How to show that $Bbb C/(-infty ,0]$ is simply connected space












0












$begingroup$


I need to show that for $Bbb C/(-infty ,0]$ that if I have arbitrary different pathes $gamma , eta :[0,1] toBbb C/(-infty ,0]$ such that $gamma(0)=eta(0)=alpha $ and $gamma(1)=eta(1)=beta $ I have continuous homotopy $H:[0,1]^2 to Bbb C/(-infty ,0]$ such that: $H(0,t) = gamma (t),H(1,t) = eta (t),H(s,0) = alpha,H(s,1) = beta$
unfortunatily my space isn`t a convex,so how do I define a proper Homotopy and proving it's continuous?






general-topology complex-analysis







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    You could show it's homeomorphic to $Bbb C$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 26 '18 at 18:29










  • $begingroup$
    we cannot show it by homeomorphism, sorry...
    $endgroup$
    – Daniel Vainshtein
    Dec 26 '18 at 18:30






  • 1




    $begingroup$
    It's not convex, but still star-shaped.
    $endgroup$
    – Jacky Chong
    Dec 26 '18 at 18:31








  • 4




    $begingroup$
    Note that $Bbb C/(-infty ,0]$ is star-shaped with center $1$ for example. You could use the straight line homotopy on each of the paths to obtain homotopies to the center, then concatenate one with the reverse of the other and use the Pasting lemma to show continuity.
    $endgroup$
    – greelious
    Dec 26 '18 at 18:32










  • $begingroup$
    Iv'e tried to build homotopy such that: $H(s,t)=gamma(t)*(1-2s)+2s , 0le s le 0.5$ and $(2-2s)+eta(t)(2s-1) ,0.5 le s le 1$ does it`s ok?
    $endgroup$
    – Daniel Vainshtein
    Dec 26 '18 at 18:59


















0












$begingroup$


I need to show that for $Bbb C/(-infty ,0]$ that if I have arbitrary different pathes $gamma , eta :[0,1] toBbb C/(-infty ,0]$ such that $gamma(0)=eta(0)=alpha $ and $gamma(1)=eta(1)=beta $ I have continuous homotopy $H:[0,1]^2 to Bbb C/(-infty ,0]$ such that: $H(0,t) = gamma (t),H(1,t) = eta (t),H(s,0) = alpha,H(s,1) = beta$
unfortunatily my space isn`t a convex,so how do I define a proper Homotopy and proving it's continuous?






general-topology complex-analysis







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    You could show it's homeomorphic to $Bbb C$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 26 '18 at 18:29










  • $begingroup$
    we cannot show it by homeomorphism, sorry...
    $endgroup$
    – Daniel Vainshtein
    Dec 26 '18 at 18:30






  • 1




    $begingroup$
    It's not convex, but still star-shaped.
    $endgroup$
    – Jacky Chong
    Dec 26 '18 at 18:31








  • 4




    $begingroup$
    Note that $Bbb C/(-infty ,0]$ is star-shaped with center $1$ for example. You could use the straight line homotopy on each of the paths to obtain homotopies to the center, then concatenate one with the reverse of the other and use the Pasting lemma to show continuity.
    $endgroup$
    – greelious
    Dec 26 '18 at 18:32










  • $begingroup$
    Iv'e tried to build homotopy such that: $H(s,t)=gamma(t)*(1-2s)+2s , 0le s le 0.5$ and $(2-2s)+eta(t)(2s-1) ,0.5 le s le 1$ does it`s ok?
    $endgroup$
    – Daniel Vainshtein
    Dec 26 '18 at 18:59
















0












0








0


0



$begingroup$


I need to show that for $Bbb C/(-infty ,0]$ that if I have arbitrary different pathes $gamma , eta :[0,1] toBbb C/(-infty ,0]$ such that $gamma(0)=eta(0)=alpha $ and $gamma(1)=eta(1)=beta $ I have continuous homotopy $H:[0,1]^2 to Bbb C/(-infty ,0]$ such that: $H(0,t) = gamma (t),H(1,t) = eta (t),H(s,0) = alpha,H(s,1) = beta$
unfortunatily my space isn`t a convex,so how do I define a proper Homotopy and proving it's continuous?






