Elementary clarification about complex square root mapsHow do I get the square root of a complex number?Square root of a complex numberContinuity of complex square root functionFinding square root of $-5-12i$ by formula and by De Moivre's Theoremcontinuity of the complex square root functionProperties of the principal square root of a complex numbercomplex modulus and square rootDefining a principal square root in the complex space : Wolfram caseComplex square root analytic with Cauchy-Riemann equationsOn the real square root and branches of the complex square root.Choice of a square root

Why isn't P and P/poly trivially the same?

Inorganic chemistry handbook with reaction lists

Has a sovereign Communist government ever run, and conceded loss, on a fair election?

Insult for someone who "doesn't know anything"

The (Easy) Road to Code

Can multiple states demand income tax from an LLC?

Unfamiliar notation in Diabelli's "Duet in D" for piano

Was this cameo in Captain Marvel computer generated?

How to make sure I'm assertive enough in contact with subordinates?

What is the purpose of a disclaimer like "this is not legal advice"?

Averaging over columns while ignoring zero entries

Should we avoid writing fiction about historical events without extensive research?

Why would /etc/passwd be used every time someone executes `ls -l` command?

Create chunks from an array

Why is there an extra space when I type "ls" on the Desktop?

Rationale to prefer local variables over instance variables?

How spaceships determine each other's mass in space?

Vector-transposing function

ESPP--any reason not to go all in?

How to recover against Snake as a heavyweight character?

Is this Paypal Github SDK reference really a dangerous site?

Precision notation for voltmeters

What is the oldest European royal house?

Short story about cities being connected by a conveyor belt



Elementary clarification about complex square root maps


How do I get the square root of a complex number?Square root of a complex numberContinuity of complex square root functionFinding square root of $-5-12i$ by formula and by De Moivre's Theoremcontinuity of the complex square root functionProperties of the principal square root of a complex numbercomplex modulus and square rootDefining a principal square root in the complex space : Wolfram caseComplex square root analytic with Cauchy-Riemann equationsOn the real square root and branches of the complex square root.Choice of a square root













1












$begingroup$


Let $p:mathbbC-0to mathbbC-0$ be $zto z^2$, wich is $a+ibmapsto (a^2-b^2)+i(2ab)$. By fundamental theorem of algebra we know that $p$ is surjective, and if $w$ is a square root of $z$, also $-w$ is. Thus for each $zne 0$ we have two distinct square roots of $z$.



I want to prove that the map $p$ is a covering map.



Now, I'm not acquainted of Complex Analysis, but I have read something here on MSE and on Wikipedia.



If we let $zin mathbbC-0$, then we can represent $z$ in polar coordinates in unique way as $z=[r,theta]$ with $r>0$ and $thetain [0,2pi)$ and by de Moivre formula I have that $w_1=[sqrtr,theta/2]$ and $w_2=[sqrtr,theta/2+pi]$ are the two square roots of $z$.



Now, in order to define a function, I have to make a choice on who I want as square root of $z$.



So, let's say that my square root map is $mathbbC-0to mathbbC-0,quad[r,theta] mapsto [sqrtr,theta/2]$, so I'm taking the root that lies in the upper half space.



The problem now is that I want a continuous map. So this map is not good since points of the form $[r,theta], [r,2pi-theta]$ are close to each other when $theta$ goes to $0^+$, but their images are not. But if I restrict my map to $mathbbC-([0,+infty)times 0)$, then it shoud be continuous.



But now, how can I formally prove the that map is truly continuous? By definition the topology on $mathbbC$ is the one that came with the set-indentification with $mathbbR^2$, so by definition a map $mathbbCto mathbbC$ is continuous if and only if it is continuous as a map $mathbbR^2to mathbbR^2$.



For example the map $p:mathbbC-0to mathbbC-0$ , $zto z^2$ is continuous beacuse as a map $p:mathbbR^2-0to mathbbR^2-0$ , $(a,b)to (a^2-b^2,2ab)$ it is continuous.




So to show that $q:mathbbC-([0,+infty)times 0)to mathbbC, quad [r,theta]mapsto [sqrtr,theta/2]$ should I write it as a map from a certain subset of $mathbbR^2$ to a certain subset of $mathbbR^2$ and then control that it is continuous? (As done for $p$, so this include adap the defyning formula of $q$ to "cartesian coordinates")




EDIT




Then, how can i prove that $p$ is a covering map with a strategy like this: pick $zin mathbbC-0$, let $U$ be a certain set. Observe that $U$ is open neighborhood of $z$ and is connected. Now observe that $p^-1(U)$ has the following components, each of which is mapped homeomorphically to $U$ by $p$











share|cite|improve this question











$endgroup$











  • $begingroup$
    The issue becomes moot if we consider $[0,2pi)$ as $mathbb R/2pi mathbb R$, i.e. as $[0,2pi)$ being a partitioning of equivalences classes of $mathbb R$ where we have the equivalence relation $yequiv x iff exist k in mathbb Z|y-x=2kpi$.
    $endgroup$
    – fleablood
    yesterday










  • $begingroup$
    or to put it another way... consider $[0,2pi)$ as a metric space with $d(x,y) = min (|x-y|, |x-y + 2pi|, |x-y - 2pi|$. This basically creates a new metric space and $d(e, 2pi -d) = |e+d|$ which is essentially claiming that $2pi = 0$.
    $endgroup$
    – fleablood
    yesterday















1












$begingroup$


Let $p:mathbbC-0to mathbbC-0$ be $zto z^2$, wich is $a+ibmapsto (a^2-b^2)+i(2ab)$. By fundamental theorem of algebra we know that $p$ is surjective, and if $w$ is a square root of $z$, also $-w$ is. Thus for each $zne 0$ we have two distinct square roots of $z$.



I want to prove that the map $p$ is a covering map.



Now, I'm not acquainted of Complex Analysis, but I have read something here on MSE and on Wikipedia.



If we let $zin mathbbC-0$, then we can represent $z$ in polar coordinates in unique way as $z=[r,theta]$ with $r>0$ and $thetain [0,2pi)$ and by de Moivre formula I have that $w_1=[sqrtr,theta/2]$ and $w_2=[sqrtr,theta/2+pi]$ are the two square roots of $z$.



Now, in order to define a function, I have to make a choice on who I want as square root of $z$.



So, let's say that my square root map is $mathbbC-0to mathbbC-0,quad[r,theta] mapsto [sqrtr,theta/2]$, so I'm taking the root that lies in the upper half space.



