# Is $f(x)=x+\sin x$ a homeomorphic function?

I want to determine $$f(x) = x+\sin x$$ is homeomorphic or not on $$\mathbb{R}$$?

A bijective continuous function is homeomorphic if its inverse is also continuous. I know that $$f$$ is bijective. Also $$f$$ is continuous being the sum of two continuous functions.

How to look for the continuity of $$f^{-1}$$.

• Which properties? Commented Nov 22, 2018 at 18:32
• continuity of both sides.
– PAMG
Commented Nov 22, 2018 at 18:34
• The statement follows from the inverse function theorem. Commented Nov 22, 2018 at 18:38
• @ José Carlos Santos see the edited part.
– PAMG
Commented Nov 22, 2018 at 18:38
• @freakish Not quite: the derivative vanishes at $x=(2k+1)\pi$, $k\in\mathbb Z$. Commented Nov 22, 2018 at 18:42

The function $$f$$ is continuous and strictly increasing, because its derivative is $$\ge0$$ and is positive on the intervals $$(\pi+2k\pi,\pi+2(k+1)\pi)$$.
This its inverse function exists and is strictly increasing as well and defined over $$\mathbb{R}$$ because $$f$$ is neither upper nor lower bounded. In particular $$f^{-1}$$ has left and right limit at every point, because $$\lim_{x\to c^-}f^{-1}(x)=\sup\{f(x):xc\}$$ Suppose that at some point $$c$$ the two limits are different, say $$\lim_{x\to c^-}f^{-1}(x)=\sup\{f(x):xc\}=b$$ with $$a. Then $$(a+b)/2$$ doesn't belong to the codomain of $$f^{-1}$$, which is the domain of $$f$$. Contradiction.
Once you know that $$f$$ is strictly increasing, just use this.
If you want to know some regularity of $$f^{-1}$$, continue reading.
Let $$N=\{(2k+1)\pi:k\in\mathbb Z\}$$.
In $$\mathbb R\setminus N$$ you have $$f'\neq 0$$, so $$f$$ is actually a diffeomorphism. Now we have to investigate what happens near $$N$$. By translation, it is sufficient to examine what happens around $$\pi$$. By Taylor expansion, we have that for $$\epsilon>0$$ small enough $$|f(x)-f(\pi)| = |x+\sin(x)-\pi| \geq \left|\frac{(x-\pi)^3}{12}\right| \qquad \forall x\in(\pi-\epsilon,\pi+\epsilon)$$ so the inverse $$f^{-1}$$ is $$\tfrac13$$-Holder near $$\pi$$.