Showing there exists $g:\mathbb{S}^1 \rightarrow \mathbb{S}^1$ given odd $f:\mathbb{S}^1 \rightarrow \mathbb{S}^1$. In John M. Lee's Introduction to Topological Manifolds on pg 303, problem 11-4, the general idea is to show that every odd map $f:\mathbb{S}^1 \rightarrow \mathbb{S}^1$ has odd degree.  There are several answers here in stack exchange that seem helpful in the overall theme of the question and leading into Borsuk Ulam itself.  However, while I suspect its a fundamental misunderstanding on my part, I find myself stuck at the first hurdle.
Lee suggests in part (a) to show that if f is odd, there exists a continuous map $g:\mathbb{S}^1 \rightarrow \mathbb{S}^1$ s.t. $deg(f) = deg(g)$ and where $p_2$ is the covering map s.t. $p_2(z) = z^2$, where $p_2:\mathbb{S}^1 \rightarrow \mathbb{S}^1$  makes $p_2(f) = g(p_2)$.
My initial intuition is to use a unique lifting property, but I already have the "lift" here and am looking for the original map I would be "lifting."  Another thought is to use covering homomorphisms, but once again I feel like I'd be missing information.
Any helpful thoughts?
 A: Let's first see how to construct the function $g$ without worrying about continuity. Given $z \in \Bbb S^1$ (the $\Bbb S^1$ in the bottom-left corner of the diagram), the fiber of $p_2$ over $z$ contains exactly two points, which are opposites; that is, we have $z' \in \Bbb S^1$ such that $p_2(z') = p_2(-z') = z$. Since $f$ is odd, we have $f(-z') = -f(z')$, and then $p_2(f(-z')) = p_2(-f(z')) = p_2(f(z'))$. In order for the diagram to commute, we have to set $g(z) := p_2(f(z'))$. (If $f$ weren't odd, $g$ would have to send $z$ to both $p_2(f(z'))$ and $p_2(f(-z'))$, which might be different.)
Now to show that this is continuous, it's enough to check continuity locally. For any $z \in \Bbb S^1$ (again, the bottom-left $\Bbb S^1$), we can find an open neighborhood $U$ of $z$ and a local section of $p_2$, that is, a continuous map $s : U \to \Bbb S^1$ such that $p_2(s(z')) = z'$ for all $z' \in U$. (For instance, take $U := \Bbb S^1 \setminus \{-z\}$.) We defined our map $g$ in such a way that $p_2 \circ f \circ s = g|_U$. Since $g$ restricted to such open sets is continuous, $g$ is continuous.