The problem now is that I want a continuous map. So this map is not good since points of the form $[r,theta], [r,2pi-theta]$ are close to each other when $theta$ goes to $0^+$, but their images are not. But if I restrict my map to $mathbbC-([0,+infty)times 0)$, then it shoud be continuous.



But now, how can I formally prove the that map is truly continuous? By definition the topology on $mathbbC$ is the one that came with the set-indentification with $mathbbR^2$, so by definition a map $mathbbCto mathbbC$ is continuous if and only if it is continuous as a map $mathbbR^2to mathbbR^2$.



For example the map $p:mathbbC-0to mathbbC-0$ , $zto z^2$ is continuous beacuse as a map $p:mathbbR^2-0to mathbbR^2-0$ , $(a,b)to (a^2-b^2,2ab)$ it is continuous.




So to show that $q:mathbbC-([0,+infty)times 0)to mathbbC, quad [r,theta]mapsto [sqrtr,theta/2]$ should I write it as a map from a certain subset of $mathbbR^2$ to a certain subset of $mathbbR^2$ and then control that it is continuous? (As done for $p$, so this include adap the defyning formula of $q$ to "cartesian coordinates")




EDIT




Then, how can i prove that $p$ is a covering map with a strategy like this: pick $zin mathbbC-0$, let $U$ be a certain set. Observe that $U$ is open neighborhood of $z$ and is connected. Now observe that $p^-1(U)$ has the following components, each of which is mapped homeomorphically to $U$ by $p$











share|cite|improve this question











$endgroup$











  • $begingroup$
    The issue becomes moot if we consider $[0,2pi)$ as $mathbb R/2pi mathbb R$, i.e. as $[0,2pi)$ being a partitioning of equivalences classes of $mathbb R$ where we have the equivalence relation $yequiv x iff exist k in mathbb Z|y-x=2kpi$.
    $endgroup$
    – fleablood
    yesterday










  • $begingroup$
    or to put it another way... consider $[0,2pi)$ as a metric space with $d(x,y) = min (|x-y|, |x-y + 2pi|, |x-y - 2pi|$. This basically creates a new metric space and $d(e, 2pi -d) = |e+d|$ which is essentially claiming that $2pi = 0$.
    $endgroup$
    – fleablood
    yesterday













1












1








1





$begingroup$


Let $p:mathbbC-0to mathbbC-0$ be $zto z^2$, wich is $a+ibmapsto (a^2-b^2)+i(2ab)$. By fundamental theorem of algebra we know that $p$ is surjective, and if $w$ is a square root of $z$, also $-w$ is. Thus for each $zne 0$ we have two distinct square roots of $z$.



I want to prove that the map $p$ is a covering map.



Now, I'm not acquainted of Complex Analysis, but I have read something here on MSE and on Wikipedia.



If we let $zin mathbbC-0$, then we can represent $z$ in polar coordinates in unique way as $z=[r,theta]$ with $r>0$ and $thetain [0,2pi)$ and by de Moivre formula I have that $w_1=[sqrtr,theta/2]$ and $w_2=[sqrtr,theta/2+pi]$ are the two square roots of $z$.



Now, in order to define a function, I have to make a choice on who I want as square root of $z$.



So, let's say that my square root map is $mathbbC-0to mathbbC-0,quad[r,theta] mapsto [sqrtr,theta/2]$, so I'm taking the root that lies in the upper half space.



The problem now is that I want a continuous map. So this map is not good since points of the form $[r,theta], [r,2pi-theta]$ are close to each other when $theta$ goes to $0^+$, but their images are not. But if I restrict my map to $mathbbC-([0,+infty)times 0)$, then it shoud be continuous.



But now, how can I formally prove the that map is truly continuous? By definition the topology on $mathbbC$ is the one that came with the set-indentification with $mathbbR^2$, so by definition a map $mathbbCto mathbbC$ is continuous if and only if it is continuous as a map $mathbbR^2to mathbbR^2$.



For example the map $p:mathbbC-0to mathbbC-0$ , $zto z^2$ is continuous beacuse as a map $p:mathbbR^2-0to mathbbR^2-0$ , $(a,b)to (a^2-b^2,2ab)$ it is continuous.




So to show that $q:mathbbC-([0,+infty)times 0)to mathbbC, quad [r,theta]mapsto [sqrtr,theta/2]$ should I write it as a map from a certain subset of $mathbbR^2$ to a certain subset of $mathbbR^2$ and then control that it is continuous? (As done for $p$, so this include adap the defyning formula of $q$ to "cartesian coordinates")




EDIT




Then, how can i prove that $p$ is a covering map with a strategy like this: pick $zin mathbbC-0$, let $U$ be a certain set. Observe that $U$ is open neighborhood of $z$ and is connected. Now observe that $p^-1(U)$ has the following components, each of which is mapped homeomorphically to $U$ by $p$











share|cite|improve this question











$endgroup$




Let $p:mathbbC-0to mathbbC-0$ be $zto z^2$, wich is $a+ibmapsto (a^2-b^2)+i(2ab)$. By fundamental theorem of algebra we know that $p$ is surjective, and if $w$ is a square root of $z$, also $-w$ is. Thus for each $zne 0$ we have two distinct square roots of $z$.



I want to prove that the map $p$ is a covering map.



Now, I'm not acquainted of Complex Analysis, but I have read something here on MSE and on Wikipedia.



If we let $zin mathbbC-0$, then we can represent $z$ in polar coordinates in unique way as $z=[r,theta]$ with $r>0$ and $thetain [0,2pi)$ and by de Moivre formula I have that $w_1=[sqrtr,theta/2]$ and $w_2=[sqrtr,theta/2+pi]$ are the two square roots of $z$.



Now, in order to define a function, I have to make a choice on who I want as square root of $z$.



So, let's say that my square root map is $mathbbC-0to mathbbC-0,quad[r,theta] mapsto [sqrtr,theta/2]$, so I'm taking the root that lies in the upper half space.



The problem now is that I want a continuous map. So this map is not good since points of the form $[r,theta], [r,2pi-theta]$ are close to each other when $theta$ goes to $0^+$, but their images are not. But if I restrict my map to $mathbbC-([0,+infty)times 0)$, then it shoud be continuous.



