coverage mapping at covering spaces
$begingroup$
the professor at university asked us:
is a coverage mapping like P from X to Y a closed mapping or not. Also; is p an open mapping?
i could prove that P is an open mapping but for proving that P is not closed mapping; i should use the fact that if X equals to real line and Y equals to $S^1$ and the P domain order equal $n+1/n$ for $n>3$ but i cant prove it .would you help me please?
algebraic-topology covering-spaces closed-map
asked Dec 7 '18 at 18:42
pershina oladpershina olad
7710
$endgroup$
add a comment |
$begingroup$
the professor at university asked us:
is a coverage mapping like P from X to Y a closed mapping or not. Also; is p an open mapping?
i could prove that P is an open mapping but for proving that P is not closed mapping; i should use the fact that if X equals to real line and Y equals to $S^1$ and the P domain order equal $n+1/n$ for $n>3$ but i cant prove it .would you help me please?
algebraic-topology covering-spaces closed-map
asked Dec 7 '18 at 18:42
pershina oladpershina olad
7710
$endgroup$
add a comment |
$begingroup$
the professor at university asked us:
is a coverage mapping like P from X to Y a closed mapping or not. Also; is p an open mapping?
i could prove that P is an open mapping but for proving that P is not closed mapping; i should use the fact that if X equals to real line and Y equals to $S^1$ and the P domain order equal $n+1/n$ for $n>3$ but i cant prove it .would you help me please?
algebraic-topology covering-spaces closed-map
asked Dec 7 '18 at 18:42
pershina oladpershina olad
7710
$endgroup$
the professor at university asked us:
is a coverage mapping like P from X to Y a closed mapping or not. Also; is p an open mapping?
i could prove that P is an open mapping but for proving that P is not closed mapping; i should use the fact that if X equals to real line and Y equals to $S^1$ and the P domain order equal $n+1/n$ for $n>3$ but i cant prove it .would you help me please?
algebraic-topology covering-spaces closed-map
algebraic-topology covering-spaces closed-map
asked Dec 7 '18 at 18:42
pershina oladpershina olad
7710
asked Dec 7 '18 at 18:42
pershina oladpershina olad
7710
asked Dec 7 '18 at 18:42
pershina oladpershina olad
7710
asked Dec 7 '18 at 18:42
pershina oladpershina olad
7710
asked Dec 7 '18 at 18:42
pershina oladpershina olad
7710
7710
add a comment |
add a comment |
1 Answer
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You didn't tell us which map you are talking about, but I suspect that it is$$begin{array}{rccc}Pcolon&mathbb R&longrightarrow&S^1\&x&mapsto&(cos2pi x,sin2pi x).end{array}$$If that's so and if $S=left{n+frac1n,middle|,ninmathbb Nsetminus{1,2}right}$, then $S$ is a closed subset of $mathbb R$, but $P(S)$ is not a closed subset of $S^1$, since$$lim_{ntoinfty}Pleft(n+frac1nright)=0,$$but $0notin P(S)$.
answered Dec 7 '18 at 18:50
José Carlos SantosJosé Carlos Santos
155k22124227
$endgroup$
$begingroup$
Thank you ,sorry for the way i asked my question,since i cant write formulas in this site yet.Your answer was exactly what i needed.Thank you
$endgroup$
– pershina olad
Dec 7 '18 at 20:41
add a comment |
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1 Answer
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1 Answer
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$begingroup$
You didn't tell us which map you are talking about, but I suspect that it is$$begin{array}{rccc}Pcolon&mathbb R&longrightarrow&S^1\&x&mapsto&(cos2pi x,sin2pi x).end{array}$$If that's so and if $S=left{n+frac1n,middle|,ninmathbb Nsetminus{1,2}right}$, then $S$ is a closed subset of $mathbb R$, but $P(S)$ is not a closed subset of $S^1$, since$$lim_{ntoinfty}Pleft(n+frac1nright)=0,$$but $0notin P(S)$.
