# Topological Equivalence between $(- \pi/2, \pi/2)$ to $R$

I know that the key is to use $\tan$ and $\arctan$ to do it. Take any $(a,b) \subset \mathbb{R}$, $(a,b)$ is open. Now I want to show $\tan^{-1}(a,b)$ is open. I need a hint for the next step (just a hint suffices) Thanks.

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Well once I wrote the question I was thinking the tangent and arctan function are inverses and the arctan should map an open interval into an open interval since they are continuous. – Daniel Oct 10 '12 at 19:04
It's not clear what you're allowed to take on faith for solving this exercise. Proving that $\tan$ and $\arctan$ are continuous from scratch won't be simple. I imagine your book expects you to just quote this fact from a real analysis/calculus course you've previously taken. – user29743 Oct 10 '12 at 19:28
@jsk to prove that $f^{-1}(O)$ is an open in $\mathbb{R}$, whit $O$ is an open in $\mathbb{R}$, you should notice that $f^{-1}$ ought to be continous. then prove that $f^{-1}(O)$ is included in an open set in $\mathbb{R}$. Since we are dealing whit a real valued function here. – Mohamez Oct 10 '12 at 19:46
@countinghaus yes I was thinking about the continuity of $\tan$ and $arctan$ too. I think I will take it as granted they are continuous since I have never actually proved the continuity of trigonometric functions before – Daniel Oct 10 '12 at 21:35

## 1 Answer

Look at the definition of continuous function in terms of open sets.

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The topological equivalence follows directly from the continuity of $\tan$ and $\arctan$? (well I think so...) – Daniel Oct 10 '12 at 19:10
@jsk, well, you have to note that they are bijective on that interval. Then it is an homeomorphism since tan is continuous and arctan is also continuous (both are differentiable, hence continuous) – sxd Oct 10 '12 at 19:27