Show the identity map and the map $g(x,y)=(-x,-y)$ are homotopic maps from $S^1$ to itself This is a homework problem for a point-set topology course.
$\DeclareMathOperator{\img}{im}$
View $S^1$ as the unit circle $S^1 := \{(x,y)\in \mathbb R^2 \mid x^2 +y^2 = 1\}$. Let $f,g:S^1 \to S^1$ be the maps $f(x,y)=(x,y)$ and $g(x,y)=(-x,-y)$. Prove that $f$ and $g$ are homotopic.
Question. Is the below proof of this statement correct? (Edit: In particular, I am unsure about proving continuity rigorously.)
Proof. We proceed to show $f$ and $g$ are homotopic by construction of a homotopy ${F:S^1 \times I\to S^1}$ from $f$ to $g$.
For each $(x,y)\in S^1$, define the map $\underset{x,y}{\Psi}:I\to S^1$ to be the clockwise path in $S^1$ from $(x,y)$ to $(-x,-y)$. Now let $F$ be given by $F\left( (x,y),t \right) = \underset{x,y}{\Psi}(t)$, and we claim $F$ is a homotopy from $f$ to $g$.
When $t=0,1$ it is easy to see that $F$ is equivalent to $f$ and $g$, respectively. It remains to show that $F$ is continuous, or equivalently that the preimage of a basic open set is open.
[Side note: I guess it should be obvious that $F$ is continuous? But for sake of completeness here is my attempt at proving it.]
Let $U\subseteq S^1$ be a basic open set with respect to the basis of $\mathbb R^2$ comprised of open balls. Then $U=S^1\cap B$ for some open ball $B\subseteq \mathbb R^2$. If $U=S^1$, then $F^{-1}(U)=S^1 \times I$ and we are done. Assuming $U\neq S^1$, then $U$ is homeomorphic to an open interval, and $S^1 \setminus U$ to a closed interval. Let $(a,b)$ and $(c,d)$ be the endpoints of the arc $S^1\setminus U$, and then its preimage is given by
\begin{equation}
F^{-1}(S^1 \setminus U) = \left( \img \underset{-a,-b}{\Psi} \cup \img \underset{-c,-d}{\Psi} \right) \times I.
\end{equation}
Since $\img \underset{x,y}{\Psi}$ is closed for any $(x,y)\in S^1$, then the union in the above equation is closed, so $F^{-1}(S^1 \setminus U)$ is the product of closed sets and hence closed. Therefore $F^{-1}(U)$ is open, so $F$ is continuous and it follows that $f\simeq g$. $\Box$
Edit 1: Per some comments here is a more explicit version, viewing $S^1\subset \mathbb C$.
For each $x\in S^1$ define the function $h_x:[a,b]\to S^1$ via $h_x (t)=e^{2\pi i t}$, where $a:= \min (\{t\in I \mid e^{2\pi i t} =x \})$, and $b= a+1/2$, if $\operatorname{Im}(x)\geq 0$, or $a-1/2$ otherwise.
Then define $F$ via $F(x,t)=h_x (a+(b-a)t)$ and up to homeomorphism $F$ is a homotopy from $f$ to $g$.
 A: The map that sends each point $(x,y)$ to its antipodal point $(-x,-y)$ in the circle is the same as rotation by 180 degrees.  The identity map is rotation by o degrees.  The homotopy consists of all the rotations by intermediate amounts, from $0$ to $\pi$ radians.
I'll leave it up to you to formalize, but it's a lot easier than your long winded proof.
A: As I commented some hours before the two previous answers, viewing $S^1$ as a subset of $\Bbb C,$ the (obviously continuous) map
$$F:S^1 \times I\to S^1,\quad(z,t)\mapsto e^{i\pi t}z$$
is a homotopy from $f={\rm id}_{S^1}$ to $g=-{\rm id}_{S^1}.$
If you insist on staying in $\Bbb R^2,$ rewrite it
$$F\left((x,y),t\right)=(x\cos(\pi t)-y\sin(\pi t),x\sin(\pi t)+y\cos(\pi t)).$$
A: Thanks to @PatrickR for suggesting to use trigonometric functions and a simple rotation.
Proof. We proceed to show $f$ and $g$ are homotopic by construction of a homotopy ${F:S^1 \times I\to S^1}$ from $f$ to $g$.
For each $(x,y)\in S^1$, we note that $(x,y)=(\cos \theta, \sin \theta)$ for some angle $\theta \in [0,2\pi)$ with respect to the positive $x$-axis. In view of this fact, define $F$ via ${F(x,y,t)=(\cos(\theta+\pi t),\sin(\theta + \pi t))}$, and we claim $F$ is a homotopy from $f$ to $g$. When $t=0$, we obtain the identity map $f$. When $t=1$, the map $F$ represents a rotation counterclockwise by $\pi$ radians or $180^\circ$, which is equivalent to the rotation $(x,y)\mapsto (-x,-y)$ defined by $g$. Finally, since $F$ is composed of elementary trignometric functions, it is continuous. Hence $F$ is a homotopy and $f\simeq g$ as desired. $\Box$