answered Dec 7 '18 at 18:50
José Carlos SantosJosé Carlos Santos
155k22124227
$endgroup$
$begingroup$
Thank you ,sorry for the way i asked my question,since i cant write formulas in this site yet.Your answer was exactly what i needed.Thank you
$endgroup$
– pershina olad
Dec 7 '18 at 20:41
add a comment |
$begingroup$
You didn't tell us which map you are talking about, but I suspect that it is$$begin{array}{rccc}Pcolon&mathbb R&longrightarrow&S^1\&x&mapsto&(cos2pi x,sin2pi x).end{array}$$If that's so and if $S=left{n+frac1n,middle|,ninmathbb Nsetminus{1,2}right}$, then $S$ is a closed subset of $mathbb R$, but $P(S)$ is not a closed subset of $S^1$, since$$lim_{ntoinfty}Pleft(n+frac1nright)=0,$$but $0notin P(S)$.
answered Dec 7 '18 at 18:50
José Carlos SantosJosé Carlos Santos
155k22124227
$endgroup$
$begingroup$
Thank you ,sorry for the way i asked my question,since i cant write formulas in this site yet.Your answer was exactly what i needed.Thank you
$endgroup$
– pershina olad
Dec 7 '18 at 20:41
add a comment |
$begingroup$
You didn't tell us which map you are talking about, but I suspect that it is$$begin{array}{rccc}Pcolon&mathbb R&longrightarrow&S^1\&x&mapsto&(cos2pi x,sin2pi x).end{array}$$If that's so and if $S=left{n+frac1n,middle|,ninmathbb Nsetminus{1,2}right}$, then $S$ is a closed subset of $mathbb R$, but $P(S)$ is not a closed subset of $S^1$, since$$lim_{ntoinfty}Pleft(n+frac1nright)=0,$$but $0notin P(S)$.
answered Dec 7 '18 at 18:50
José Carlos SantosJosé Carlos Santos
155k22124227
$endgroup$
You didn't tell us which map you are talking about, but I suspect that it is$$begin{array}{rccc}Pcolon&mathbb R&longrightarrow&S^1\&x&mapsto&(cos2pi x,sin2pi x).end{array}$$If that's so and if $S=left{n+frac1n,middle|,ninmathbb Nsetminus{1,2}right}$, then $S$ is a closed subset of $mathbb R$, but $P(S)$ is not a closed subset of $S^1$, since$$lim_{ntoinfty}Pleft(n+frac1nright)=0,$$but $0notin P(S)$.
answered Dec 7 '18 at 18:50
José Carlos SantosJosé Carlos Santos
155k22124227
answered Dec 7 '18 at 18:50
José Carlos SantosJosé Carlos Santos
155k22124227
answered Dec 7 '18 at 18:50
José Carlos SantosJosé Carlos Santos
155k22124227
answered Dec 7 '18 at 18:50
José Carlos SantosJosé Carlos Santos
155k22124227
155k22124227
$begingroup$
Thank you ,sorry for the way i asked my question,since i cant write formulas in this site yet.Your answer was exactly what i needed.Thank you
$endgroup$
– pershina olad
Dec 7 '18 at 20:41
add a comment |
$begingroup$
Thank you ,sorry for the way i asked my question,since i cant write formulas in this site yet.Your answer was exactly what i needed.Thank you
$endgroup$
– pershina olad
Dec 7 '18 at 20:41
$begingroup$
Thank you ,sorry for the way i asked my question,since i cant write formulas in this site yet.Your answer was exactly what i needed.Thank you
$endgroup$
– pershina olad
Dec 7 '18 at 20:41
$begingroup$
Thank you ,sorry for the way i asked my question,since i cant write formulas in this site yet.Your answer was exactly what i needed.Thank you
$endgroup$
– pershina olad
Dec 7 '18 at 20:41
add a comment |
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