Homotopy from $f(x,y) = (x,y)$ to $g(x,y) = -(x,y)$ 
Let $\hat{f} : \Bbb S^1 \to \Bbb R^2$, $\hat{f}(x,y) = (x,y)$ and $\hat{g} : \Bbb S^1 \to \Bbb R^2$, $\hat{g}(x,y) = -(x,y)$. Show that there exists a Homotopy $\hat{H} : \Bbb S^1 \times [0,1] \to \Bbb R^2$ from $\hat{f}$ to $\hat{g}$.

So both $\hat{f}$ and $\hat{g}$ are mapping points from the unit disk to the plane. Isn't the image of both maps just the disk back itself? I'm confused on how to get an intuition for the problem here. The definition of Homotopy is that I would need to construct $\hat{H}$ such that $$\hat{H}(x,y,0) = \hat{f}$$ and that $$\hat{H}(x,y,1) = \hat{g}.$$ Certainly $$\hat{H}(x,y,t) = (tx, ty)$$ doesn't work since $\hat{H}(x,y,0) = (0,0) \ne (-x,-y).$ What can I do here?
 A: Here is an approach that doesn't use any parametrisation of $S^1$: Define
$$
\hat{H} : S^1 \times [0,1] \to \mathbb{R}^2, \qquad ((x,y),t) \mapsto ((-2t+1)x,(-2t+1)y).
$$
This map is continuous and you can check that it indeed defines the desired homotopy between $\hat{f}$ and $\hat{g}$. Hope this helps!
A: $\hat{f}$ and $\hat{g}$ map from the unit circle ($S^1$) to the plane. Intuitively, one can imagine that $t\in[0,1]$ parameterizes a rotation along the circle. In particular, at $t=0$, we rotate $(x,y)$ an angle of $0$, and at $t=1$, we rotate $(x,y)$ an angle of $\pi$. In general, we can imagine the function $h_t(x,y)=\hat{H}(x,y,t)$ as rotating the point $(x,y)\in S^1$ by an angle of $\pi t$ and then throwing the resulting point to $\mathbb{R}^2$ via the canonical inclusion $S^1\hookrightarrow\mathbb{R}^2$.
So if we took coordinates on $S^1$ to be $(\cos{\theta},\sin{\theta})$ instead of $(x,y)$, a homotopy would be given by
$$\hat{H}(\cos{\theta},\sin{\theta},t)=\left(\cos{(\theta+\pi t)},\sin{(\theta+\pi t)}\right).$$
It is easy to see that this is indeed a homotopy from $\hat{f}$ to $\hat{g}$.
