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If f is a real-valued function defined on a connected open set of real numbers, and f is injective and continuous, is the inverse of f continuous?

I realize that in general the inverse is not necessarily continuous, but I believe that in this case it is.

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up vote 5 down vote accepted

Since the continuous image of a connected set is connected, and since the connected sets in $\Bbb R$ are the intervals (possibly degenerate), $f$ maps intervals to intervals. Since $f$ is 1-1, it follows that $f$ is an open mapping. That is, $f$ maps open sets to open sets (in particular, open intervals to open intervals). Thus, the inverse $f^{-1}$ is continuous.


For completeness, I'll show that $f$ maps open sets to open sets (the other facts used above are all standard theorems):

Let $O=\text{ dom}(f)$ and let $U$ be an open subset of $O$.

Let $y$ be in $f(U)$. Choose $x\in U$ and an open interval $(a,b)$ containing $x$ with $[a,b]\subset U$ and $f(x)=y$. Then, since $f$ is 1-1 and continuous, $f\bigl(\,[a,b]\,\bigr)$ is either $\bigl[\,f(a),f(b)\,\bigr]$ or $[\,f(b),f(b)\,\bigr]$.

Without loss of generality, assume $f\bigl(\,[a,b]\,\bigr)=\bigl[\,f(a),f(b)\,\bigr]$. Then, since $y\in(\,f(a),f(b)\,\bigr)\subset f(U)$, it follows that $y$ is in the interior of $f(U)$.

Thus, every point of $f(U)$ is an interior point of $f(U)$; whence $f(U)$ is open.

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You need to prove first that it is strictly monotone, then the inverse is also monotone and its range is connected - it follows that it is continuous. See e.g. http://www.math.utah.edu/~taylor/3_Continuous p.74.

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