I'm trying understand the article "Curvature bound for curve shortening flow via distance comparison and a direct proof of Grayson's theorem" by Ben Andrews and Paul Bryan and they stated that

Define $a(p, q) = \inf \ \{e^t : d(p, q) ≥ f(ℓ(p, q), −t) \}$ for $p \neq q$ in $S^1$. Then by the implicit function theorem $a$ is continuous, and smooth and positive where $0 < d < 2 \sin(ℓ/2)$ where it is defined by the identity

$$d(p, q) = f(ℓ(p, q), − log(a(p, q))).$$

I'm stuck in understanding why the Implicit Function Theorem ensures that $a$ is smooth and why the identity $d(p, q) = f(ℓ(p, q), − log(a(p, q)))$ is valid for $0 < d < 2 \sin(ℓ/2)$? I think that the Implicit Function Theorem is applied on the $C^k$ function

$$F: S^1 \times S^1 \times \mathbb{R}\ - \left( \{ (p,q) \in S^1 \times S^1 \ ; \ p = q \} \times \mathbb{R} \right) \longrightarrow \mathbb{R}$$

defined by $F(p,q,t) := d(p, q) - f(ℓ(p, q), −t)$, because $\frac{\partial F}{\partial t} = \frac{\partial f}{\partial t} > 0$ as computed in the article and exist $(p_0, q_0, t_0) \in \text{Dom} \ F$ such that $F(p_0,q_0,t_0) = 0$ (we can ensure this because we choose $t_0$ so that $d(p_0, q_0) = f(ℓ(p_0, q_0), −t_0)$), then exist open sets $U$ containing $(p,q)$ and $V$ containing $t$ so that $g: U \longrightarrow V$ is a function $C^k(U)$ and $F\left(p, \ q, \ g(p,q)\right) = 0$ for all $p,q \in U$.

The Implicit Function Theorem ensure that exist a function $g$, but the theorem doesn't ensure that the function $a$ is the function $g$. I didn't understand why the smoothness of $a$ is ensured by Implicit Function Theorem. If anyone can help me in understand this and why $d(p, q) = f(ℓ(p, q), − log(a(p, q)))$ for $0 < d < 2 \sin(ℓ/2)$, I will be grateful!