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I found this comment

I would also like to point out that invariance of domain and the statement that every continuous bijection from $\mathbf{R}^n$ to $\mathbf{R}^n$ is a homeomorphism are equivalent. Therefore I would not expect there to be a proof that avoids this machinery.

but couldn't prove it myself using fundamental methods, i.e. no algebraic topology. As far as I can tell, invariance of domain is equivalent to the statement restricted to open balls in $\mathbf{R}^n$. But from there, I don't see how to prove this using only the statement about continuous bijections and no algebraic topology. Clearly, the preimage of an open ball is open, but how can I conclude that it's homeomorphic to an open ball?

fweth
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  • Yes, but which direction of the equivalence did you show? For me it's clear that invariance of domain implies the other thing, but curious, how do I get invariance of domain only from knowing that every continuous bijection $\mathbf{R}^n\to\mathbf{R}^n$ is a homeomorphism? Here my idea was, given $f:U\to V$ a continuous bijection with $U$ open, to find some family of open balls ${B_i}_i$ such that $V=\cup_i B_i$ and $f(B_i)$ is homeomorphic to an open ball so that I can use the preimise... – fweth Jul 24 '22 at 15:58
  • Oh. Sorry I was talking about the same. The other one I am not sure. Didn't realise at first that it might be harder. There is already an answer. Let's hope it's correct and helps. – Rishi Jul 24 '22 at 16:01
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    Invariance of domain is really the statement that every continuous injection $\mathbb{R}^n\to\mathbb{R}^n$ is open. I don't see any easy way to deduce this from the corresponding statement only about bijections. I would not take the comment you quoted too seriously. – Eric Wofsey Jul 24 '22 at 17:53
  • There are analytical proofs of IDT which avoid AT. For instance, Terry Tao wrote one. But his proof does not use the property that every continuous bijection is a homeomorphism. MSE is full of erroneous claims, I would not take the linked comment seriously. – Moishe Kohan Jul 24 '22 at 18:48

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Let's call a topological space $X$ domain invariant, if it fullfills the usual condition of the theorem of domain invariance, i.e. if $U, V$ are homeomorphis subsets of $X$, and $U$ is open, then also $V$ is open.

$X$ fullfills (*), if every continuous bijection $f: X \rightarrow X$ is a homeomorphism (sorry, no better notation came up to my mind).

By a standard argument we have: If X is doamin invariant, T2 and locally compact, then $X$ fullfills (*).

Proof. Let $f: X \rightarrow X$ be a continuous bijection. Of course, it suffices to show that $f$ is open: Let $U$ be open in $X$, $x \in U$. By local compactness pick an open $V$ such that $x \in V \subset \overline{V} \subset U$ and $\overline{V}$ compact. Then $f|\overline{V}: \overline{V} \rightarrow f(\overline{V})$ is continuous, bijective, hence a homeomorphism. Thus, $f|V: V \rightarrow f(V)$ is a homeomorhism, and $ V \cong f(V)$. Since $X$ is domain invariant, $f(V)$ is open. Therefore $f(U)$ is open.

This might be interpreted as "$\Rightarrow$" of the above statement (if $X$ is T2, locally compact). However, "$\Leftarrow$" is not true in general: Of course, if $X$ is compact, T2 then $X$ fullfills (*), but need not be domain invariant, as the unit interval shows.

Ulli
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  • Thanks a lot for the answer! So, also for $\mathbf{R}^n$ specifically, there is no fundamental way to show the equivalence? If so, is the statement from which I took the comment still true, that there is no fundamental (without algebraic topology) proof of (*)? – fweth Jul 24 '22 at 16:02
  • @fweth: I merely answered your question of the title. In order to answer your other question, one could perhaps think of other conditions, which are (easy) to proof for $\mathbb{R}^n$ and provide domain invariance together with (*). I thought about homogenity, but $[0,1]^\omega$ is homogeneous, compact, but not domain invariant. No, sorry, I don't have a concrete proposal. – Ulli Jul 24 '22 at 16:06
  • No worries! I also think that even (*) is hard for $\mathbf{R}^n$. Someone on IRC said its easy, but didn't provide a proof, so I thought I'd ask here again. (Tbf, another person on IRC also said there is no easy proof.) – fweth Jul 24 '22 at 16:10
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    Well, now I understand what your essential question was. Since my answer does not really answer that, you should probably remove its acceptance in order to get a better answer. – Ulli Jul 24 '22 at 16:18