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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?

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Let $s_i \to s$ and $y_i \to y$. We have to show that $f_{s_i}^{-1}(y_i) \to f_s^{-1}(y)$.

Let us first show that the sequence $(f_{s_i}^{-1}(y_i))_{i \in \mathbb{N}}$ is bounded. Since $y_i$ converges, there are numbers $c$ and $d$ and an $\varepsilon > 0$ such that the set $\{y_i,y\colon i \in \mathbb{N}\}$ is contained in the open interval $(c-\varepsilon, d + \varepsilon)$. Since the function $s' \mapsto f(f_s^{-1}(d),s')$ is continuous, there exists a $\delta > 0$ such that $|f_{s'}(f_s^{-1}(d)) - d| < \varepsilon$ for all $s'$ with $|s' - s| < \delta$. A similar statement holds for $c$ instead of $d$. Since each $f_{s'}^{-1}$ is monotone (since it is continuous and bijective), we conclude $\{f_{s'}^{-1}(y_i)\colon i \in \mathbb{N}\} \subset [f_s^{-1}(c),f_s^{-1}(d)]$ (or in $[f_s^{-1}(d),f_s^{-1}(c)]$ if it is decreasing) for all $s'$ with $|s'-s| < \delta$. This proves the claim.

Since the sequence $(f_{s_i}^{-1}(y_i))_{i \in \mathbb{N}}$ is bounded, it has an accumulation point $\bar x$. Let $x_{i_k} = f_{s_{i_k}}^{-1}(y_{i_k})$ be a subsequence converging to it. Then we have $$f_s(\bar x) = f(\lim_{k \to \infty}(x_{i_k},s_{i_k})) \stackrel{!}= \lim_{k \to \infty} f(x_{i_k},s_{i_k}) = \lim_{k \to \infty} f_{s_{i_k}}(f_{s_{i_k}}^{-1}(y_{i_k})) = y = f_s(f_s^{-1}(y))\,,$$ where at the marked equality we used continuity of $f$. Since $f_s$ is injective, we conclude that $\bar x = f_s^{-1}(y)$. But $\bar x$ was the limit of $f_{s_{i_k}}^{-1}(y_{i_k})$. All in all, we conclude that the sequence $(f_{s_i}^{-1}(y_i))_{i \in \mathbb{N}}$ is bounded and only has one accumulation point, namely $\bar x = f_s^{-1}(y)$. It follows that the sequence $(f_{s_i}^{-1}(y_i))_{i \in \mathbb{N}}$ converges to it, proving continuity of our family of inverse functions.

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