general-topology complex-analysis







share|cite|improve this question









$endgroup$




I need to show that for $Bbb C/(-infty ,0]$ that if I have arbitrary different pathes $gamma , eta :[0,1] toBbb C/(-infty ,0]$ such that $gamma(0)=eta(0)=alpha $ and $gamma(1)=eta(1)=beta $ I have continuous homotopy $H:[0,1]^2 to Bbb C/(-infty ,0]$ such that: $H(0,t) = gamma (t),H(1,t) = eta (t),H(s,0) = alpha,H(s,1) = beta$
unfortunatily my space isn`t a convex,so how do I define a proper Homotopy and proving it's continuous?






general-topology complex-analysis




general-topology complex-analysis






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 26 '18 at 18:27









Daniel VainshteinDaniel Vainshtein

19011




19011








  • 1




    $begingroup$
    You could show it's homeomorphic to $Bbb C$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 26 '18 at 18:29










  • $begingroup$
    we cannot show it by homeomorphism, sorry...
    $endgroup$
    – Daniel Vainshtein
    Dec 26 '18 at 18:30






  • 1




    $begingroup$
    It's not convex, but still star-shaped.
    $endgroup$
    – Jacky Chong
    Dec 26 '18 at 18:31








  • 4




    $begingroup$
    Note that $Bbb C/(-infty ,0]$ is star-shaped with center $1$ for example. You could use the straight line homotopy on each of the paths to obtain homotopies to the center, then concatenate one with the reverse of the other and use the Pasting lemma to show continuity.
    $endgroup$
    – greelious
    Dec 26 '18 at 18:32










  • $begingroup$
    Iv'e tried to build homotopy such that: $H(s,t)=gamma(t)*(1-2s)+2s , 0le s le 0.5$ and $(2-2s)+eta(t)(2s-1) ,0.5 le s le 1$ does it`s ok?
    $endgroup$
    – Daniel Vainshtein
    Dec 26 '18 at 18:59
















  • 1




    $begingroup$
    You could show it's homeomorphic to $Bbb C$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 26 '18 at 18:29










  • $begingroup$
    we cannot show it by homeomorphism, sorry...
    $endgroup$
    – Daniel Vainshtein
    Dec 26 '18 at 18:30






  • 1




    $begingroup$
    It's not convex, but still star-shaped.
    $endgroup$
    – Jacky Chong
    Dec 26 '18 at 18:31








  • 4




    $begingroup$
    Note that $Bbb C/(-infty ,0]$ is star-shaped with center $1$ for example. You could use the straight line homotopy on each of the paths to obtain homotopies to the center, then concatenate one with the reverse of the other and use the Pasting lemma to show continuity.
    $endgroup$
    – greelious
    Dec 26 '18 at 18:32










  • $begingroup$
    Iv'e tried to build homotopy such that: $H(s,t)=gamma(t)*(1-2s)+2s , 0le s le 0.5$ and $(2-2s)+eta(t)(2s-1) ,0.5 le s le 1$ does it`s ok?
    $endgroup$
    – Daniel Vainshtein
    Dec 26 '18 at 18:59










1




1




$begingroup$
You could show it's homeomorphic to $Bbb C$.
$endgroup$
– Lord Shark the Unknown
Dec 26 '18 at 18:29




$begingroup$
You could show it's homeomorphic to $Bbb C$.
$endgroup$
– Lord Shark the Unknown
Dec 26 '18 at 18:29












$begingroup$
we cannot show it by homeomorphism, sorry...
$endgroup$
– Daniel Vainshtein
Dec 26 '18 at 18:30




$begingroup$
we cannot show it by homeomorphism, sorry...
$endgroup$
– Daniel Vainshtein
Dec 26 '18 at 18:30




