My question is pretty simple to state, although I did not find any satisfying answer on the internet or books. I assume that the answer has to be well-known, but I can not figure any proof or counterexample.

Suppose that $B \subset \mathbb{R}^n$ is a connected topological space (with the relative topology) and we have a map $f:B \to \mathbb{R}$ which is assumed to be a continuous bijection. The question is if under these simple hypotheses it is enough to conclude that $f$ an homeomorphism.

Some remarks:

1) I am already aware about the invariance of domain theorem, that would give a YES answer for the case $B=\mathbb{R}$.

2) I know that the statement would be false if $\mathbb{R}$ was in the domain and $B$ in the codomain (classical counterexample where $B=\mathbb{S}^1$).

My intuitive guess is that it is true.

One idea is to induce an order in $B$ from the order in $\mathbb{R}$. Therefore, if there is some result that ensures me that $B$ is homeomorphic to a subset of $\mathbb{R}$ it is everything done, because connectedness forces this homeomorphic set to $B$ to be an interval and it is immediate to conclude that has to be an open one (and from there is obvious). Any ideas?

  • 1
    $\begingroup$ What happens for instance if we take $B=\mathbb{R}$ endowed with the discrete topology, and $f$ the identity map ? $\endgroup$ – user120527 Nov 22 '18 at 10:59
  • $\begingroup$ It is OK for the general case, but not enough if we assume B to be connected. $\endgroup$ – DCao Nov 22 '18 at 11:01
  • $\begingroup$ I have just seen this. I am not quite familiar with the order topology, but it seems that it would be enough to show that the order induced in $B$ induces an order topology equivalent to the one that $B$ already has? But maybe I suspect it is no simplification at all. at.yorku.ca/cgi-bin/… $\endgroup$ – DCao Nov 22 '18 at 11:03

You have the function $f:B \to \mathbb{R}$ that it is known to be biyective and continuous. Furthermore, we assume that $B$ is Hausdorff and path-connected. We want to see that $f^{-1}$ is continuous. I separate my answer in claims.

1) It is possible to induce a total order in $B$. Given $x,y \in B$ just define $x>y$ iff $f(x)>f(y)$.

2) The preimage of $[f(x),f(y)]$ is the interval $[x,y] \subset B$, meaning all the elements in $B$ that lie between $f(x)$ and $f(y)$ (trivial). Moreover, $[x,y] \subset B$ is closed since it is the continuous preimage of a closed set.

3) Since $B$ is path-connected there is a continuous map $h: [0,1] \to B$, with $f(0)=x$ and $f(1)=y$. We have that $f \circ h:[0,1] \to \mathbb{R}$ is a continuous path from $f(x)$ to $f(y)$ and, by Bolzano, it contains $[f(x),f(y)]$. Therefore, since $f$ is bijective, $[x,y] \subset h[0,1]$. Since $h[0,1]$ is compact, $[x,y]$ is closed and $B$ is Hausdorff, we have that $[x,y]$ is compact.

4) Therefore, the restriction of $f$ to the compact $[x,y]$ is an homeomorphism between $[x,y]$ and $[f(x),f(y)]$. In particular, the same happens for $(x,y)$ with image $(f(x),f(y))$. Changing $x$ for a sequence $x_n$ that goes to $-\infty$ and $y$ for a sequence $y_n$ that goes to $+\infty$ gives the result from a well-known lemma of prolongation of functions that have open domains and coincide in the intersection.

Remark: Point $4)$ holds for $\mathbb{R}$ and not for $\mathbb{S}^1$ (because you can not put $\mathbb{S}^1$ as an increasing union of open sets homeomorphic to segments).


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