A contiuous function which cannot be liftedto the canonical covering of the real projective space.

Question

Let $p:S^n \longrightarrow \mathbb{R}P^n, \, x\mapsto \{x,-x\}$ be the canonical covering of the real projective space $\mathbb{R}P^n$. Denote by $j:\mathbb{R}^2 \longrightarrow \mathbb{R}^{n+1}$ the function $j(x_1,x_2)= (x_1,x_2,0,\ldots,0)$ and by $i:\mathbb{R}P^1\longrightarrow \mathbb{R}P^n,$ the function $i(\{x,-x\})=j(\{x,-x\})=\{j(x),-j(x)\}$.

Prove that $i$ is a continuous function which can not be lifted to the covering $(S^n,p)$.

If a lifting for $i$ exists, say $\tilde{i}$, then $p\circ \tilde{i}=i$. So, for every $x\in S^1, \tilde{i}(\{x,-x\})\in \{j(x),-j(x)\}$.

I am trying to figure out why $\tilde{i}$ can not be continuous. Any suggestion would be greatly appreciated.

Answer 1

For $n = 1$, I think you can probably work this out for yourself. Think "fundamental group."

Let's look at $n > 1$.

Suppose that there's a map $I: RP^1 \to S^n$ such that $p\circ I = i$.

Then on the fundamental-group level, you have

$$p_{*} ([I]) = [i],$$

where brackets indicate homotopy classes. Since $[i]$ is a generator of $\pi_1(RP^n) \approx \Bbb Z / 2\Bbb Z$, but $[I]$ is necessarily zero, because the fundamental group of the sphere is trivial (for $n > 1$). And since $p_{*}$ is a homomorphism, this is impossible, for a homomorphism never carries the identity element of the domain to a non-identity element of the codomain.