0

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.

Martin Sleziak
  • 56,060
Skorpion
  • 458
  • 1
    Yes, it does follow immediately. No, it doesn’t have to be justified, it’s obvious. Consider even the natural generalization that homeomorphisms of topological spaces $X → X'$ and $Y → Y'$ induce a homeomorphism $X × Y → X' × Y'$ – can you see that this is true? Contemplate if you can’t see it immediately until you can. You should. – k.stm Oct 09 '20 at 05:34
  • 1
    Okay, maybe what's missing is having internalized the following basic facts about topological products of spaces $Z$ and $Z'$:

    (1) The projections $Z × Z' → Z,~(z,z') ↦ z$ and $Z × Z' → Z',~(z,z') ↦ z'$ are continuous. (2) For spaces $T$ and continuous maps $f \colon T → Z$ and $f' \colon T →Z'$ the map $(f, f')\colon T →Z × Z',~t ↦ (f(t), f'(t))$ is continuous.

    From this, conclude that for continuous maps $X → Y$ and $X' →Y'$ the induced map $X × X' → Y × Y'$ is continuous. From here, the rest should be clear.

     – k.stm Oct 09 '20 at 05:43
  • @k.stm What I'm thinking about is the following: If there exist homeos $f:X\to X'$ and $g: Y\to Y'$, then $f$ and $g$ are bicontinuous and bijective. Suppose sequences $(x_n)\in X, (y_n)\in Y$ converge to $x\in X, y\in Y$ respectively, then the function defined by $(a,b)\mapsto (f(a), g(b))$ preserves the sequential continuity of the sequence $( (x_n), (y_n))\in X \times Y$. Likewise for the other direction. This means we have a bicontinuous function from $X \times Y$ to $X' \times Y'$. The bijectivity immediately follows from the bijectivity of $f$ and $g$. So $X\times Y \cong X'\times Y'$. – Skorpion Oct 09 '20 at 05:45
  • This reasoning works whenever continuity can be characterized in terms of converging sequences (that is for instance for all maps between metric spaces). The reasoning can also be generalized using nets. However, you don’t even have to look into the spaces and follow definitions of continuity – provided you acknowledge the two basic facts about topological spaces I have mentioned. The statement then follows synthetically. You’re being unnesseccarily analytical about it. – k.stm Oct 09 '20 at 05:50
  • I've just begun to take an introductory analysis course this fall, using Pugh's Real Mathematical Analysis. I haven't seen the two facts that you stated before, but I agree they are `intuitive', and yes, my original question follows right away. My college does offer topology as a separate course, and hopefully I'll be taking it very soon. Thank you for the help! – Skorpion Oct 09 '20 at 05:54

1 Answers1

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.

freakish
  • 47,446
  • I agree with the general line that everything has to be justified and that there is danger in simply saying “This is obvious.” and then not caring about a justification – however, the problem with this generally occurs when people talk about intuitive statements being seemingly obvious. They say “This is obvious,” but merely mean “This is intuitive,” so they confuse being obvious with being intuitive. However, some statements are truly obvious, namely if they have obvious reasons to be true or an obvious justification, something which you can see immediately. – k.stm Oct 09 '20 at 08:23
  • Or maybe people mean “This is apparently true.” when saying “This is obvious.”, the distinction being that the former rely on the appearance of the statement instead of seeing the statement immediately with its justification. So again: I concur that getting in the habit of calling statements obvious is truly problematic. However, pretending that there are no such things as obvious statements probably just stifles intellectual drive. Of course, what is “obvious” depends on your level of mathematical maturity. – k.stm Oct 09 '20 at 08:27