Let $f:\mathbb{R}\to\mathrm{S}$, where $\mathrm{S}\subseteq\mathbb{R}$, be a strictly increasing (decreasing) function. I wish to find out whether its inverse $f^{-1}$ is also strictly increasing (decreasing) or not. While a proof for this exists online, I will provide a counterexample that seems to contradict the conclusions in the proof.
The proof can be found in: https://proofwiki.org/wiki/Inverse_of_Strictly_Monotone_Function
I will for simplicity only consider a strictly increasing function. From the provided link above we have that: \begin{align*} x < y & \quad\Leftrightarrow\quad f(x) < f(y) \\ f^{-1}(x) < f^{-1}(y) & \quad\Leftrightarrow\quad f^{-1}(f(x)) < f^{-1}(f(y)) \\ f^{-1}(x) < f^{-1}(y) & \quad\Leftrightarrow\quad x < y \\ \end{align*} Thus, if $f$ is a strictly increasing function, then so is $f^{-1}$.
Now for the counterexample. With some loss of generality, let $f:\mathbb{R}\to\mathbb{R}_{++}$ be a strictly positive function, where $\mathbb{R}_{++}=\{ x\in\mathbb{R}: x>0\}$. Furthermore, for an inverse it follows that $f^{-1}(x)f(x)=f(x)f^{-1}(x)=1$. Then from \begin{align*} x < y & \quad\Leftrightarrow\quad f(x) < f(y) \end{align*} we can rewrite the right-hand side as \begin{align*} f^{-1}(y)=\frac{1}{f(y)}<\frac{1}{f(x)}=f^{-1}(x) \end{align*} which implies that if $f$ is a strictly increasing function, then its inverse $f^{-1}$ is strictly decreasing.
Please provide comments if and why my counterexample is invalid. Thank you in advance.
It appears that you may be confusing the reciprocal function $\frac{1}{f(x)}$ with the inverse function (denoted $f^{-1}(x)$).
This latter notation, $f^{-1}$, is usually reserved for, and understood to be, the inverse function rather than the reciprocal function.
Edit:
In response to index's comment: I would prefer to structure the proof as:
We know $x<y \Leftrightarrow f(x)<f(y)$.
Hence $f^{-1}(x)<f^{-1}(y) \Leftrightarrow f(f^{-1}(x))<f(f^{-1}(y)) \Leftrightarrow x<y$.