But now, how can I formally prove the that map is truly continuous? By definition the topology on $mathbbC$ is the one that came with the set-indentification with $mathbbR^2$, so by definition a map $mathbbCto mathbbC$ is continuous if and only if it is continuous as a map $mathbbR^2to mathbbR^2$.



For example the map $p:mathbbC-0to mathbbC-0$ , $zto z^2$ is continuous beacuse as a map $p:mathbbR^2-0to mathbbR^2-0$ , $(a,b)to (a^2-b^2,2ab)$ it is continuous.




So to show that $q:mathbbC-([0,+infty)times 0)to mathbbC, quad [r,theta]mapsto [sqrtr,theta/2]$ should I write it as a map from a certain subset of $mathbbR^2$ to a certain subset of $mathbbR^2$ and then control that it is continuous? (As done for $p$, so this include adap the defyning formula of $q$ to "cartesian coordinates")




EDIT




Then, how can i prove that $p$ is a covering map with a strategy like this: pick $zin mathbbC-0$, let $U$ be a certain set. Observe that $U$ is open neighborhood of $z$ and is connected. Now observe that $p^-1(U)$ has the following components, each of which is mapped homeomorphically to $U$ by $p$








general-topology complex-analysis complex-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited yesterday







Minato

















asked yesterday









MinatoMinato

545313




545313











  • $begingroup$
    The issue becomes moot if we consider $[0,2pi)$ as $mathbb R/2pi mathbb R$, i.e. as $[0,2pi)$ being a partitioning of equivalences classes of $mathbb R$ where we have the equivalence relation $yequiv x iff exist k in mathbb Z|y-x=2kpi$.
    $endgroup$
    – fleablood
    yesterday










  • $begingroup$
    or to put it another way... consider $[0,2pi)$ as a metric space with $d(x,y) = min (|x-y|, |x-y + 2pi|, |x-y - 2pi|$. This basically creates a new metric space and $d(e, 2pi -d) = |e+d|$ which is essentially claiming that $2pi = 0$.
    $endgroup$
    – fleablood
    yesterday
















  • $begingroup$
    The issue becomes moot if we consider $[0,2pi)$ as $mathbb R/2pi mathbb R$, i.e. as $[0,2pi)$ being a partitioning of equivalences classes of $mathbb R$ where we have the equivalence relation $yequiv x iff exist k in mathbb Z|y-x=2kpi$.
    $endgroup$
    – fleablood
    yesterday










  • $begingroup$
    or to put it another way... consider $[0,2pi)$ as a metric space with $d(x,y) = min (|x-y|, |x-y + 2pi|, |x-y - 2pi|$. This basically creates a new metric space and $d(e, 2pi -d) = |e+d|$ which is essentially claiming that $2pi = 0$.
    $endgroup$
    – fleablood
    yesterday















$begingroup$
The issue becomes moot if we consider $[0,2pi)$ as $mathbb R/2pi mathbb R$, i.e. as $[0,2pi)$ being a partitioning of equivalences classes of $mathbb R$ where we have the equivalence relation $yequiv x iff exist k in mathbb Z|y-x=2kpi$.
$endgroup$
– fleablood
yesterday




$begingroup$
The issue becomes moot if we consider $[0,2pi)$ as $mathbb R/2pi mathbb R$, i.e. as $[0,2pi)$ being a partitioning of equivalences classes of $mathbb R$ where we have the equivalence relation $yequiv x iff exist k in mathbb Z|y-x=2kpi$.
$endgroup$
– fleablood
yesterday












$begingroup$
or to put it another way... consider $[0,2pi)$ as a metric space with $d(x,y) = min (|x-y|, |x-y + 2pi|, |x-y - 2pi|$. This basically creates a new metric space and $d(e, 2pi -d) = |e+d|$ which is essentially claiming that $2pi = 0$.
$endgroup$
– fleablood
yesterday




$begingroup$
or to put it another way... consider $[0,2pi)$ as a metric space with $d(x,y) = min (|x-y|, |x-y + 2pi|, |x-y - 2pi|$. This basically creates a new metric space and $d(e, 2pi -d) = |e+d|$ which is essentially claiming that $2pi = 0$.
$endgroup$
– fleablood
yesterday










2 Answers
2






active

oldest

votes


















1












$begingroup$

What you ask in the last box is correct, but one needs to say exactly which subsets, and they need to be open subsets.



For example, you can restrict the domain of $p$ to the open subset
$$mathbb C_x+ = z=x+iy mid x > 0
$$

and you can restrict the range of $p$ to the open subset
$$mathbb C - ((-infty,0] times 0)
$$

and the result is a map
$$p_x+ : mathbb C_x+ tomathbb C - ((-infty,0] times 0)
$$

which, of course, is given by the formula $p_x+(z)=z^2$. This map is a continuous bijection with a continuous inverse, i.e. it is a homeomorphism.



If you carefully write a few more such restrictions, then you'll get all the pieces that you need to prove that $p$ is a covering map.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    But should I convert the "formula" of $p^-1$ in "cartesian coordinates" (as done for $p$ in the post) for concluding the continuity? Or are there " other characterizations" which I can use? Because writing the square root map in cartesian coordintes is ugly, so i thought there should be a more elegant way to see the matter.
    $endgroup$
    – Minato
    yesterday











  • $begingroup$
    In other terms, what is the most rigorous (and possibly elegant) way to prove that $p$ is a covering map? (Assuming that for the moment the only thing which is evident to me is the continuity and surjectivity of $p$ and the fact its fibers have 2 elements)
    $endgroup$
    – Minato
    yesterday










  • $begingroup$
    For example, this should be of the form: pick $zin mathbbC-0$, let $U$ be a certain set. Observe that $U$ is open neighborhood of $z$ and is connected. Now observe that $p^-1(U)$ has the following components, each of which is mapped homeomorphically onto $U$ by $p$.
    $endgroup$
    – Minato
    yesterday











  • $begingroup$
    It's straightforward enough to write the formula for the inverse, using what you know from analytic geometry; one can decide for oneself which formula is the most elegant. But the inverse will certainly have different formulas depending on the domain and range restriction.
    $endgroup$
    – Lee Mosher
    yesterday







  • 1




    $begingroup$
    Do you know that $p : S^1 to S^1, p(z) = z^2$ is a covering map? If yes, I shall give you a very simple proof that also your map is one.
    $endgroup$
    – Paul Frost
    yesterday


















0












$begingroup$

For $M subset mathbbC$ define $-M = -z mid z in M $. It is easy to verify that $p^-1(p(M)) = M cup -M$.



