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Could someone help me to show that if $X\subset \mathbb{R}^m$ is compact, then every continuous open map $f:X\to S^n$ is surjective?

This question was taken of an Analysis book (the subject of section is connectedness)

Thanks.

Ben Grossmann
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Pedro
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    What do you know about connectedness? Do you understand why you will be done if you show $f(X)$ is both open and closed in $S^n$? – Mike F Jul 19 '13 at 01:38
  • Is $S^n$ a sphere of dimension n? – user86828 Jul 19 '13 at 01:43
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    @Mike I know the only subsets of a connected set $Y$ which are open and closed in $Y$ simultaneously are $Y$ and $\varnothing$. Hence, if $f(X)$ is both open and closed in $S^n$ then $f(X)=S^n$ (because $S^n$ is connected). Is it right? Can you give me more details about how to do this? – Pedro Jul 19 '13 at 02:16
  • @user86828 I think the dimension is not important in this case (because $n$ is arbitrary). – Pedro Jul 19 '13 at 02:17
  • Yes what you said in your comment is correct. Do you see why $f(X) \subset S^n$ is compact? – Mike F Jul 19 '13 at 02:22
  • @Mike $f(X)$ is compact because $X$ is compact and $f$ is continuos. – Pedro Jul 19 '13 at 02:31
  • Yup! And don't forget that a compact subset of a metric space (or Hausdorff space if you like) is closed. Not much further now... – Mike F Jul 19 '13 at 02:38
  • @Mike As $f$ is open, $f(X)$ is open in $S^n$ (because $X$ is open in $X$). Is it correct? – Pedro Jul 19 '13 at 03:00
  • Mhm. If you want, you can answer your own question below. That way this question won't stay classified as unanswered, and people can comment on your writeup. – Mike F Jul 19 '13 at 03:04
  • @Mike Thank you very much. One last question: Is it correct that we can replace $S^n$ by any connected closed set? – Pedro Jul 19 '13 at 03:35
  • Yes, that's correct. In fact, closedness is not needed either. – Mike F Jul 19 '13 at 03:46

1 Answers1

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As $f$ is open and $X$ is open in $X$, we conclude that $f(X)$ is open in $S^n$.

As $f$ is continuous and $X$ is compact, we conclude that $f(X)$ is compcat. But a subset of $\mathbb{R}^{n+1}$ is compct if, and only if, it is closed and bounded. So, $f(X)$ is closed (in $\mathbb{R}^{n+1}$ and) in $S^n$.

Hence, $f(X)$ is both open and closed in $S^n$. As $S^n$ is connected, the only subsets of $S^n$ which satisfies this condition are $S^n$ and $\varnothing$. Then, $f(X)=S^n$ and $f$ is a surjection.

Remark: in this problem we can replace $S^n$ by any connected closed set.

Pedro
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  • Looks good to me. The only real issue (and it's a minor one) is that technically $S^n$ sits in $\mathbb{R}^{n+1}$, not $\mathbb{R}^n$. See here. – Mike F Jul 19 '13 at 03:48
  • Also, it seems like you have some mild confusion about what is needed for a set to be "relatively closed". If $C,Y \subset \mathbb{R}^n$ and $C$ is closed in $\mathbb{R}^n$, then $C \cap Y$ should be closed in $Y$ (by whatever definition you are using). In particular, if $C \subset Y \subset \mathbb{R}^n$ and $C$ is closed in $\mathbb{R}^n$, then $C$ is closed in $Y$. – Mike F Jul 19 '13 at 03:53
  • @Mike You're right. It's not necessary $S^n$ be closed to $f(x)$ be closed in $S^n$. – Pedro Jul 19 '13 at 04:03