# Continuous dependence of inverse functions on parameters

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?

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.