1




1




$begingroup$
It's not convex, but still star-shaped.
$endgroup$
– Jacky Chong
Dec 26 '18 at 18:31






$begingroup$
It's not convex, but still star-shaped.
$endgroup$
– Jacky Chong
Dec 26 '18 at 18:31






4




4




$begingroup$
Note that $Bbb C/(-infty ,0]$ is star-shaped with center $1$ for example. You could use the straight line homotopy on each of the paths to obtain homotopies to the center, then concatenate one with the reverse of the other and use the Pasting lemma to show continuity.
$endgroup$
– greelious
Dec 26 '18 at 18:32




$begingroup$
Note that $Bbb C/(-infty ,0]$ is star-shaped with center $1$ for example. You could use the straight line homotopy on each of the paths to obtain homotopies to the center, then concatenate one with the reverse of the other and use the Pasting lemma to show continuity.
$endgroup$
– greelious
Dec 26 '18 at 18:32












$begingroup$
Iv'e tried to build homotopy such that: $H(s,t)=gamma(t)*(1-2s)+2s , 0le s le 0.5$ and $(2-2s)+eta(t)(2s-1) ,0.5 le s le 1$ does it`s ok?
$endgroup$
– Daniel Vainshtein
Dec 26 '18 at 18:59






$begingroup$
Iv'e tried to build homotopy such that: $H(s,t)=gamma(t)*(1-2s)+2s , 0le s le 0.5$ and $(2-2s)+eta(t)(2s-1) ,0.5 le s le 1$ does it`s ok?
$endgroup$
– Daniel Vainshtein
Dec 26 '18 at 18:59












1 Answer
1






active

oldest

votes


















1












$begingroup$

Consider the complex plane using polar coordinate system.
Every points in $mathbb{C}setminus(-inf,0]$ can be written as $(rho,theta)$, $rhoin(0,+inf)$, $thetain(-pi,pi)$.
The function $fcolonmathbb{C}setminus(-inf,0]to{zinmathbb{C}, Im(z)>0}$ defined as $f(rho,theta)=(rho,frac{theta}{2})$ is an homeomorphism, and Imf is a convess space, so the domain is simply connected.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    I noticed that you're looking for an explicit homotopy. The homeomorphism helps you in this search: write the two curves in polar coordinates as $gamma(t)=(gamma_{rho}(t),gamma_{theta}(t))$ and $eta(t)=(eta_{rho}(t),eta_{theta}(t))$, then the homotopy is $H(s,t)=(1-s)(gamma_{rho}(t),gamma_{theta}(t))+s(eta_{rho}(t),eta_{theta}(t))$.
    $endgroup$
    – ecrin
    Dec 26 '18 at 19:09












  • $begingroup$
    Iv'e tried to build homotopy such that: $H(s,t)=γ(t)∗(1−2s)+2s,0≤s≤0.5$ and $(2−2s)+η(t)(2s−1),0.5≤s≤1$ does it`s ok?
    $endgroup$
    – Daniel Vainshtein
    Dec 27 '18 at 8:57












  • $begingroup$
    No, it's not. In fact your homotopy isn't rel{0,1}: $H(s,0)=gamma(0)*(1-2s)+2snegamma(0)$ for $sne 0$.
    $endgroup$
    – ecrin
    Dec 27 '18 at 9:48










  • $begingroup$
    so how do I find a literal homotopy between 2 pathes with the same start and end?
    $endgroup$
    – Daniel Vainshtein
    Dec 27 '18 at 20:32










  • $begingroup$
    I showed it in the previous comment, you have to use polar coordinates
    $endgroup$
    – ecrin
    Dec 27 '18 at 22:14











