Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
3  
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 '13 at 1:38
    
Is $S^n$ a sphere of dimension n? –  user86828 Jul 19 '13 at 1:43
1  
@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 2:16
    
@user86828 I think the dimension is not important in this case (because $n$ is arbitrary). –  Pedro Jul 19 '13 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 '13 at 2:22
show 6 more comments

1 Answer

up vote 1 down vote accepted

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.

share|improve this answer
    
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 '13 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 '13 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 '13 at 4:03
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.