $x_1<x_3<x_4$ forms  a $\vee$. 