Let us prove that $p$ is an open map. Assume that there exists an open $U subset mathbbC^* = mathbbC setminus 0 $ such that $p(U)$ is not open. This means that there exists a sequence $(z_n )$ in $mathbbC^* setminus p(U)$ converging to some $z in p(U)$. Let $(w_n)$ be a sequence in $mathbbC^*$ such that $w_n^2 = z_n$. Since $(z_n)$ is bounded, also $(w_n)$ must be bounded and thus has subsequence converging to some $w in mathbbC$. W.l.o.g. we may assume that $w_n to w$. This shows $p(w) = lim p(w_n) = lim z_n = z$. Hence $w$ is a square root of $z$. We have $w in p^-1(p(U)) = U cup -U$. If $w in -U$, we replace $(w_n)$ by the sequence $(-w_n)$. This satisfies $(-w_n)^2 = z_n$ and converges to $-w in U$ which is the other squre root of $z$. Therefore we may assume w.l.o.g. that $w in U$. Hence $w_n in U$ for $n ge n_0$, hence $z_n = p(w_n) in p(U)$ for $n ge n_0$. This contradicts the fact that all $z_n notin p(U)$.



We now show that each $z in mathbbC^* = mathbbC setminus 0 $ has an open neighborhood $U$ which is evenly covered by $p$.



Let us first consider $z = -1$. $U = mathbbC setminus [0,infty)$ is an open neighborhood of $-1$ in $mathbbC^*$. We have $p^-1(U) = mathbbC setminus p^-1([0,infty)) = mathbbC setminus mathbbR$. Let $V_+$ and $V_-$ denote the open halfplanes $textIm(z) > 0$ and $textIm(z) < 0$, respectively. They are disjoint and their union is $mathbbC setminus mathbbR$. Obviously $p$ maps both $V_pm$ bijectively onto $U$. Since $p$ is an open map, we see that $U$ is evenly covered with sheets $V_pm$.



For $z in mathbbC^*$ define $h_z : mathbbC^* to mathbbC^*, h_z (zeta) = -z zeta$. This is a homeomorphism with inverse $h_-1/z$. We have $h_z(-1) = z$.



Now consider an arbitrary $z in mathbbC^*$. Let $w$ be a square root of $-z$. $U' = h_z(U)$ is an open neighborhood of $z$ and the sets $V'_pm = h_w(V_pm)$ are open and disjoint. We have $p(h_w(zeta)) = w^2 zeta^2 = -z zeta^2 =h_z(p(zeta))$, therefore $p(V'_pm) = p(h_w(V_pm)) = h_z(p(V_pm)) = h_z(U) = V'$. Let $p(z_1) = p(z_2)$ with $z_1, z_2$ either in $V'_+$ or in $V'_-$. Let $zeta_i = h_w^-1(z_i) in V_pm$. We have $p(zeta_i) = p(h_w(z_i)) = h_z(p(z_i))$, hence $p(zeta_1) = p(zeta_2)$ and we conclude $zeta_1 = zeta_2$ and thus $z_1 = z_2$. This shows that $p$ is injective on $V'_pm$. Therefore $p$ maps $V'_pm$ homeomorphically onto $U'$. It remains to show that $p^-1(U') = V'_+ cup V'_-$. So let $zeta in p^-1(U')$, i.e. $p(zeta) in U' = h_z(U)$. Thus $zeta^2 = -z zeta' = w^2 zeta'$ with some $zeta' in U$. We conclude $p(-zeta/w) = (-zeta/w)^2 in U$, hence $-zeta/w in V_pm$. This means $zeta = h_w(-zeta/w) in h_w(V_pm) = V'_pm$.



Remark:



This proof is completely elementary. It does not use any theorems of complex analysis. The only ingredients are the continuity of complex multplication and the fact that each complex number has a root (which can be explictly calculated, see How do I get the square root of a complex number?
).






share|cite|improve this answer











$endgroup$












    Your Answer





    StackExchange.ifUsing("editor", function ()
    return StackExchange.using("mathjaxEditing", function ()
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    );
    );
    , "mathjax-editing");

    StackExchange.ready(function()
    var channelOptions =
    tags: "".split(" "),
    id: "69"
    ;
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function()
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled)
    StackExchange.using("snippets", function()
    createEditor();
    );

    else
    createEditor();

    );

    function createEditor()
    StackExchange.prepareEditor(
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader:
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    ,
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    );



    );













    draft saved

    draft discarded


















    StackExchange.ready(
    function ()
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3139028%2felementary-clarification-about-complex-square-root-maps%23new-answer', 'question_page');

    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    What you ask in the last box is correct, but one needs to say exactly which subsets, and they need to be open subsets.



    For example, you can restrict the domain of $p$ to the open subset
    $$mathbb C_x+ = z=x+iy mid x > 0
    $$

    and you can restrict the range of $p$ to the open subset
    $$mathbb C - ((-infty,0] times 0)
    $$

    and the result is a map
    $$p_x+ : mathbb C_x+ tomathbb C - ((-infty,0] times 0)
    $$

    which, of course, is given by the formula $p_x+(z)=z^2$. This map is a continuous bijection with a continuous inverse, i.e. it is a homeomorphism.



    If you carefully write a few more such restrictions, then you'll get all the pieces that you need to prove that $p$ is a covering map.






    share|cite|improve this answer









    $endgroup$












    • $begingroup$
      But should I convert the "formula" of $p^-1$ in "cartesian coordinates" (as done for $p$ in the post) for concluding the continuity? Or are there " other characterizations" which I can use? Because writing the square root map in cartesian coordintes is ugly, so i thought there should be a more elegant way to see the matter.
      $endgroup$
      – Minato
      yesterday











    • $begingroup$
      In other terms, what is the most rigorous (and possibly elegant) way to prove that $p$ is a covering map? (Assuming that for the moment the only thing which is evident to me is the continuity and surjectivity of $p$ and the fact its fibers have 2 elements)
      $endgroup$
      – Minato
      yesterday










    • $begingroup$
      For example, this should be of the form: pick $zin mathbbC-0$, let $U$ be a certain set. Observe that $U$ is open neighborhood of $z$ and is connected. Now observe that $p^-1(U)$ has the following components, each of which is mapped homeomorphically onto $U$ by $p$.
      $endgroup$
      – Minato
      yesterday











    • $begingroup$
      It's straightforward enough to write the formula for the inverse, using what you know from analytic geometry; one can decide for oneself which formula is the most elegant. But the inverse will certainly have different formulas depending on the domain and range restriction.
      $endgroup$
      – Lee Mosher
      yesterday







    • 1




      $begingroup$
      Do you know that $p : S^1 to S^1, p(z) = z^2$ is a covering map? If yes, I shall give you a very simple proof that also your map is one.
      $endgroup$
      – Paul Frost
      yesterday















    1












    $begingroup$

    What you ask in the last box is correct, but one needs to say exactly which subsets, and they need to be open subsets.



