Question: Let a map $e \colon \mathbb{R} \rightarrow S^1$. $f_0,f_1$ are paths and $a \in \mathbb{R}$ such that $e(a) = f_0(0) = f_1(0)$. Prove that if $f_0 \sim f_1$ then $\ell_a f_0(1) = \ell_a f_1(1)$, with $e \colon \mathbb{R} \rightarrow S^1$ defined by $e(t) = e^{2 \pi i t} = \cos 2 \pi t + i \sin 2 \pi t, \forall t \in \mathbb{R}$ and the lifting of a $f$ towards $e$ with the origin at $a$ denoted by $\ell_a f$
My attempt: Let $H \colon f_0 \simeq f_1$, then exists unique a $\widetilde{H} \colon I \times I \rightarrow \mathbb{R}$ such that $\widetilde{H}(0,0) = a$ and $e \widetilde{H} = H$. We have $e\widetilde{H}(t,0) = H(t,0) = f_0(t)$, $\widetilde{H}(0,0) = a$ so $\widetilde{H}(t,0)= \ell_a f_0(t)$, for all $t \in I$; and $e\widetilde{H}(t,1) = H(t,1) = f_1(t), \widetilde{H}(0,1) = b$ so $\widetilde{H}(t,1) = \ell_b f_1(t)$, for all $t \in I$. Moreover, $e\widetilde{H}(1,s) = H(1,s) = f_0(1) = f_1(1)$, for all $s \in I$.
Let $\widetilde{H}(1,0) = c \in \mathbb{R}$, consider the constant line $c_c \colon I \rightarrow \mathbb{R}$. Since $e c_c(s) = e(c) = f_0(1) = f_1(1)$ so $\widetilde{H}(1,s)$ and $c_c(s)$ are lifting of constant line $c_d$, with $d = f_0(1) = f_1(1)$ and $\widetilde{H}(1,0) = c = c_c(0)$. Hence, $\widetilde{H}(1,s) = c_c(s) = c$, for all $s \in I$. So $\ell_a f_0(1) = \widetilde{H}(1,0) = \widetilde{H}(1,1) = \ell_a f_1(1)$.