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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Consider the complex plane using polar coordinate system.
Every points in $mathbb{C}setminus(-inf,0]$ can be written as $(rho,theta)$, $rhoin(0,+inf)$, $thetain(-pi,pi)$.
The function $fcolonmathbb{C}setminus(-inf,0]to{zinmathbb{C}, Im(z)>0}$ defined as $f(rho,theta)=(rho,frac{theta}{2})$ is an homeomorphism, and Imf is a convess space, so the domain is simply connected.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    I noticed that you're looking for an explicit homotopy. The homeomorphism helps you in this search: write the two curves in polar coordinates as $gamma(t)=(gamma_{rho}(t),gamma_{theta}(t))$ and $eta(t)=(eta_{rho}(t),eta_{theta}(t))$, then the homotopy is $H(s,t)=(1-s)(gamma_{rho}(t),gamma_{theta}(t))+s(eta_{rho}(t),eta_{theta}(t))$.
    $endgroup$
    – ecrin
    Dec 26 '18 at 19:09












  • $begingroup$
    Iv'e tried to build homotopy such that: $H(s,t)=γ(t)∗(1−2s)+2s,0≤s≤0.5$ and $(2−2s)+η(t)(2s−1),0.5≤s≤1$ does it`s ok?
    $endgroup$
    – Daniel Vainshtein
    Dec 27 '18 at 8:57












  • $begingroup$
    No, it's not. In fact your homotopy isn't rel{0,1}: $H(s,0)=gamma(0)*(1-2s)+2snegamma(0)$ for $sne 0$.
    $endgroup$
    – ecrin
    Dec 27 '18 at 9:48










  • $begingroup$
    so how do I find a literal homotopy between 2 pathes with the same start and end?
    $endgroup$
    – Daniel Vainshtein
    Dec 27 '18 at 20:32










  • $begingroup$
    I showed it in the previous comment, you have to use polar coordinates
    $endgroup$
    – ecrin
    Dec 27 '18 at 22:14
















1












$begingroup$

Consider the complex plane using polar coordinate system.
Every points in $mathbb{C}setminus(-inf,0]$ can be written as $(rho,theta)$, $rhoin(0,+inf)$, $thetain(-pi,pi)$.
The function $fcolonmathbb{C}setminus(-inf,0]to{zinmathbb{C}, Im(z)>0}$ defined as $f(rho,theta)=(rho,frac{theta}{2})$ is an homeomorphism, and Imf is a convess space, so the domain is simply connected.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    I noticed that you're looking for an explicit homotopy. The homeomorphism helps you in this search: write the two curves in polar coordinates as $gamma(t)=(gamma_{rho}(t),gamma_{theta}(t))$ and $eta(t)=(eta_{rho}(t),eta_{theta}(t))$, then the homotopy is $H(s,t)=(1-s)(gamma_{rho}(t),gamma_{theta}(t))+s(eta_{rho}(t),eta_{theta}(t))$.
    $endgroup$
    – ecrin
    Dec 26 '18 at 19:09












  • $begingroup$
    Iv'e tried to build homotopy such that: $H(s,t)=γ(t)∗(1−2s)+2s,0≤s≤0.5$ and $(2−2s)+η(t)(2s−1),0.5≤s≤1$ does it`s ok?
    $endgroup$
    – Daniel Vainshtein
    Dec 27 '18 at 8:57












  • $begingroup$
    No, it's not. In fact your homotopy isn't rel{0,1}: $H(s,0)=gamma(0)*(1-2s)+2snegamma(0)$ for $sne 0$.
    $endgroup$
    – ecrin
    Dec 27 '18 at 9:48










  • $begingroup$
    so how do I find a literal homotopy between 2 pathes with the same start and end?
    $endgroup$
    – Daniel Vainshtein
    Dec 27 '18 at 20:32










  • $begingroup$
    I showed it in the previous comment, you have to use polar coordinates
    $endgroup$
    – ecrin
    Dec 27 '18 at 22:14














1












1








1





$begingroup$

Consider the complex plane using polar coordinate system.
Every points in $mathbb{C}setminus(-inf,0]$ can be written as $(rho,theta)$, $rhoin(0,+inf)$, $thetain(-pi,pi)$.
The function $fcolonmathbb{C}setminus(-inf,0]to{zinmathbb{C}, Im(z)>0}$ defined as $f(rho,theta)=(rho,frac{theta}{2})$ is an homeomorphism, and Imf is a convess space, so the domain is simply connected.






