# Continuous bijection equivalent to homeomorphism under suitable hypotheses

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?

• What happens for instance if we take $B=\mathbb{R}$ endowed with the discrete topology, and $f$ the identity map ? – user120527 Nov 22 '18 at 10:59
• It is OK for the general case, but not enough if we assume B to be connected. – DCao Nov 22 '18 at 11:01
• 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/… – 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).