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Given a continuous bijection between manifolds of the same dimension, does it have to be a homeomorphism?

I know that this has a straightforward proof for compact Hausdorff space, be they manifolds or not.

I also know that there are continuous bijections from non-compact manifolds to subsets of higher-dimensional manifolds, the easiest example being non-periodic curves on the torus as images of the real line with dense image. In that case the continuous bijection is not homeomorphism, but the subspace topology of the image is not that of a manifold anyway.

I think that this should be true for the following reason. Since continuity of the inverse map is a local property, the question should reduce (in chart neighborhoods of an arbitrary point and its image) to showing that a continuous bijection between n-balls is a homeomorphism. But this is true because the n-ball is a compact Hausdorff space. Anyway I would prefer to have a citeable reference.

user39082
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  • Doesn't this boil down to the question whether the inverse of a bijective continuous function is again continuous? Which is false. –  Dec 04 '18 at 10:17
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    This is a direct consequence of the invariance of domain theorem. Can you produce a simple proof using that?

    https://en.wikipedia.org/wiki/Invariance_of_domain

     – Gustavo Dec 04 '18 at 10:29
  • I see. This was what I was looking for. If you write it as an answer, I will accept it. – user39082 Dec 04 '18 at 11:08
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    If you consider manifolds with boundary, then it is not true. The map $e : ´[0,2\pi) \to S^1, e(t) = e^{it}$, is a continuous bijection. – Paul Frost Dec 04 '18 at 18:05

2 Answers2

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Let $X$ and $Y$ be $m$-manifolds (without boundary) and $ \ f:X \to Y \ $ be a continuous bijection. Let $ \ p \in X \ $. By definition there exists open sets $ \ A \subset X \ $ and $ \ B \subset Y \ $ and charts $ \ x: A \to \mathbb{R}^m \ $ and $ \ y: B \to \mathbb{R}^m \ $ such that $ \ p \in A \ $ and $ \ f(p) \in B \ $. Let $ \ f_{xy} : im(x) \to \mathbb{R}^m \ $ be such that $ \ f_{xy} (q) = y \Big( f \big( x^{-1} (q) \big) \Big) \ $, $\forall q \in im(x) \ $. That is $ \ f_{xy} = (y|_{f[A]}) \circ (f|_A) \circ x^{-1} \ $. Now $ \ im(x) \subset \mathbb{R}^m \ $ is open (since $ \ x \ $ is a homeomorphism) and $ \ f_{xy} \ $ is injective continuous (since $ \ x^{-1}$, $f|_A \ $ and $ \ y|_{f[A]} \ $ are injective continuous). By invariance of domain we have that $ \ f_{xy}[im(x)] = im(f_{xy}) = y \big[ f[A] \big] \ $ is open and $ \ f_{xy} \ $ is a homeomorphism between $ \ im(x) \ $ and $ \ y \big[ f[A] \big] \ $. Therefore $ \ D = B \cap f[A] = y^{-1} \Big[ y \big[ f[A] \big] \Big] \ $ and $ \ C = A \cap f^{-1} [B] = f^{-1} \big[ B \cap f[A] \big] \ $ are open, $ \ x[C] = f_{xy}^{-1} \big[ y[D] \big] \ $, $ \ y \big[ f[A] \big] = y[D] \ $, $ \ y|_{f[A]} = y|_D \ $ and $ \ p \in C \ $. We are left with a continuous bijection $ \ f|_C : C \to D \ $ whose inverse is $ \ f|_C^{-1} = \big( x^{-1}|_{x[C]} \big) \circ \big( f_{xy}^{-1}|_{y[D]} \big) \circ (y|_D) : D \to C \ $. Thus $ \ f|_C^{-1} \ $ is continuous because $ \ x^{-1}|_{x[C]} \, $, $f_{xy}^{-1}|_{y[D]} \ $ and $ \ y|_D \ $ are continuous. Thus $ \ p \in C$, $C \subset X \ $ is open, $D \subset Y \ $ is open and $ \ f|_C :C \to D \ $ is a homeomorphism. Since $p$ is arbitrary, $f: X \to Y \ $ is a local homeomorphism and since bijective local homeomorphisms are global homeomorphisms, we are done.

Gustavo
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  • This answer does not make any sense to me — half-open interval have obvious bijective map to a circle, and it is clearly not a homeomorphism. – xsnl Dec 05 '18 at 15:01
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    The answer covers the case of manifolds without boundary. I edited to make it clear. – Gustavo Dec 05 '18 at 16:56
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It may happens that your manifolds are not homeomorphic but there is a continuous bijection between them. However, if we assume that in your question that they are homeomorphic, then in Theorem 2 of the paper

  1. P.H. Doyle and J.G. Hocking, Continuous bijections on manifolds, J. Austral. Math. Soc. 22 (1976), 257-263.

it is proved that if $M$ is a manifold with compact boundary $\partial M$, then any continuous bijection $f:M\to M$ is a homeomorphism.

The proof follows from invariance domain theorem as mentioned in Gustavo's answer.

That paper also contains a lot of different results when a self bijection is a homeomorphism as well as counterexamples.

Sergiy Maksymenko
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