    For example, you can restrict the domain of $p$ to the open subset
    $$mathbb C_x+ = z=x+iy mid x > 0
    $$

    and you can restrict the range of $p$ to the open subset
    $$mathbb C - ((-infty,0] times 0)
    $$

    and the result is a map
    $$p_x+ : mathbb C_x+ tomathbb C - ((-infty,0] times 0)
    $$

    which, of course, is given by the formula $p_x+(z)=z^2$. This map is a continuous bijection with a continuous inverse, i.e. it is a homeomorphism.



    If you carefully write a few more such restrictions, then you'll get all the pieces that you need to prove that $p$ is a covering map.






    share|cite|improve this answer









    $endgroup$












    • $begingroup$
      But should I convert the "formula" of $p^-1$ in "cartesian coordinates" (as done for $p$ in the post) for concluding the continuity? Or are there " other characterizations" which I can use? Because writing the square root map in cartesian coordintes is ugly, so i thought there should be a more elegant way to see the matter.
      $endgroup$
      – Minato
      yesterday











    • $begingroup$
      In other terms, what is the most rigorous (and possibly elegant) way to prove that $p$ is a covering map? (Assuming that for the moment the only thing which is evident to me is the continuity and surjectivity of $p$ and the fact its fibers have 2 elements)
      $endgroup$
      – Minato
      yesterday










    • $begingroup$
      For example, this should be of the form: pick $zin mathbbC-0$, let $U$ be a certain set. Observe that $U$ is open neighborhood of $z$ and is connected. Now observe that $p^-1(U)$ has the following components, each of which is mapped homeomorphically onto $U$ by $p$.
      $endgroup$
      – Minato
      yesterday











    • $begingroup$
      It's straightforward enough to write the formula for the inverse, using what you know from analytic geometry; one can decide for oneself which formula is the most elegant. But the inverse will certainly have different formulas depending on the domain and range restriction.
      $endgroup$
      – Lee Mosher
      yesterday







    • 1




      $begingroup$
      Do you know that $p : S^1 to S^1, p(z) = z^2$ is a covering map? If yes, I shall give you a very simple proof that also your map is one.
      $endgroup$
      – Paul Frost
      yesterday













    1












    1








    1





    $begingroup$

    What you ask in the last box is correct, but one needs to say exactly which subsets, and they need to be open subsets.



    For example, you can restrict the domain of $p$ to the open subset
    $$mathbb C_x+ = z=x+iy mid x > 0
    $$

    and you can restrict the range of $p$ to the open subset
    $$mathbb C - ((-infty,0] times 0)
    $$

    and the result is a map
    $$p_x+ : mathbb C_x+ tomathbb C - ((-infty,0] times 0)
    $$

    which, of course, is given by the formula $p_x+(z)=z^2$. This map is a continuous bijection with a continuous inverse, i.e. it is a homeomorphism.



    If you carefully write a few more such restrictions, then you'll get all the pieces that you need to prove that $p$ is a covering map.






    share|cite|improve this answer









    $endgroup$



    What you ask in the last box is correct, but one needs to say exactly which subsets, and they need to be open subsets.



    For example, you can restrict the domain of $p$ to the open subset
    $$mathbb C_x+ = z=x+iy mid x > 0
    $$

    and you can restrict the range of $p$ to the open subset
    $$mathbb C - ((-infty,0] times 0)
    $$

    and the result is a map
    $$p_x+ : mathbb C_x+ tomathbb C - ((-infty,0] times 0)
    $$

    which, of course, is given by the formula $p_x+(z)=z^2$. This map is a continuous bijection with a continuous inverse, i.e. it is a homeomorphism.



    If you carefully write a few more such restrictions, then you'll get all the pieces that you need to prove that $p$ is a covering map.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered yesterday









    Lee MosherLee Mosher

    50.2k33787




    50.2k33787











    • $begingroup$
      But should I convert the "formula" of $p^-1$ in "cartesian coordinates" (as done for $p$ in the post) for concluding the continuity? Or are there " other characterizations" which I can use? Because writing the square root map in cartesian coordintes is ugly, so i thought there should be a more elegant way to see the matter.
      $endgroup$
      – Minato
      yesterday











    • $begingroup$
      In other terms, what is the most rigorous (and possibly elegant) way to prove that $p$ is a covering map? (Assuming that for the moment the only thing which is evident to me is the continuity and surjectivity of $p$ and the fact its fibers have 2 elements)
      $endgroup$
      – Minato
      yesterday










    • $begingroup$
      For example, this should be of the form: pick $zin mathbbC-0$, let $U$ be a certain set. Observe that $U$ is open neighborhood of $z$ and is connected. Now observe that $p^-1(U)$ has the following components, each of which is mapped homeomorphically onto $U$ by $p$.
      $endgroup$
      – Minato
      yesterday











    • $begingroup$
      It's straightforward enough to write the formula for the inverse, using what you know from analytic geometry; one can decide for oneself which formula is the most elegant. But the inverse will certainly have different formulas depending on the domain and range restriction.
      $endgroup$
      – Lee Mosher
      yesterday







    • 1




      $begingroup$
      Do you know that $p : S^1 to S^1, p(z) = z^2$ is a covering map? If yes, I shall give you a very simple proof that also your map is one.
      $endgroup$
      – Paul Frost
      yesterday
















    • $begingroup$
      But should I convert the "formula" of $p^-1$ in "cartesian coordinates" (as done for $p$ in the post) for concluding the continuity? Or are there " other characterizations" which I can use? Because writing the square root map in cartesian coordintes is ugly, so i thought there should be a more elegant way to see the matter.
      $endgroup$
      – Minato
      yesterday











    • $begingroup$
      In other terms, what is the most rigorous (and possibly elegant) way to prove that $p$ is a covering map? (Assuming that for the moment the only thing which is evident to me is the continuity and surjectivity of $p$ and the fact its fibers have 2 elements)
      $endgroup$
      – Minato
      yesterday










