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Is the following statement true？

Let $U,V \subset R^n$ and let $f:U \to V$ be a continuous bijection. If $U$ is an open set in the topology induced by the Euclidean norm, then $V$ is also an open set.

I think, it has something to do with the fact that preimage of any open set under $f$ is open too (one of the equivalent definitions of continuity). However, it also has to do with some specific topological properties of $\mathbb{R}^n$, as the generalization of this statement to the arbitrary topological space is clearly false (take the Sierpinski space $(\{0, 1\}, \{\emptyset , \{1\}, \{0, 1\}\}$ and function $f(x)$ that maps $1$ to $0$, and thus is a continuous bijection of two one-element sets)

Chain Markov
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Saito
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    This is true, but the proof is non-trivial. Here's a reference: https://en.wikipedia.org/wiki/Invariance_of_domain. – user729424 Dec 23 '19 at 14:21
  • A proof can be found on page 126 of Allan Hatcher's book "Algebraic Topology", which is available for free on his website: http://pi.math.cornell.edu/~hatcher/AT/AT.pdf. – user729424 Dec 23 '19 at 14:25
  • Thank you very much – Saito Dec 23 '19 at 15:34
  • A different proof (using an analytic approach to the mapping degree) can be found in "Nonlinear functional analysis" by Deimling. Yet another source is https://terrytao.wordpress.com/2011/06/13/brouwers-fixed-point-and-invariance-of-domain-theorems-and-hilberts-fifth-problem/ – PhoemueX Dec 23 '19 at 15:40
  • @Saito: No problem! – user729424 Dec 24 '19 at 00:19
  • See https://math.stackexchange.com/questions/956753/a-valid-proof-for-the-invariance-of-domain-theorem, https://math.stackexchange.com/questions/475933/is-there-some-elementary-proof-of-invariance-of-domain and many other posts. It is known as "invariance of domain". – Paul Frost Dec 24 '19 at 11:16

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