Let $F: I \times I \rightarrow \mathbb{S}^1$ a continuous function such that $F(0,t)=F(1,t)=p \in \mathbb{S}^1$ for a fixed $p$ (i.e. a homotopy between two loops in the circle based at $p$).
Show there exists a continuous $G:I\times I \rightarrow \mathbb{R}$ such that $\epsilon \circ G=F$, where $\epsilon: \mathbb{R} \rightarrow \mathbb{S}^1$ is the exponential map $t \mapsto e^{2\pi i t}$.
My try: fix $x \in \epsilon ^{-1}(p)$, and define $G$ as $G(s,t)=\tilde{\gamma}_t(s)$, where $\tilde{\gamma}_t$ is the unique lifting (with initial point $x$) of the closed loop $\gamma_t(s):=F(s,t)$. $G$ is well defined and lifts $F$ by definition.
How do I prove $G$ is continuous?
I tried the following approach with which I'm stuck. Using the Lebesgue number lemma and the properties of $\epsilon$, subdivide $I\times I$ in squares $R_{ij}$, for $i,j=1,...$ some $N$, with the following properties:
- for each $i,j$ there exists $V_{ij}$ open in $\mathbb{S}^1$, such that $\epsilon^{-1}(V_{ij})=\bigsqcup_{k \in \mathbb{Z}} U_{ij}^k$ for open sets $U_{ij}^k \subseteq \mathbb{R}$ and $\epsilon|_{U_{ij}^k}^{V_{ij}}$ is a homeomorphism
- $F(R_{ij})\subseteq V_{ij}$
By the gluing lemma, if I show $G|_{R_{ij}}$ to be continuous I'm done. Supposing $G(R_{ij})\subseteq U_{ij}^k$ for some $k$ I could write the implication $\epsilon\circ G|_{R_{ij}}=F \implies G|_{R_{ij}}=(\epsilon|_{U_{ij}^k}^{V_{ij}})^{-1} \circ F$. The problem is that I wasn't able to prove $G(R_{ij})\subseteq U_{ij}^k$. Is it true in the first place, and if so, how can you show this?