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I have this problem: If $f: \mathbb{R} \rightarrow \mathbb{R}$ is a continuous, open surjection, must it be a homeomorphism? What about if $f$ was defined from $S \rightarrow S$ instead, where S is the unit circle?

Now, I know that if $f$ is a continuous, open bijection, then $f$ is a homeomorphism, so my plan is to show that $f$ is injective. Would I use the property of continuity that convergent sequences are mapped to convergent sequences? How would I handle the second part of the question?

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2 Answers 2

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Assuming that you are using the standard metric, injectivity can be shown as follows: Take $a,b\in\mathbb R,a<b$, such that $f(a)=f(b)$. Then $f$ attains its minimum $m$ and maximum $M$ in $[a,b]$, at $x_m,x_M\in[a,b]$, respectively. If $x_m,x_M\in\{a,b\}$, then $f$ is constant on $[a,b]$ and hence not open, contradiction. Thus, $x_m\in(a,b)$ or $x_M\in(a,b)$, and we assume the former. Then $f((a,b))=[m,M]$ (if $x_M\in(a,b))$ or $f((a,b))=[m,M)$ (if $x_M\in\{a,b\})$, but in any case, $f((a,b))$ is not open, again a contradiction. Thus, $f$ is injective and therefore strictly monotone. Now it is clear that $f$ is a homeomorphism.

For the second question consider $f:S^1\to S^1,\ z\mapsto z^2$. Then $f$ is clearly surjective. It is open, since if we take $I:=\{e^{it}\ |\ t\in(a,b)\}$, then $f(I)=\{e^{2it}\ |\ t\in(a,b)\}$. But it is not injective, since $f(1)=1=f(-1)$.

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  • $\begingroup$ How about $ \mathbb R^n \to \mathbb R^n $. Is then a surjective open and continuous map also injective? $\endgroup$
    – Nemesis
    Mar 1, 2019 at 17:42
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Hint.

You're right to try to prove that $f$ is injective. In order order to prove it I would use following argument. If this was not the case, then you will be able to find two reals $x<y$ with $f(x)=f(y)$. If $f$ is constant on $(x,y)$ you're done: $f((x,y))$ is a singleton, hence close.

If $f$ is not constant, then $f$ reaches its extrema $m,M$ as it is continuous so $f((x,y))$ is not open.

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