Application of intermediate value theorem or Borsuk-Ulam theorem on cylinder? Consider  three distinct points $a,b,c$ on the top of the cylinder with anticlockwise direction. Consider other three distinct points $a',c',b'$ in the bottom of the cylinder. There are continuous paths  connecting $a$ with $a'$,$b$ with $b'$ and $c$ with $c'$. As we can see in the figure that  anticlockwise orientation of points $a',b',c'$ is changed, that is, $c'$ got in between $a'$ and $b'$ (all transformations are continuous). I do not want this happen, that is, I do not want the orientation of $a',b',c'$ to get changed. This claim will follow if I would be able to prove that we cannot connect $c$ with $c'$ by a continuous path unless we cross one of the other two paths. This will lead me to a certain contradiction and I am done. 
As we can see in the diagram that the continous red path between $c$ and $c'$ must intersect one of the other either paths connecting $a$ with $a'$ or $b$ with $b'$. How to formulate this result mathematically (with equations and all) and how to prove it. I am wondering if it is the application of a generalized Intermediate value theorem or Borsuk-Ulum theorem on cylinder? I rewrite the question:

Prove that $c$ and $c'$ cannot be connected by a continuous path $C$ unless this path $C$ intersects any of the other two paths on the cylinder.

EDIT: The path A from $a$ to $a′$ and the path B from $b$ to $b′$ do not intersect.

 A: Hint Assuming that the $z$-coordinate of each path is a strictly monotone function of $t$ as illustrated in the drawing, we can regard the paths as maps $\alpha, \beta, \gamma : I \to \Bbb S^1 \subset \Bbb R^2$ satisfying $\alpha(0) = a', \alpha(1) = a$, etc. At the top of the cylinder, $a, b, c$ are in anticlockwise order, so the quantity $$\det\pmatrix{b - a & c - a} = (b_1 - a_1)(c_2 - a_2) - (c_1 - a_1)(b_2 - a_2)$$ is positive (its value is just twice the area of the triangle $\triangle abc$), but at the bottom of the cylinder the reverse is true. 

Additional hint This suggests considering the quantity $$f(t) := \det \pmatrix{\beta(t) - \alpha(t) & \gamma(t) - \alpha(t)} .$$ At $t = 0, 1$, we have $$f(0) = \det \pmatrix{b' - a' & c' - a'} < 0 , \qquad \det \pmatrix{b - a & c - a} > 0,$$ so the Intermediate Value Theorem gives that there is some time $t_0 \in [0, 1]$ such that $f(t_0) = 0$. But this implies that $b(t_0) - a(t_0)$ and $c(t_0) - a(t_0)$ are parallel, so either one is zero---which implies $a(t_0) = b(t_0)$ or $a(t_0) = c(t_0)$---or they are equal---in which case $b(t_0) = c(t_0)$.

A: I'll also assume that:
1) The path $A$ from $a$ to $a'$ and the path $B$ from $b$ to $b'$ do not intersect, and
2) The $z$-coordinate of each path is monotone.
Assumption 1) is clearly necessary; assumption 2) is almost certainly unnecessary but makes my life a lot easier.
Suppose that $C$ does not intersect $A$. Since $B$ also does not intersect $A$, $C$ and $B$ are both paths in $S^1 \times [0,1]-A$. Because $A$ is monotone, $S^1 \times [0,1]-A$ is homeomorphic to $(0,1) \times [0,1]$.
Because $B$ and $C$ are monotone, they correspond under this homeomorphism to functions $f,g$ from $[0,1]$ to $(0,1)$. Because the orientations on the two edges are opposite, we have $f(0)<g(0)$ and $f(1)>g(1)$. So by the intermediate value theorem we have $f(c)=g(c)$ for some $c$, which gives an intersection between $B$ and $C$.
A: Here's an argument using the difficult Jordan Curve Theorem that applies to the case that at least two of the curve are simple (i.e., do not have any self-intersections), say, the curves $\alpha, \beta$ respectively corresponding to $a, b$. Even failing that condition, the result can probably be recovered by replacing (with some care) the curves with suitable simple curves.
Map the cylinder $S^1 \times [0, 1] \subset \Bbb C \times [0, 1]$ to the annulus $A := \{1 \leq \sqrt{x^2 + y^2} \leq 2\} \subset \Bbb C$ via the homeomorphism $$\Phi: (z, s) \mapsto (s + 1) z .$$  Then construct a simple path $\delta$ from $\Phi(a)$ to $\Phi(b)$ not intersecting $A$ elsewhere and a simple path $\epsilon$ from $\Phi(a')$ to $\Phi(b')$ not intersecting $A$ elsewhere. By construction the concatenated curve $$\zeta := \Phi(\alpha) \cdot \delta \cdot \Phi(\beta^{-1}) \cdot (\epsilon^{-1})$$ is a Jordan curve in $\Bbb C$ and hence partitions $\Bbb C$ into a bounded open set $U$, the image $Z$ of $\zeta$, and an unbounded open set $V$. In particular, intersecting these three sets with $A$ gives a partition of $A$ into two open sets and the union of the images of $\alpha$ and $\beta$. But $a, b, c$ and $a', b', c'$ having the relative orders indicated in the diagram implies that $\Phi(c) \in A \cap V$ but $\Phi(c') \in A \cap U$. So, if $\gamma$ is the curve from $c$ to $c'$, $\Phi \circ \gamma$ intersects the union of the images of $\Phi \circ \alpha$ and $\Phi \circ \beta$, and applying the inverse homeomorphism gives that $\gamma$ intersects the image of $\alpha$ or the image of $\beta$.

At any rate, I still strongly suspect that there must be a short algebraic-topological proof (and certainly one that does not rely on as much machinery as this argument does via Jordan Curve Theorem).
