This an example from a book I am reading. The author claims that there are two ways to see that the bijection $f : (-1,1) \rightarrow \mathbb{R}$ defined by $f(x) = \frac{x}{1-x^2}$ is a homeomorphism. The simpler one is to note that both $f$ and its inverse are continuous functions, according to basis principles of calculus. The second way is to note that $f$ takes basis elements to basis elements and vice versa, where $(-1,1)$ and $\mathbb{R}$ are both endowed with the order topology.

How does one show that? What does he mean by "vice-versa," that the inverse function $f^{-1}$ takes a basis element to a basis element? I have tried many things, but I could not figure it out. E.g., I tried showing that $f((a,b)) = (f(a),f(b))$, but I had difficulty proving this.


Okay I have a few related follow up questions. The proof seems to rely on the following claim:

Let $(X, \tau_X)$ and $(Y,\tau_Y)$ be topology spaces generated by the bases $\mathcal{B}$ and $\mathcal{C}$, respectively. If $f : X \rightarrow Y$ is a bijection such that $f(B) \in \mathcal{C}$ and $f^{-1}(C) \in \mathbb{B}$ for all $B \in \mathcal{B}$ and $C \in \mathcal{C}$, then $f$ is a homeomorphism.

Is this right?

My next question is,

If the function $g$ is an order preserving bijection between the ordered topogical spaces $X$ and $Y$, will $g$ always map open intervals to open intervals? Do we have some sort of generalized intermediate value theorem for maps between general ordered spaces?

  • $\begingroup$ Are you familiar with the intermediate value theorem? $\endgroup$ Nov 11 '16 at 18:50
  • $\begingroup$ @MoisheCohen Yes, I am. And if I recall correctly, once can use that to show the image of an interval under a continuous function is an interval. But my issue is, what does the interval look like? Is it of the form $(a,b)$? $\endgroup$
    – user193319
    Nov 11 '16 at 18:57
  • $\begingroup$ For the intermediate value theorem, you might need that your ordered spaces are both connected and (order-)complete. The rationals are neither connected, nor order-complete. The integers are order-complete, but not connected. $\endgroup$
    – Mirko
    Nov 11 '16 at 20:24

You have the right idea: you want to show that both $f$ and $f^{-1}$ take basic open sets $(a,b)$ to basic open sets. One way to show this is to note that your function is monotone (necessarily, lest it not be invertible!). Because it is monotone increasing, it means it is 'order preserving' so $a<b$ if and only if $f(a) < f(b)$.

So you should show two things (this is the vice-versa): $f^{-1}\big((a,b)\big)$ is some interval of the form $(a', b')$, and $f\big((c',d')\big)$ is some interval $(c,d)$. Moishe's comment is a great way to do this.

  • $\begingroup$ Thanks for the response. I will try to prove what you suggested, and if I have any difficulties---I will be back! In the meantime, I have edited my OP and would appreciate if you could take a look at my new (but related) questions. $\endgroup$
    – user193319
    Nov 11 '16 at 19:16
  • $\begingroup$ @user193319 To answer your latter question, nope not always! What about $f(x) = x$ for $x<0$ and $f(x) = x+1$ for $x\ge 0$. Need some sort of continuity hypothesis! $\endgroup$ Nov 11 '16 at 19:17
  • $\begingroup$ Ah, so $f$ has to be an order preserving bijection that is continuous with respect to the given order topology? Another silly question: is it ever true that $f((a,b)) = (f(a),f(b))$? $\endgroup$
    – user193319
    Nov 11 '16 at 19:20
  • $\begingroup$ Sure! The identity will do as a nice simple example of when $f((a,b) = (f(a),f(b))$. In fact, what if $f$ is strictly monotone increasing? Then from the formal statement I gave in my response above, no point in $(a,b)$ can get mapped to a higher value than $b$ does, and similarly no point in $(a,b)$ can get mapped to lower than $a$ does. Thus your result holds so long as the image of an interval is an interval (rather than, say, a union of intervals). $\endgroup$ Nov 11 '16 at 19:22
  • $\begingroup$ With regard to that last bit, if you're just starting out with point-set topology, a good exercise is to prove that the continuous image of a connected set is connected. $\endgroup$ Nov 11 '16 at 19:25

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