Correspondence $(\mathbb{R}P^2,q) \rightarrow (\mathbb{R}^2,(0,0))$ and $(\mathbb{R}P^2,q) \rightarrow (S^1 \times S^1, x_0)$

Question

edit: my solution is garbage, correct solution in comments.

Let $p: \mathbb{R}^2 \rightarrow S^1 \times S^1$ be the standard universal covering map of the torus. Let $x_0=p(0,0)$. Suppose $q \in \mathbb{R}P^2$. Show that composition with $p$ induces a one-to-one correspodnence between the continuous maps $(\mathbb{R}P^2,q) \rightarrow (\mathbb{R}^2,(0,0))$ and the continuous maps $(\mathbb{R}P^2,q) \rightarrow (S^1 \times S^1, x_0)$.

Proof: my idea for the proof is to show that these sets of continuous maps are both contained in each-other. Therefore there will be a bijection between the sets.

If $f \in (\mathbb{R}P^2,q) \rightarrow (\mathbb{R}^2,(0,0))$, then composing this with $p$ will give a map $pf: (\mathbb{R}P^2,q) \rightarrow (S^1 \times S^1, x_0)$.

If $f \in (\mathbb{R}P^2,q) \rightarrow (S^1 \times S^1, x_0)$ then this map will lift uniquely to to $\tilde{f} \in (\mathbb{R}P^2,q) \rightarrow (\mathbb{R}^2,(0,0))$ such that $q\tilde{f}=f$.

I'm not sure if I answered the question, If someone could show me how to do this that would be great.

Answer 1

Define $E$ (resp. $F$) to be the collection of continuous maps $(\Bbb RP^2,q)\to(S^1\times S^1,x_0)$ (resp. $(\Bbb RP^2,q)\to(\Bbb R^2,(0,0))$).

" my idea for the proof is to show that these sets of continuous maps are both contained in each-other."

There is no inclusion between $E$ and $F$ so you can't really do this. Just do as the problem asks: define $\phi:F\to E$ by $\Phi(f)=p\circ f$. You have to show two things:

• $\Phi$ is one-to-one.

This comes from uniqueness of the lift of a map. If $f_1,f_2\in F$ are such that $\Phi(f_1)=\Phi(f_2)$, i.e $p\circ f_1=p\circ f_2$, then $f_1$ and $f_2$ are two lifts of $p\circ f_1$ which coincide at $q$, so they must be equal.

• $\Phi$ is onto.

For this you have to show that any $f\in E$ has a lift $\tilde{f}\in F$. You said this was true but you didn't actually prove it. To prove it you just have to use the lifting criteria, i.e prove that if $f\in E$ then $$f_*\left(\pi_1(\Bbb RP^2,q)\right)\subset p_*\left(\pi_1(\Bbb R^2,(0,0))\right)=\{0\}.$$ This is true because $f_*:\Bbb Z/2\Bbb Z\to \Bbb Z\times \Bbb Z$ must be trivial, as any group homomorphism $\Bbb Z/2\Bbb Z\to \Bbb Z\times \Bbb Z$ must be trivial.