    • $begingroup$
      For example, this should be of the form: pick $zin mathbbC-0$, let $U$ be a certain set. Observe that $U$ is open neighborhood of $z$ and is connected. Now observe that $p^-1(U)$ has the following components, each of which is mapped homeomorphically onto $U$ by $p$.
      $endgroup$
      – Minato
      yesterday











    • $begingroup$
      It's straightforward enough to write the formula for the inverse, using what you know from analytic geometry; one can decide for oneself which formula is the most elegant. But the inverse will certainly have different formulas depending on the domain and range restriction.
      $endgroup$
      – Lee Mosher
      yesterday







    • 1




      $begingroup$
      Do you know that $p : S^1 to S^1, p(z) = z^2$ is a covering map? If yes, I shall give you a very simple proof that also your map is one.
      $endgroup$
      – Paul Frost
      yesterday















    $begingroup$
    But should I convert the "formula" of $p^-1$ in "cartesian coordinates" (as done for $p$ in the post) for concluding the continuity? Or are there " other characterizations" which I can use? Because writing the square root map in cartesian coordintes is ugly, so i thought there should be a more elegant way to see the matter.
    $endgroup$
    – Minato
    yesterday





    $begingroup$
    But should I convert the "formula" of $p^-1$ in "cartesian coordinates" (as done for $p$ in the post) for concluding the continuity? Or are there " other characterizations" which I can use? Because writing the square root map in cartesian coordintes is ugly, so i thought there should be a more elegant way to see the matter.
    $endgroup$
    – Minato
    yesterday













    $begingroup$
    In other terms, what is the most rigorous (and possibly elegant) way to prove that $p$ is a covering map? (Assuming that for the moment the only thing which is evident to me is the continuity and surjectivity of $p$ and the fact its fibers have 2 elements)
    $endgroup$
    – Minato
    yesterday




    $begingroup$
    In other terms, what is the most rigorous (and possibly elegant) way to prove that $p$ is a covering map? (Assuming that for the moment the only thing which is evident to me is the continuity and surjectivity of $p$ and the fact its fibers have 2 elements)
    $endgroup$
    – Minato
    yesterday












    $begingroup$
    For example, this should be of the form: pick $zin mathbbC-0$, let $U$ be a certain set. Observe that $U$ is open neighborhood of $z$ and is connected. Now observe that $p^-1(U)$ has the following components, each of which is mapped homeomorphically onto $U$ by $p$.
    $endgroup$
    – Minato
    yesterday





    $begingroup$
    For example, this should be of the form: pick $zin mathbbC-0$, let $U$ be a certain set. Observe that $U$ is open neighborhood of $z$ and is connected. Now observe that $p^-1(U)$ has the following components, each of which is mapped homeomorphically onto $U$ by $p$.
    $endgroup$
    – Minato
    yesterday













    $begingroup$
    It's straightforward enough to write the formula for the inverse, using what you know from analytic geometry; one can decide for oneself which formula is the most elegant. But the inverse will certainly have different formulas depending on the domain and range restriction.
    $endgroup$
    – Lee Mosher
    yesterday





    $begingroup$
    It's straightforward enough to write the formula for the inverse, using what you know from analytic geometry; one can decide for oneself which formula is the most elegant. But the inverse will certainly have different formulas depending on the domain and range restriction.
    $endgroup$
    – Lee Mosher
    yesterday





    1




    1




    $begingroup$
    Do you know that $p : S^1 to S^1, p(z) = z^2$ is a covering map? If yes, I shall give you a very simple proof that also your map is one.
    $endgroup$
    – Paul Frost
    yesterday




    $begingroup$
    Do you know that $p : S^1 to S^1, p(z) = z^2$ is a covering map? If yes, I shall give you a very simple proof that also your map is one.
    $endgroup$
    – Paul Frost
    yesterday











    0












    $begingroup$

    For $M subset mathbbC$ define $-M = -z mid z in M $. It is easy to verify that $p^-1(p(M)) = M cup -M$.



    Let us prove that $p$ is an open map. Assume that there exists an open $U subset mathbbC^* = mathbbC setminus 0 $ such that $p(U)$ is not open. This means that there exists a sequence $(z_n )$ in $mathbbC^* setminus p(U)$ converging to some $z in p(U)$. Let $(w_n)$ be a sequence in $mathbbC^*$ such that $w_n^2 = z_n$. Since $(z_n)$ is bounded, also $(w_n)$ must be bounded and thus has subsequence converging to some $w in mathbbC$. W.l.o.g. we may assume that $w_n to w$. This shows $p(w) = lim p(w_n) = lim z_n = z$. Hence $w$ is a square root of $z$. We have $w in p^-1(p(U)) = U cup -U$. If $w in -U$, we replace $(w_n)$ by the sequence $(-w_n)$. This satisfies $(-w_n)^2 = z_n$ and converges to $-w in U$ which is the other squre root of $z$. Therefore we may assume w.l.o.g. that $w in U$. Hence $w_n in U$ for $n ge n_0$, hence $z_n = p(w_n) in p(U)$ for $n ge n_0$. This contradicts the fact that all $z_n notin p(U)$.



    We now show that each $z in mathbbC^* = mathbbC setminus 0 $ has an open neighborhood $U$ which is evenly covered by $p$.



    Let us first consider $z = -1$. $U = mathbbC setminus [0,infty)$ is an open neighborhood of $-1$ in $mathbbC^*$. We have $p^-1(U) = mathbbC setminus p^-1([0,infty)) = mathbbC setminus mathbbR$. Let $V_+$ and $V_-$ denote the open halfplanes $textIm(z) > 0$ and $textIm(z) < 0$, respectively. They are disjoint and their union is $mathbbC setminus mathbbR$. Obviously $p$ maps both $V_pm$ bijectively onto $U$. Since $p$ is an open map, we see that $U$ is evenly covered with sheets $V_pm$.



    For $z in mathbbC^*$ define $h_z : mathbbC^* to mathbbC^*, h_z (zeta) = -z zeta$. This is a homeomorphism with inverse $h_-1/z$. We have $h_z(-1) = z$.



