Why is the inverse function of a monotonous function $f:\mathbb{R}\rightarrow \mathbb{R}$ the reflection on the $0$-diagonal of the coordinate system? Every monotonous function $f:\mathbb{R}\rightarrow\mathbb{R}$ is injective, therefore $f:\mathbb{R}\rightarrow f(\mathbb{R})$ is bijective. I.e. it is invertible. The function $\bar{f}:f(\mathbb{R})\rightarrow\mathbb{R}$ is the inversefunction. Where is the Connection with the reflection on the diagonal intersecting $0$ of the coordinate System, or also the $id$ function.

As per my comment, let's define $Gr_f =\left\\{(x,y)\in \mathbb{R}^2 \mid f(x)=y\right\\}$ (from **Gr** aphic).

Now, if $(x,y)\in Gr_f \iff f(x)=y \iff f^{-1}(y)=x \iff (y,x) \in Gr_{f^{-1}}$, $f^{-1}$ is the notation for the $f$'s inverse. As a result

> $(x,y) \in Gr_f \iff (y,x) \in Gr_{f^{-1}}$

But $(x,y)$ and $(y,x)$ are symmetric to each other with respect to $id_{\mathbb{R}}$

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