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Assume we have the following map $f: \mathbb{R}^{k} \to \mathbb{R}^{k}$, which is continuous and also there exists the inverse map $f^{-1}$. Is it enough to conclude that $f$ is bijection?

amWhy
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AnTlr
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    The existence of an inverse alone is equivalent to being a bijection - we do not need continuity. – jpmacmanus Jan 13 '22 at 13:31
  • I do not agree... look here https://math.stackexchange.com/questions/1384258/if-a-function-has-an-inverse-then-it-is-bijective?noredirect=1&lq=1 – AnTlr Jan 13 '22 at 13:33
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    We're disagreeing on what we mean by inverse. The link you provided discusses one-sided inverses, whereas I was referring to a two-sided inverse. I think the latter is more standard. Either way, continuity will play no role. – jpmacmanus Jan 13 '22 at 13:36
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    In particular, if we assume that the claimed inverse $f^{-1}$ has domain $\mathbb R^k$, then this is enough to deduce that $f$ is bijective. – jpmacmanus Jan 13 '22 at 13:40

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