Twists and half-twists on ribbons in $\mathbb{R}^4$ I've heard that in a manifold of dimension four or higher, if one takes a two-dimensional ribbon and gives it two half-twists (ie. two $180$-degree twists), it is the same as if the ribbon has no twists. Thus in $\mathbb{R}^3$, a ribbon can have $z$ half-twists where $z \in \mathbb{Z}$ because it can have any number of half-twists in either direction, whereas in $\mathbb{R}^4$, it can only have $z$ half-twists where $z \in \mathbb{Z} / 2\mathbb{Z}$.
I don't know the theory behind this, and it doesn't seem obvious. Can anyone offer a proof or some intuition for this fact?
 A: $\newcommand{\Reals}{\mathbf{R}}$Consider the parallel unit circles in $\Reals^{3} \times \{0\} \subset \Reals^{4}$ parametrized by
$$
c_{0}(u) = (\cos u, \sin u, 0, 0),\qquad
c_{1}(u) = (\cos u, \sin u, 1, 0),\quad 0 \leq u \leq 2\pi,
$$
and think of these as the boundary of an untwisted ribbon. "Tip" the second so that only two points lies in $\Reals^{3} \times \{0\}$. Slide the center appropriately, then tip the circle back into $\Reals^{3} \times \{0\}$ so that the two circles are now linked, e.g., via
\begin{align*}
  c(t, u)
  &= \bigl(t + \cos u, (1 - t)\sin u, 1 - t + t\sin u, \sqrt{2t(1 - t)}\sin u\bigr) \\
  &= \bigl(t + \cos u, 0, 1 - t, 0\bigr) + \sin u\bigl(0, 1 - t, t, \sqrt{2t(1 - t)}\bigr),\quad 0 \leq t \leq 1,
\end{align*}
whose image for each $t$ is a unit circle centered at $(t, 0, 1 - t, 0)$, and for which $c(0, u) = c_{1}(u)$ and $c(1, u) = (1 + \cos u, 0, \sin u, 0)$.
The mapping $H:[0, 1] \times [0, 2\pi] \times [0, 1] \to \Reals^{4}$ defined by
$$
H_{t}(u, v) = (1 - v)c_{0}(u) + vc(t, u),\quad 0 \leq t \leq 1,\ 0 \leq u \leq 2\pi,\ 0 \leq v \leq 1
$$
is a smooth homotopy through embeddings from the cylinder
\begin{align*}
  H_{0}(u, v)
  &= (1 - v)c(0, u) + vc(0, u) \\
  &= (\cos u, \sin u, v, 0)
\end{align*}
to the twice-twisted ribbon
\begin{align*}
  H_{1}(u, v)
  &= (1 - v)c_{0}(u) + vc(1, u) \\
  &= (v + \cos u, (1 - v)\sin u, v\sin u, 0).
\end{align*}
The animation shows its projection to $\Reals^{3}$.

