Showing $\mathbb{R}^2 \cong \mathbb{R}\times (-\frac{\pi}{2}, \frac{\pi}{2})$ using $\tan^{-1}$

Question

just a quick, if not obvious, question here -- since the $\arctan$ function from $\mathbb{R}$ to $(-\pi /2, \pi /2)$ is a homeomorphism, does it naturally follow that the function $f: \mathbb{R}^2 \to \mathbb{R}\times (-\frac{\pi}{2}, \frac{\pi}{2})$ defined by $$ (x,y)\mapsto (x, \tan^{-1}y)$$ is a homeomorphism, because the first component remains constant and the second gets mapped by a homeomorphism? Or does it need to be further justified? I am aware of this post here, but I think my case is simpler and no further justification is needed. Thanks.

Answer 1

Everything in maths has to be justified. Especially when in doubt. Even the greatest minds made the "this is obvious" mistake. The most recent example that comes to my mind is Michael Atiyah's proof attempt of the Riemann hypothesis (2018 I think).

Anyway there's a general construction: if $f:X\to Y$ and $g:X'\to Y'$ are two continuous functions (between any topological spaces) then

$$f\times g:X\times X'\to Y\times Y'$$ $$f\times g(x,x')=\big(f(x), g(x')\big)$$

is a continuous function as well. Note that this is unrelated to "component wise continuity" that you've mentioned. It follows directly from the construction of product topology, i.e. it is enough to consider preimages of open sets of the form $U\times V$.

Now if both $f,g$ are homeomorphisms, then $f\times g$ is a homeomorphism as well with the explicit inverse $f^{-1}\times g^{-1}$.

And so if you already know that $arctan:\mathbb{R}\to(-\frac{\pi}{2},\frac{\pi}{2})$ is a homeomorphism (note the different symbol, since technically $tan$ is not invertible) then the rest follows from the simple observation that the identity $\mathbb{R}\to\mathbb{R}$, $x\mapsto x$ is a homeomorphism as well.