A question about injective continuous function and homeomorphism
Let $f:I rightarrow S^2$ be a continuous injective function.Does $I$ homeomorphic to $f(I)$?
$I=[0,1]$.$S^2$ is a 2-sphere.
I think it's right,but I don't know how to prove that $f(U)$ is a open subset in $f(I)$ if $U$ is a open subset in $I$.If it's wrong, I also cannot come up with counterexample...
general-topology
asked Jan 4 at 9:47
MaxwellMaxwell
264
add a comment |
Let $f:I rightarrow S^2$ be a continuous injective function.Does $I$ homeomorphic to $f(I)$?
$I=[0,1]$.$S^2$ is a 2-sphere.
I think it's right,but I don't know how to prove that $f(U)$ is a open subset in $f(I)$ if $U$ is a open subset in $I$.If it's wrong, I also cannot come up with counterexample...
general-topology
asked Jan 4 at 9:47
MaxwellMaxwell
264
add a comment |
Let $f:I rightarrow S^2$ be a continuous injective function.Does $I$ homeomorphic to $f(I)$?
$I=[0,1]$.$S^2$ is a 2-sphere.
I think it's right,but I don't know how to prove that $f(U)$ is a open subset in $f(I)$ if $U$ is a open subset in $I$.If it's wrong, I also cannot come up with counterexample...
general-topology
asked Jan 4 at 9:47
MaxwellMaxwell
264
Let $f:I rightarrow S^2$ be a continuous injective function.Does $I$ homeomorphic to $f(I)$?
$I=[0,1]$.$S^2$ is a 2-sphere.
I think it's right,but I don't know how to prove that $f(U)$ is a open subset in $f(I)$ if $U$ is a open subset in $I$.If it's wrong, I also cannot come up with counterexample...
general-topology
general-topology
asked Jan 4 at 9:47
MaxwellMaxwell
264
asked Jan 4 at 9:47
MaxwellMaxwell
264
asked Jan 4 at 9:47
MaxwellMaxwell
264
asked Jan 4 at 9:47
MaxwellMaxwell
264
asked Jan 4 at 9:47
MaxwellMaxwell
264
264
add a comment |
add a comment |
1 Answer
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$f(I)$ is compact and any bijective continuous map between compact Hausdorff spaces is a homeomorphism. Instead of proving that $f(U)$ is open for any open set $U$ you can prove the equivalent statement that $f(C)$ is closed for any closed set $C$. Use the fact that $C$ is compact, so $f(C)$ is compact.
answered Jan 4 at 9:49
Kavi Rama MurthyKavi Rama Murthy
51.8k32055
1
Thank you very much!!
– Maxwell
Jan 4 at 10:02
add a comment |
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$f(I)$ is compact and any bijective continuous map between compact Hausdorff spaces is a homeomorphism. Instead of proving that $f(U)$ is open for any open set $U$ you can prove the equivalent statement that $f(C)$ is closed for any closed set $C$. Use the fact that $C$ is compact, so $f(C)$ is compact.
answered Jan 4 at 9:49
Kavi Rama MurthyKavi Rama Murthy
51.8k32055
1
Thank you very much!!
– Maxwell
Jan 4 at 10:02
add a comment |
$f(I)$ is compact and any bijective continuous map between compact Hausdorff spaces is a homeomorphism. Instead of proving that $f(U)$ is open for any open set $U$ you can prove the equivalent statement that $f(C)$ is closed for any closed set $C$. Use the fact that $C$ is compact, so $f(C)$ is compact.
answered Jan 4 at 9:49
Kavi Rama MurthyKavi Rama Murthy
51.8k32055
1
Thank you very much!!
– Maxwell
Jan 4 at 10:02
add a comment |
$f(I)$ is compact and any bijective continuous map between compact Hausdorff spaces is a homeomorphism. Instead of proving that $f(U)$ is open for any open set $U$ you can prove the equivalent statement that $f(C)$ is closed for any closed set $C$. Use the fact that $C$ is compact, so $f(C)$ is compact.
answered Jan 4 at 9:49
Kavi Rama MurthyKavi Rama Murthy
51.8k32055
$f(I)$ is compact and any bijective continuous map between compact Hausdorff spaces is a homeomorphism. Instead of proving that $f(U)$ is open for any open set $U$ you can prove the equivalent statement that $f(C)$ is closed for any closed set $C$. Use the fact that $C$ is compact, so $f(C)$ is compact.
answered Jan 4 at 9:49
Kavi Rama MurthyKavi Rama Murthy
51.8k32055
answered Jan 4 at 9:49
Kavi Rama MurthyKavi Rama Murthy
51.8k32055
answered Jan 4 at 9:49
Kavi Rama MurthyKavi Rama Murthy
51.8k32055
answered Jan 4 at 9:49
Kavi Rama MurthyKavi Rama Murthy
51.8k32055
51.8k32055
1
Thank you very much!!
– Maxwell
Jan 4 at 10:02
add a comment |
1
Thank you very much!!
– Maxwell
Jan 4 at 10:02
1
1
Thank you very much!!
– Maxwell
Jan 4 at 10:02
Thank you very much!!
– Maxwell
Jan 4 at 10:02
add a comment |
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