share|cite|improve this answer









$endgroup$



Consider the complex plane using polar coordinate system.
Every points in $mathbb{C}setminus(-inf,0]$ can be written as $(rho,theta)$, $rhoin(0,+inf)$, $thetain(-pi,pi)$.
The function $fcolonmathbb{C}setminus(-inf,0]to{zinmathbb{C}, Im(z)>0}$ defined as $f(rho,theta)=(rho,frac{theta}{2})$ is an homeomorphism, and Imf is a convess space, so the domain is simply connected.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 26 '18 at 18:50









ecrinecrin

3477




3477








  • 1




    $begingroup$
    I noticed that you're looking for an explicit homotopy. The homeomorphism helps you in this search: write the two curves in polar coordinates as $gamma(t)=(gamma_{rho}(t),gamma_{theta}(t))$ and $eta(t)=(eta_{rho}(t),eta_{theta}(t))$, then the homotopy is $H(s,t)=(1-s)(gamma_{rho}(t),gamma_{theta}(t))+s(eta_{rho}(t),eta_{theta}(t))$.
    $endgroup$
    – ecrin
    Dec 26 '18 at 19:09












  • $begingroup$
    Iv'e tried to build homotopy such that: $H(s,t)=γ(t)∗(1−2s)+2s,0≤s≤0.5$ and $(2−2s)+η(t)(2s−1),0.5≤s≤1$ does it`s ok?
    $endgroup$
    – Daniel Vainshtein
    Dec 27 '18 at 8:57












  • $begingroup$
    No, it's not. In fact your homotopy isn't rel{0,1}: $H(s,0)=gamma(0)*(1-2s)+2snegamma(0)$ for $sne 0$.
    $endgroup$
    – ecrin
    Dec 27 '18 at 9:48










  • $begingroup$
    so how do I find a literal homotopy between 2 pathes with the same start and end?
    $endgroup$
    – Daniel Vainshtein
    Dec 27 '18 at 20:32










  • $begingroup$
    I showed it in the previous comment, you have to use polar coordinates
    $endgroup$
    – ecrin
    Dec 27 '18 at 22:14














  • 1




    $begingroup$
    I noticed that you're looking for an explicit homotopy. The homeomorphism helps you in this search: write the two curves in polar coordinates as $gamma(t)=(gamma_{rho}(t),gamma_{theta}(t))$ and $eta(t)=(eta_{rho}(t),eta_{theta}(t))$, then the homotopy is $H(s,t)=(1-s)(gamma_{rho}(t),gamma_{theta}(t))+s(eta_{rho}(t),eta_{theta}(t))$.
    $endgroup$
    – ecrin
    Dec 26 '18 at 19:09












  • $begingroup$
    Iv'e tried to build homotopy such that: $H(s,t)=γ(t)∗(1−2s)+2s,0≤s≤0.5$ and $(2−2s)+η(t)(2s−1),0.5≤s≤1$ does it`s ok?
    $endgroup$
    – Daniel Vainshtein
    Dec 27 '18 at 8:57












  • $begingroup$
    No, it's not. In fact your homotopy isn't rel{0,1}: $H(s,0)=gamma(0)*(1-2s)+2snegamma(0)$ for $sne 0$.
    $endgroup$
    – ecrin
    Dec 27 '18 at 9:48










  • $begingroup$
    so how do I find a literal homotopy between 2 pathes with the same start and end?
    $endgroup$
    – Daniel Vainshtein
    Dec 27 '18 at 20:32










  • $begingroup$
    I showed it in the previous comment, you have to use polar coordinates
    $endgroup$
    – ecrin
    Dec 27 '18 at 22:14








