Continuous Bijection From Locally Compact to Hausdorff is Homeomorphism
Continuous bijection on open interval is homeomorphism
Solution 1
It is easy to prove by the intermediate value theorem that any continuous injection from an interval to $\mathbb{R}$ must be monotone (for instance, if $c<d<e$ and $f(c)<f(e)<f(d)$ then $f(e')=f(e)$ for some $e'\in (c,d)$; other failures of monotonicity can be handled similarly). Thus $f$ is either an order-preserving bijection or an order-reversing bijection $(a,b)\to\mathbb{R}$, which maps open intervals to open intervals and is hence a homeomorphism.
Solution 2
Another possibility is is to use the fact that the continuous image of a compact set is compact. If $C\subset (a,b)$ is closed, then it compact since it is bounded. It follows that $f(C)$ is compact, hence closed and bounded. Since $f$ sends closed sets to closed sets (and it is bijective), it also sends open sets to open sets. Hence, for every $U$ open in $(a,b)$, $f(U)$ is open. Then $(f^{-1})^{-1}(U)=f(U)$ is open and $f^{-1}$ is continuous.
Solution 3
There is an elementary, general, textbook result which answers your question, and which relies on the concept of a proper map $f : X \to Y$, meaning that the pre-image of every compact subset of $Y$ is compact in $X$.
Theorem: If $f : X \to Y$ is a continuous, proper bijection and $Y$ is a compactly generated Hausdorff space then $f$ is a closed map and hence a homeomorphism.
One might not prefer this as a proof for your specific problem, though, because one still needs to verify properness of your map $f : (a,b) \to \mathbb{R}$, and my guess is that any argument for properness would be similar to one of the existing answers by others.
This theorem is an exercise, for example, in Munkres' topology book.
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Comments
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Suppose that $a,b\in\mathbb R$ with $a<b$ and that $f:(a,b)\to\mathbb R$ is continuous and bijective. I would like to prove that $f$ is a homeomorphism using elementary methods (no resort to invariance of domain, for instance).
To my surprise, I have not found a straightforward, standard (textbook) reference for this result. If you know any or you have a clever hint in mind, I would be grateful if you could share it.
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This seems to work more generally bounded open sets in $U\subset \mathbb{R}^n$ and a continuous bijection $U\to \mathbb{R}^n$. I think I recall some more general theorem from point set topology about continuous bijections from compact Hausdorff spaces to Hausdorff spaces
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You don't need compact Hausdorff: A continuous injection from a compact space to a Hausdorff space is a homeomorphism onto its image.
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Ah, ok. As I implied, I couldn't quite recall what exactly the weakest conditions were. Thanks.
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I'm a little wary of this approach. For example, if $c\in(a,b)$, then the half-interval $(a,c]$ is closed in $(a,b)$ (with respect to the relative topology) but not compact, so I cannot see a priori whether $f(a,c]$ is compact or even closed...
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My problem with showing that $f$ is proper is similar to my concern with the approach suggested by user43687. Clearly, the pre-image of any compact (and hence closed) subset of $\mathbb R$ under $f$ is closed in $(a,b)$ because of continuity. However, since the relevant topology on $(a,b)$ is the relative one, this does not immediately imply compactness (half-open intervals starting from or ending at either endpoint are closed in the relative topology but not compact). Anyway, thank you for this suggestion.
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I was hoping there was a way around showing monotonicity (simple but inelegantly tedious proofs are my pet peeves...), but in the light of the other answers, this seems to be the most straightforward elementary approach, indeed. Thank you for confirming!
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Yes, as triple_sec said, this doesn't work. It only shows that $f(U)$ is open if $U$ is a cocompact open set, but those do not generate the entire topology.
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Indeed I agree with your comment, as my third paragraph suggests. And while monotonicity is also available as a tool for verifying properness, this is more complicated than how monotonicity is used in the answer of @EricWolsey. However, I disagree with your comment to that answer regarding the inelegant tediousness of the use of monotonicity. In 1 dimension monotonicity is equivalent to local injectivity, and in higher dimensions local injectivity (of smooth maps between open sets) becomes a powerful tool for verifying properness among other uses.
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Yes, your right. I forgot to respect the relative topology...it appears that at least 4 people did not read this answer carefully!
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