    Now consider an arbitrary $z in mathbbC^*$. Let $w$ be a square root of $-z$. $U' = h_z(U)$ is an open neighborhood of $z$ and the sets $V'_pm = h_w(V_pm)$ are open and disjoint. We have $p(h_w(zeta)) = w^2 zeta^2 = -z zeta^2 =h_z(p(zeta))$, therefore $p(V'_pm) = p(h_w(V_pm)) = h_z(p(V_pm)) = h_z(U) = V'$. Let $p(z_1) = p(z_2)$ with $z_1, z_2$ either in $V'_+$ or in $V'_-$. Let $zeta_i = h_w^-1(z_i) in V_pm$. We have $p(zeta_i) = p(h_w(z_i)) = h_z(p(z_i))$, hence $p(zeta_1) = p(zeta_2)$ and we conclude $zeta_1 = zeta_2$ and thus $z_1 = z_2$. This shows that $p$ is injective on $V'_pm$. Therefore $p$ maps $V'_pm$ homeomorphically onto $U'$. It remains to show that $p^-1(U') = V'_+ cup V'_-$. So let $zeta in p^-1(U')$, i.e. $p(zeta) in U' = h_z(U)$. Thus $zeta^2 = -z zeta' = w^2 zeta'$ with some $zeta' in U$. We conclude $p(-zeta/w) = (-zeta/w)^2 in U$, hence $-zeta/w in V_pm$. This means $zeta = h_w(-zeta/w) in h_w(V_pm) = V'_pm$.



    Remark:



    This proof is completely elementary. It does not use any theorems of complex analysis. The only ingredients are the continuity of complex multplication and the fact that each complex number has a root (which can be explictly calculated, see How do I get the square root of a complex number?
    ).






    share|cite|improve this answer











    $endgroup$

















      0












      $begingroup$

      For $M subset mathbbC$ define $-M = -z mid z in M $. It is easy to verify that $p^-1(p(M)) = M cup -M$.



      Let us prove that $p$ is an open map. Assume that there exists an open $U subset mathbbC^* = mathbbC setminus 0 $ such that $p(U)$ is not open. This means that there exists a sequence $(z_n )$ in $mathbbC^* setminus p(U)$ converging to some $z in p(U)$. Let $(w_n)$ be a sequence in $mathbbC^*$ such that $w_n^2 = z_n$. Since $(z_n)$ is bounded, also $(w_n)$ must be bounded and thus has subsequence converging to some $w in mathbbC$. W.l.o.g. we may assume that $w_n to w$. This shows $p(w) = lim p(w_n) = lim z_n = z$. Hence $w$ is a square root of $z$. We have $w in p^-1(p(U)) = U cup -U$. If $w in -U$, we replace $(w_n)$ by the sequence $(-w_n)$. This satisfies $(-w_n)^2 = z_n$ and converges to $-w in U$ which is the other squre root of $z$. Therefore we may assume w.l.o.g. that $w in U$. Hence $w_n in U$ for $n ge n_0$, hence $z_n = p(w_n) in p(U)$ for $n ge n_0$. This contradicts the fact that all $z_n notin p(U)$.



      We now show that each $z in mathbbC^* = mathbbC setminus 0 $ has an open neighborhood $U$ which is evenly covered by $p$.



      Let us first consider $z = -1$. $U = mathbbC setminus [0,infty)$ is an open neighborhood of $-1$ in $mathbbC^*$. We have $p^-1(U) = mathbbC setminus p^-1([0,infty)) = mathbbC setminus mathbbR$. Let $V_+$ and $V_-$ denote the open halfplanes $textIm(z) > 0$ and $textIm(z) < 0$, respectively. They are disjoint and their union is $mathbbC setminus mathbbR$. Obviously $p$ maps both $V_pm$ bijectively onto $U$. Since $p$ is an open map, we see that $U$ is evenly covered with sheets $V_pm$.



      For $z in mathbbC^*$ define $h_z : mathbbC^* to mathbbC^*, h_z (zeta) = -z zeta$. This is a homeomorphism with inverse $h_-1/z$. We have $h_z(-1) = z$.



      Now consider an arbitrary $z in mathbbC^*$. Let $w$ be a square root of $-z$. $U' = h_z(U)$ is an open neighborhood of $z$ and the sets $V'_pm = h_w(V_pm)$ are open and disjoint. We have $p(h_w(zeta)) = w^2 zeta^2 = -z zeta^2 =h_z(p(zeta))$, therefore $p(V'_pm) = p(h_w(V_pm)) = h_z(p(V_pm)) = h_z(U) = V'$. Let $p(z_1) = p(z_2)$ with $z_1, z_2$ either in $V'_+$ or in $V'_-$. Let $zeta_i = h_w^-1(z_i) in V_pm$. We have $p(zeta_i) = p(h_w(z_i)) = h_z(p(z_i))$, hence $p(zeta_1) = p(zeta_2)$ and we conclude $zeta_1 = zeta_2$ and thus $z_1 = z_2$. This shows that $p$ is injective on $V'_pm$. Therefore $p$ maps $V'_pm$ homeomorphically onto $U'$. It remains to show that $p^-1(U') = V'_+ cup V'_-$. So let $zeta in p^-1(U')$, i.e. $p(zeta) in U' = h_z(U)$. Thus $zeta^2 = -z zeta' = w^2 zeta'$ with some $zeta' in U$. We conclude $p(-zeta/w) = (-zeta/w)^2 in U$, hence $-zeta/w in V_pm$. This means $zeta = h_w(-zeta/w) in h_w(V_pm) = V'_pm$.



      Remark:



      This proof is completely elementary. It does not use any theorems of complex analysis. The only ingredients are the continuity of complex multplication and the fact that each complex number has a root (which can be explictly calculated, see How do I get the square root of a complex number?
      ).






      share|cite|improve this answer











      $endgroup$















        0












        0








        0





        $begingroup$

        For $M subset mathbbC$ define $-M = -z mid z in M $. It is easy to verify that $p^-1(p(M)) = M cup -M$.



