If we have a bijective, continuous function $f\colon \mathbb{R} \to \mathbb{R}$, then it is proven in every introductory course to analysis that the inverse function $f^{-1}\colon \mathbb{R} \to \mathbb{R}$ will be also continuous.
Recently I needed this fact for continuous families of such functions, i.e., assume that $f\colon \mathbb{R} \times U \to \mathbb{R}$ is a continuous function, where $U \subset \mathbb{R}^r$ is an open subset (the parameter space), such that for each $s \in U$ the function $f_s\colon \mathbb{R} \to \mathbb{R}, x \mapsto f(x,s)$ is bijective. Since each $f_s$ is also continuous, we get a family of continuous inverse functions $\{f_s^{-1}\}_{s \in U}$. Does this family depend continuously on the parameter $s$, i.e., do these inverse functions assemble to a single continuous function $F\colon \mathbb{R} \times U \to \mathbb{R}, (y,s) \mapsto f_s^{-1}(y)$?
The answer is 'yes', but is there any way of proving this without an $\varepsilon$-$\delta$-massacre?