Homotopy between $id_{D^2}$ and $-id_{D^2}$ but not $id_{S^1}$ and $-id_{S^1}$.

Question

If $f:D^2 \rightarrow D^2, (x, y) \mapsto (-x, -y)$, (where $D^2$ is the unit disk including the border), is homotopic to $id_{D^2}$, via the homotopy $F((x,y), t) = (\cos(\pi t)x – \sin(\pi t)y; \sin(\pi t)x + \cos(\pi t)y)$, why is the function $f: S^1 \rightarrow S^1; (x, y) \mapsto (-x, -y)$ not homotopic to $id_{S^1}$? (Where $S^1$ is the unit circle).

Answer 1

First observe that any two maps $f_0, f_1 : D^2 \to D^2$ are homotopic. This is most simply shown using the convexity of $D^2$: Define $F : D^2 \times [0,1] \to D^2, F(\xi,t) = tf_1(\xi) +(1-t)f_0(\xi)$. Then $F(\xi,i) = f_i(\xi)$ for $i = 0,1$. But if $f_i(S^1) \subset S^1$ for $i=0,1$, we do not have $F(S^1 \times I) \subset S^1$ unless $f_0 = f_1$.

Your homotopy $F$ describes a continuous rotation from the identity $id_{D^2}$ to $-id_{D^2}$. It has the special property that if $(x,y) \in S^1$, i.e. $x^2 +y^2 = 1$, then $$(\cos(\pi t)x – \sin(\pi t)y)^2 + (\sin(\pi t)x + \cos(\pi t)y)^2 = x^2 + y^2 = 1$$ which means $F((x,y),t) \in S^1$. This shows that $id_{S^1} \simeq -id_{S^1}$.