Map $\phi: S^2 \rightarrow \mathbb{R}\mathbb{P}^2$ does not have sections [closed]

Consider the map $\phi: S^2 \rightarrow \mathbb{R}\mathbb{P}^2$ defined by $\phi(\overrightarrow{v})=[\mathbb{R} \cdot \overrightarrow{v}]$. Im trying to prove this map does not have sections. I know this means I should show there isnt any map $\gamma: \mathbb{R}\mathbb{P}^2 \rightarrow S^2$ such that $\phi \circ \gamma = id$ but I dont know how to do this.

Solution 1:

Suppose that $\gamma$ existed so that the composition $$\mathbb R P^2 \xrightarrow{\gamma} \mathbb S^2 \xrightarrow{\phi} \mathbb R P^2 $$ is the identity.

We can now apply any of our favorite algebraic topology functors: fundamental group, first homology group, second homology groups, whichever we like. Hopefully, picking the right functor will give us some kind of contradiction.

Let's try the second homology group. The composition $$H_2(\mathbb R P^2) \xrightarrow{\gamma_*} H_2(S^2) \xrightarrow{\phi_*} H_2(\mathbb R P^2) $$ is the identity. Translation: The composition $$0 \xrightarrow{\gamma_*} \mathbb Z \xrightarrow{\phi_*} 0 $$ is the identity. Well...... frankly... there's no contradiction to be had there, at all.

Ahhhh.... but the first homology group functor. The composition $$H_1(\mathbb R P^2) \xrightarrow{\gamma_*} H_1(\mathbb S^2) \xrightarrow{\phi_*} H_1(\mathbb R P^2) $$ is the identity. Translation: the composition $$\mathbb Z / 2 \mathbb Z \xrightarrow{\gamma_*} 0 \xrightarrow{\phi_*} \mathbb Z / 2 \mathbb Z $$ is the identity. Now there's a nice contradiction!

Solution 2:

The map $\phi$ is a quotient map identifying pairs of antipodal points $x, -x$ . It is an open map: Let $U \subset S^2$ be open. We have $\phi^{-1}(\phi(U)) = U \cup (-U)$, where $-U = \{-x \mid x \in U\}$. The antipodal map $a : S^2 \to S^2,a (x) = -x$, is a homeomorphism, thus $-U = a(U)$ is open. We conclude that $\phi^{-1}(\phi(U))$ is open, and therefore $\phi(U)$ is open because $\phi$ is a quotient map.

Therefore $\phi$ is a covering map with two sheets which means that each $y \in \mathbb{RP}^2$ has an open neighborhood $V$ such that $\phi^{-1}(V) = U_1 \cup U_2$ with disjoint open $U_i \subset S^2$ which are mapped by $\phi$ homeomorphically onto $V$. To see this, pick $x \in \phi^{-1}(y)$ and choose an open neigborhood $U$ of $x$ which does not contain any pair of antipodal points (you may take $U$ to be one of the six hemispheres $U_j^\pm = \{ x =(x_1,x_2,x_3) \in S^2 \mid (-1)^{\pm 1} x_j > 0 \}$; note that they cover $S^2$). Then $V = \phi(U)$ has the property $\phi^{-1} = U \cup (-U)$, where $U_1 = U,U_2 = -U$ are disjoint. Clearly $\phi : U_i \to V$ is a continuous open bijection, hence a homeomorphism.

Assume that there exists a map $\gamma: \mathbb{R}\mathbb{P}^2 \rightarrow S^2$ such that $\phi \circ \gamma = id$. Then $R= \gamma(\mathbb{RP}^2)$ is compact, hence closed in $S^2$. On the other hand $R$ must be open:

Let $x \in R$, i.e $x = \gamma(y)$ for some $y$. Let $U$ be an open connected neigborhood $U$ of $x$ in $S^2$ which does not contain any pair of antipodal points (you may take again one of the $U_j^\pm)$. Then $V = \phi(U)$ is a a connected open neigborhood of $\phi(x) = y$. Hence $W = \gamma(V)$ is connected and must be contained in $\phi^{-1}(\phi(W)) = \phi^{-1}(V) = U \cup (-U)$. Hence as a connected set it must be contained either in $U$ or in $-U$. But $x = \gamma(y) \in \gamma(V)$, thus $\gamma(V) \subset U$ since $x \in U$. Moreover, we have $\gamma(V) = U$. Otherwise there would exist $x^* \in U \setminus \gamma(V)$. But $y^* = \phi(x^*) \in V$, thus $\gamma(y^*) \in\gamma(V) \subset U$. Since $\phi(x^*) = \phi(\gamma(\phi(x^*))) = \phi(\gamma(y^*))$, we have $\gamma(y^*) = x^*$ or $\gamma(y^*) = -x^*$. The latter says that $\gamma(V)$ intersects $-U$, a contradiction. But $\gamma(V) = U$ shows that $R$ contains the open neigborhood $U$ of $x$ in $S^2$. Therefore $R$ is open in $S^2$.

Since $S^2$ is connected, we see that $R = S^2$. Thus $\gamma$ is surjective. Since it is also injective (recall $\phi \circ \gamma = id)$, it is bijective and we conclude that $\phi$ is bijective. In fact, $\phi = \gamma^{-1}$. But this is not true. Therefore $\phi$ cannot have a section.