# Continuous open maps on compact sets are surjections.

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.

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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 Jul 19 at 1:38
Is $S^n$ a sphere of dimension n? –  user86828 Jul 19 at 1:43
@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 at 2:16
@user86828 I think the dimension is not important in this case (because $n$ is arbitrary). –  Pedro Jul 19 at 2:17
Yes what you said in your comment is correct. Do you see why $f(X) \subset S^n$ is compact? –  Mike Jul 19 at 2:22
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## 1 Answer

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.

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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 Jul 19 at 3:48
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 Jul 19 at 3:53
@Mike You're right. It's not necessary $S^n$ be closed to $f(x)$ be closed in $S^n$. –  Pedro Jul 19 at 4:03
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