Consider two very simple maps from $\mathbb{R}^{2}$ to itself: the identity $I(x,y) = (x,y)$ and a reflection about the $y$-axis: $R_{y}(x,y) = (-x,y)$. For any embedded circle, $S^{1} = \{(x,y) \in \mathbb{R}^{2}, with \ x^{2} + y^{2} = r^{2} \}$, $I$ obviously preserves the orientation of of $S^{1}$ and $R_{y}$ reverses the orientation of $S^{1}$.
Using the fact that fundamental group of both $\mathbb{R}^{2} - (0,0)$ and $S^{1}$ is isomorphic to $ \mathbb{Z}$, we can show that $I$ and $R_{y}$ are not homotopic in $\mathbb{R}^{2} - (0,0)$; let $\gamma: [0,1] \rightarrow \mathbb{R}^{2} - (0,0)$, defined by $\gamma(t) = (\cos(2\pi t), \sin(2\pi t))$, be the counterclockwise circle of radius $1$, and let $[\gamma]$ denote its homotopy equivalence class in $\mathbb{R}^{2} - (0,0)$. Since $R_{y} \circ \gamma$ will be a clockwise loop in $\mathbb{R}^{2} - (0,0)$, $[R_{y} \circ \gamma] \neq [I \circ \gamma] = [\gamma]$. Since homotopies necessarily preserve the equivalence classes of loops, it follows that there is no homotopy from $I$ to $R_{y}$ in $\mathbb{R}^{2} - (0,0)$. I realize this proof was somewhat hand-wavy; it relied on the fact that there were two distinct generators (clockwise and counterclockwise) for the fundamental group of $\mathbb{R}^{2} - (0,0)$, and I did not prove this rigorously.
Transitioning from the space $\mathbb{R}^{2} - (0,0)$ to whole space $\mathbb{R}^{2}$, it is clear that $I$ and $R_{y}$ are then homotopic ; just consider $H_{t}(x,y) = ((1-2t)x, y)$; $H_{0} = I$ and $H_{1} = R_{y}$. However, how can it be proven that there is no isotopy, that is a homotopy $H: \mathbb{R}^{2} \times [0, 1] \rightarrow \mathbb{R}^{2}$ in which each $H_{t}$ is a homeomorphism from $I$ to $R_{y}$? Notice that I am not requiring that this isotopy be smooth as otherwise we could employ the use of differential forms and show that orientation must be preserved. Is there some topological invariant that an isotopy must preserve that that homotopy does not have to? My thoughts so far are that for the unit disk $D = \{ x^{2} + y^{2} \leq 1 \}$, the image $H_{t}(D)$ must always be homeomorphic to a disk if $H$ is an isotopy. Perhaps we could choose a path in the interior of $D$, $\eta: [0, 1] \rightarrow \{ x^{2} + y^{2} < 1 \}$ such that $\eta(t)$ always lives in the interior of $H_{t}(D)$. In this case, $H(D - \eta([0,1]))$ should be homeomorphic to $D - [- \frac{1}{2}, \frac{1}{2}]$. The fundamental group of $D - [-\frac{1}{2}, \frac{1}{2}]$ is identical to that of $D - (0,0)$ and $S^{1}$ (the integers $\mathbb{Z}$). Could we maybe use this to exclude the possibility of an isotopy from $I$ to $R_{y}$ since such a map would have continuously deform a counterclockwise loop in $H(D - \eta([0,1]))$ to a clockwise one? I know I am missing the details, but any suggestions would be appreciated. Thanks!
Note that there is an isotopy from $I$ to $R_{y}$ in $\mathbb{R}^{3}$; simply let $H_{t} = R_{y}(\pi * t)$, where $R_{y}(\alpha)$ is a rotation in $\mathbb{R}^{3}$ about the $y$-axis through angle $\alpha$. Then, $H: \mathbb{R}^{3} \times [0, 1] \rightarrow \mathbb{R}^{3}$ with $H_{0} = I$ and $H_{1} = R_{y}$.