Given continuous maps $f,g:S^1 \rightarrow S^1$, and the quotient map $\pi: [0,1] \rightarrow S^1$ satisfying $(f \circ \pi )(0)=(g \circ \pi )(0) $. How does this condition mean that we can lift $f$ and $g$ to maps $\tilde{f},\tilde{g}: [0,1]\rightarrow \mathbb{R}$ satisfying $\tilde{f}(0)=\tilde{g}(0)$.
Where $e: \mathbb{R} \rightarrow S^{1}$ s.t. $e(x)= (\cos(2 \pi x),\sin(2 \pi x))$.
Doubt: I understand that $e(\tilde{f}(0)) = (f \circ \pi )(0) = (g \circ \pi )(0) = e(\tilde{g}(0)) $, but how does that lead to $\tilde{f}(0)=\tilde{g}(0)$?
The map $e$ is a covering projection and therefore a fibration, i.e. has the homotopy lifting property. For a map $\varphi : [0,1] \to S^1$ this implies that for any $t \in \mathbb{R}$ such that $e(t) = \varphi(0)$ there exists a lift $\tilde{\varphi}$ such that $\tilde{\varphi}(0) = t$. Hence you obtain lifts $\tilde{f}, \tilde{g}$ as desired because you can take the same $t$ for $f, g$. However, $t$ is not unique. In fact, any $t \in e^{-1}(\varphi(0))$ will do.
Usually $\pi$ is taken as the restriction $e \mid_{[0,1]}$ though there are many other choices. In that case $(f \circ \pi)(0) = f(1)$ and any $t \in e^{-1}(f(1))$ will do.