Is the inverse of a real, continuous "1-1" function necessarily continuous itself?

Question

If so, please do provide me with an epsilon-delta proof, if possible.

Thanks in advance.

Answer 1

If the domain of $f$ is connected (i.e. an interval), then this is true. In that case the hypotheses imply that $f$ is strictly monotonic. Assume WLOG that $f$ is strictly increasing. Let $x_0$ by any point in the domain of $f$, and let $y_0 = f(x_0)$.

Let $\epsilon > 0$. Choose $\delta = \min(f(x_0)-f(x_0-\epsilon),f(x_0+\epsilon)-f(x_0))$. Then with this choice we have $f(x_0-\epsilon) < f(x_0)-\delta$ and $f(x_0)+\delta < f(x_0+\epsilon)$. This implies that any $y$ in open interval of radius $\delta$ around $f(x_0) = y_0$ satisfies $$f(x_0 - \epsilon) < y < f(x_0+\epsilon).$$

As $f^{-1}$ must also be strictly increasing, applying $f^{-1}$ to the inequalities above yields that $f^{-1}(y)$ belongs to the open interval of radius $\epsilon$ centered at $x_0 = f^{-1}(y_0)$. In other words, we have shown that $\lvert y - y_0 \rvert < \delta \implies \lvert f^{-1}(y) - f^{-1}(y_0)\rvert < \epsilon$.

Answer 2

Not without additional conditions. For example, if $f:[0,1)\cup [2,3]\to [0,2]$ with

$$f(x) = \begin{cases}x,& 0\leq x<1\\ x-1,& 2\leq x\leq 3 \end{cases}$$

then $f$ is continuous and 1-1 (and onto), but its inverse $f:[0,2]\to[0,1)\cup [2,3]$ with $$f^{-1}(x) = \begin{cases}x,& 0\leq x<1\\ x+1,& 1\leq x\leq 2 \end{cases}$$ is not continuous at $x=1$.

Answer 3

The most instructive example I can think of is not actually a function from $\mathbb{R}$ to $\mathbb{R}$. Rather it is the function $f : [0,2 \pi) \to S^1,f(t)=(\cos(t),\sin(t))$ where $S^1$ is the unit circle given the subspace topology from $\mathbb{R}^2$. The discontinuity occurs essentially because although $f$ is one-to-one, $f(0)$ and $\lim_{t \to 2 \pi^-} f(t)$ are the same. So "far apart" points are mapped "close together", even though they are not mapped to the exact same point, which creates a discontinuity in the inverse. This discontinuity is familiar from the elementary theory of complex numbers: it is the usual discontinuity in the argument function which arises from the ambiguity in the complex logarithm.

With that intuition in hand, you can make up examples in $\mathbb{R}$: take your favorite continuous, one-to-one function $f$ on $[0,2]$ and define $g : [0,1) \cup [2,3]$ by $g(x)=\begin{cases} f(x) & x \in [0,1) \\ f(x-1) & x \in [2,3] \end{cases}$. Then $\lim_{x \to 1^-} g(x)$ and $g(2)$ are the same which creates the same effect as before.