can you help me with these proofs?

Let $f: S^1 \to \mathbb{R}P^2 $ be a map from circle to real projective plane and $\pi: S^2 \to S^2/\sim$ be the quotient map where $\sim$ identifies antipodal points on the sphere. $$f(x,y)=\pi(x/2,y/2,\sqrt3/2)$$ a) Show that $f$ is homotopic with the constant map.
b) Show that $\mathbb{R}P^2-f(S^1)$ has two connected components one of which is not a contractible space.

I know that $S^2/\sim$ is homeomorphic with $\mathbb{R}P^2$ but I don't see how to take it from there.


$f$ has the form $f = \pi \circ g$ with $g : S^1 \to S^2, g(x,y) = (x/2,y/2,\sqrt{3}/2)$.

$g$ is homotopic to a constant map because $S = g(S^1) \subset S^2_+$ (= upper hemisphere of $S^2$) which is homeomorphic to a disk and thus contractible. [By the way, it is well-known that all all maps $S^1 \to S^2$ are homotopic to a constant map.]

Hence $f$ is is homotopic to a constant map.

Note that $f$ is an embedding since for $p \ne q$ the points $g(p), g(q)$ are never antipodal, i.e. we get $f(p) \ne f(q)$.

Let $-S = \{ -p \mid p \in S \}$. Then $R = \mathbb{R}P^2 \setminus f(S^1)$ is nothing else than $\pi(D)$ with $D = S^2 \setminus (S \cup -S)$.

$S$ is a copy of the circle in $S^2$ (it is the intersection of the plane $z = \sqrt {3}/2$ with $S^2$). The space $D$ has three components $C_+$ above $z = \sqrt{3}/2$, $C_-$ below $z = -\sqrt{3}/2$ and $B$ between $z = \sqrt{3}/2$ and $z = -\sqrt{3}/2$. These three sets are open in $S^2$.

$\pi(C_+) = \pi(C_-) = C'$ and $\pi(B) = B'$ are connected subsets of $R$. Both $C', B'$ are open in $\mathbb{R}P^2$ (and therefore open in $R$) because $\pi^{-1}(C') = C_+ \cup C_-$ and $\pi^{-1}(B') = B$. Moreover, they are disjoint because $\pi^{-1}(C' \cap B') = \pi^{-1}(C') \cap \pi^{-1}(B') = \emptyset$. This shows that $R$ has $C',B'$ as its two components.

Obviously $C'$ is homeomorphic to $C_+$ which is contractible.

The component $B'$ is homotopy equivalent to $S^1$ which is not contractible. For a proof see the second answer to What is the space by identifying the antipodal points of a cylinder?.


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