# Map from circle to real projective plane

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?.