Consider the continuous and injective mapping  

\begin{eqnarray\*}
\varphi:[0,1] &\rightarrow& \mathbb{R}^2, \\
t &\mapsto& (x(t),y(t)),
\end{eqnarray\*}
such that $x(0)<x(1)$, and
\begin{equation\*}
\big( (x(t)-x(s)\big)\big( y(t)-y(s)\big) \ge 0,\quad \forall t,s\in [0,1].
\end{equation\*}


My intuition is that $x(0)\le x(t)\le x(1)$ for any $0<t<1$.


I believe the key idea to solve this is to use the Intermediate Value Theorem, and the following result for univariate functions (continuity and injectivity together deduce monotonicity in, see <https://math.stackexchange.com/questions/170147/a-continuous-injective-function-f-mathbbr-to-mathbbr-is-either-strict>) to get the result, but still cannot proceed with it.


Thank you for your reading. Any help is very appreciated.



A continuous injection from $[0,1]$ to $\mathbb{R}^2$
If $x(t\_0)\notin [x(0),x(1)]$ for some $t\_0\in (0,1)$, then for a small positive $c$ we have $z(t\_0)\notin [z(0),z(1)]$ where $z=x+cy$. Thus $z$ is not monotone, therefore not injective. But two points with the same value of $z$ contradict to your second assumption (it is crucial here that $(x, y)$ is injective).