1




1




$begingroup$
I noticed that you're looking for an explicit homotopy. The homeomorphism helps you in this search: write the two curves in polar coordinates as $gamma(t)=(gamma_{rho}(t),gamma_{theta}(t))$ and $eta(t)=(eta_{rho}(t),eta_{theta}(t))$, then the homotopy is $H(s,t)=(1-s)(gamma_{rho}(t),gamma_{theta}(t))+s(eta_{rho}(t),eta_{theta}(t))$.
$endgroup$
– ecrin
Dec 26 '18 at 19:09






$begingroup$
I noticed that you're looking for an explicit homotopy. The homeomorphism helps you in this search: write the two curves in polar coordinates as $gamma(t)=(gamma_{rho}(t),gamma_{theta}(t))$ and $eta(t)=(eta_{rho}(t),eta_{theta}(t))$, then the homotopy is $H(s,t)=(1-s)(gamma_{rho}(t),gamma_{theta}(t))+s(eta_{rho}(t),eta_{theta}(t))$.
$endgroup$
– ecrin
Dec 26 '18 at 19:09














$begingroup$
Iv'e tried to build homotopy such that: $H(s,t)=γ(t)∗(1−2s)+2s,0≤s≤0.5$ and $(2−2s)+η(t)(2s−1),0.5≤s≤1$ does it`s ok?
$endgroup$
– Daniel Vainshtein
Dec 27 '18 at 8:57






$begingroup$
Iv'e tried to build homotopy such that: $H(s,t)=γ(t)∗(1−2s)+2s,0≤s≤0.5$ and $(2−2s)+η(t)(2s−1),0.5≤s≤1$ does it`s ok?
$endgroup$
– Daniel Vainshtein
Dec 27 '18 at 8:57














$begingroup$
No, it's not. In fact your homotopy isn't rel{0,1}: $H(s,0)=gamma(0)*(1-2s)+2snegamma(0)$ for $sne 0$.
$endgroup$
– ecrin
Dec 27 '18 at 9:48




$begingroup$
No, it's not. In fact your homotopy isn't rel{0,1}: $H(s,0)=gamma(0)*(1-2s)+2snegamma(0)$ for $sne 0$.
$endgroup$
– ecrin
Dec 27 '18 at 9:48












$begingroup$
so how do I find a literal homotopy between 2 pathes with the same start and end?
$endgroup$
– Daniel Vainshtein
Dec 27 '18 at 20:32




$begingroup$
so how do I find a literal homotopy between 2 pathes with the same start and end?
$endgroup$
– Daniel Vainshtein
Dec 27 '18 at 20:32












$begingroup$
I showed it in the previous comment, you have to use polar coordinates
$endgroup$
– ecrin
Dec 27 '18 at 22:14




$begingroup$
I showed it in the previous comment, you have to use polar coordinates
$endgroup$
– ecrin
Dec 27 '18 at 22:14


















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Wiesbaden

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Wappen Deutschlandkarte 50.0825 8.24 117 Koordinaten: 50° 5′  N , 8° 14′  O Basisdaten Bundesland: Hessen Regierungsbezirk: Darmstadt Höhe: 117 m ü. NHN Fläche: 203,93 km 2 Einwohner: 278.654 (31. Dez. 2017) [1] Bevölkerungsdichte: 1366 Einwohner je km 2 Postleitzahlen: 65183–65207, 55246, 55252 Vorlage:Infobox Gemeinde in Deutschland/Wartung/PLZ enthält Text Vorwahlen: 0611, 06122, 06127, 06134 Kfz-Kennzeichen: WI Gemeindeschlüssel: 06 4 14 000 LOCODE: DE WIB NUTS: DE714 Stadtgliederung: 26 Ortsbezirke Adresse der Stadtverwaltung: Schlossplatz 6 65183 Wiesbaden Webpräsenz: www.wiesbaden.de Oberbürgermeister: Sven Gerich (SPD) Lage der Landeshauptstadt Wiesbaden in Hessen und im Regierungsbezirk Darmstadt Offizielles Logo der Landeshauptstadt Wiesbaden Flagge der Landeshauptstadt Wiesbaden Wiesbaden is
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Marschland

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Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
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Dieringhausen

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Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
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