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I am looking for a reference, or a topologically/analytically rigorous way of showing the following:

Consider two injective paths in $\mathbb{R}^2$ parametrized as $p_1(t)$, $p_2(t)$, $0 \leq t \leq 1$ respectively, such that $p_1$ and $p_2$ are disjoint everywhere except at the endpoints (i.e. $p_1(x) = p_2(x)$ iff $x \in \{0,1\}$). Let $S$ be a unit square somewhere in the plane. Suppose that $p_1$ enters $S$ through one of its edges and exits through a different one, and that $p_1$ enters $S$ exactly once. Further suppose that $p_2$ never touches any point of $S$. Show that the interior or boundary of the closed curve formed by $p_1$ and $p_2$ contains at least one corner of $S$.

This seems completely obvious to me, but I am not sure how one would show this rigorously.

If it helps, one may assume $p_1$ and $p_2$ are composed of polygonal lines.

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  • $\begingroup$ The composed path $p = p_2^{-1} * p_1$, where $p_2^{-1}(t) = p_2(1-t)$, is a Jordan curve. Now we may apply the well-known Jordan curve theorem en.wikipedia.org/wiki/Jordan_curve_theorem. This is a non-trivial result although it seems to be so obvious. $\endgroup$ – Paul Frost Oct 28 '18 at 10:40

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