        Let us prove that $p$ is an open map. Assume that there exists an open $U subset mathbbC^* = mathbbC setminus 0 $ such that $p(U)$ is not open. This means that there exists a sequence $(z_n )$ in $mathbbC^* setminus p(U)$ converging to some $z in p(U)$. Let $(w_n)$ be a sequence in $mathbbC^*$ such that $w_n^2 = z_n$. Since $(z_n)$ is bounded, also $(w_n)$ must be bounded and thus has subsequence converging to some $w in mathbbC$. W.l.o.g. we may assume that $w_n to w$. This shows $p(w) = lim p(w_n) = lim z_n = z$. Hence $w$ is a square root of $z$. We have $w in p^-1(p(U)) = U cup -U$. If $w in -U$, we replace $(w_n)$ by the sequence $(-w_n)$. This satisfies $(-w_n)^2 = z_n$ and converges to $-w in U$ which is the other squre root of $z$. Therefore we may assume w.l.o.g. that $w in U$. Hence $w_n in U$ for $n ge n_0$, hence $z_n = p(w_n) in p(U)$ for $n ge n_0$. This contradicts the fact that all $z_n notin p(U)$.



        We now show that each $z in mathbbC^* = mathbbC setminus 0 $ has an open neighborhood $U$ which is evenly covered by $p$.



        Let us first consider $z = -1$. $U = mathbbC setminus [0,infty)$ is an open neighborhood of $-1$ in $mathbbC^*$. We have $p^-1(U) = mathbbC setminus p^-1([0,infty)) = mathbbC setminus mathbbR$. Let $V_+$ and $V_-$ denote the open halfplanes $textIm(z) > 0$ and $textIm(z) < 0$, respectively. They are disjoint and their union is $mathbbC setminus mathbbR$. Obviously $p$ maps both $V_pm$ bijectively onto $U$. Since $p$ is an open map, we see that $U$ is evenly covered with sheets $V_pm$.



        For $z in mathbbC^*$ define $h_z : mathbbC^* to mathbbC^*, h_z (zeta) = -z zeta$. This is a homeomorphism with inverse $h_-1/z$. We have $h_z(-1) = z$.



        Now consider an arbitrary $z in mathbbC^*$. Let $w$ be a square root of $-z$. $U' = h_z(U)$ is an open neighborhood of $z$ and the sets $V'_pm = h_w(V_pm)$ are open and disjoint. We have $p(h_w(zeta)) = w^2 zeta^2 = -z zeta^2 =h_z(p(zeta))$, therefore $p(V'_pm) = p(h_w(V_pm)) = h_z(p(V_pm)) = h_z(U) = V'$. Let $p(z_1) = p(z_2)$ with $z_1, z_2$ either in $V'_+$ or in $V'_-$. Let $zeta_i = h_w^-1(z_i) in V_pm$. We have $p(zeta_i) = p(h_w(z_i)) = h_z(p(z_i))$, hence $p(zeta_1) = p(zeta_2)$ and we conclude $zeta_1 = zeta_2$ and thus $z_1 = z_2$. This shows that $p$ is injective on $V'_pm$. Therefore $p$ maps $V'_pm$ homeomorphically onto $U'$. It remains to show that $p^-1(U') = V'_+ cup V'_-$. So let $zeta in p^-1(U')$, i.e. $p(zeta) in U' = h_z(U)$. Thus $zeta^2 = -z zeta' = w^2 zeta'$ with some $zeta' in U$. We conclude $p(-zeta/w) = (-zeta/w)^2 in U$, hence $-zeta/w in V_pm$. This means $zeta = h_w(-zeta/w) in h_w(V_pm) = V'_pm$.



        Remark:



        This proof is completely elementary. It does not use any theorems of complex analysis. The only ingredients are the continuity of complex multplication and the fact that each complex number has a root (which can be explictly calculated, see How do I get the square root of a complex number?
        ).






        share|cite|improve this answer











        $endgroup$



        For $M subset mathbbC$ define $-M = -z mid z in M $. It is easy to verify that $p^-1(p(M)) = M cup -M$.



        Let us prove that $p$ is an open map. Assume that there exists an open $U subset mathbbC^* = mathbbC setminus 0 $ such that $p(U)$ is not open. This means that there exists a sequence $(z_n )$ in $mathbbC^* setminus p(U)$ converging to some $z in p(U)$. Let $(w_n)$ be a sequence in $mathbbC^*$ such that $w_n^2 = z_n$. Since $(z_n)$ is bounded, also $(w_n)$ must be bounded and thus has subsequence converging to some $w in mathbbC$. W.l.o.g. we may assume that $w_n to w$. This shows $p(w) = lim p(w_n) = lim z_n = z$. Hence $w$ is a square root of $z$. We have $w in p^-1(p(U)) = U cup -U$. If $w in -U$, we replace $(w_n)$ by the sequence $(-w_n)$. This satisfies $(-w_n)^2 = z_n$ and converges to $-w in U$ which is the other squre root of $z$. Therefore we may assume w.l.o.g. that $w in U$. Hence $w_n in U$ for $n ge n_0$, hence $z_n = p(w_n) in p(U)$ for $n ge n_0$. This contradicts the fact that all $z_n notin p(U)$.



        We now show that each $z in mathbbC^* = mathbbC setminus 0 $ has an open neighborhood $U$ which is evenly covered by $p$.



        Let us first consider $z = -1$. $U = mathbbC setminus [0,infty)$ is an open neighborhood of $-1$ in $mathbbC^*$. We have $p^-1(U) = mathbbC setminus p^-1([0,infty)) = mathbbC setminus mathbbR$. Let $V_+$ and $V_-$ denote the open halfplanes $textIm(z) > 0$ and $textIm(z) < 0$, respectively. They are disjoint and their union is $mathbbC setminus mathbbR$. Obviously $p$ maps both $V_pm$ bijectively onto $U$. Since $p$ is an open map, we see that $U$ is evenly covered with sheets $V_pm$.



        For $z in mathbbC^*$ define $h_z : mathbbC^* to mathbbC^*, h_z (zeta) = -z zeta$. This is a homeomorphism with inverse $h_-1/z$. We have $h_z(-1) = z$.



        Now consider an arbitrary $z in mathbbC^*$. Let $w$ be a square root of $-z$. $U' = h_z(U)$ is an open neighborhood of $z$ and the sets $V'_pm = h_w(V_pm)$ are open and disjoint. We have $p(h_w(zeta)) = w^2 zeta^2 = -z zeta^2 =h_z(p(zeta))$, therefore $p(V'_pm) = p(h_w(V_pm)) = h_z(p(V_pm)) = h_z(U) = V'$. Let $p(z_1) = p(z_2)$ with $z_1, z_2$ either in $V'_+$ or in $V'_-$. Let $zeta_i = h_w^-1(z_i) in V_pm$. We have $p(zeta_i) = p(h_w(z_i)) = h_z(p(z_i))$, hence $p(zeta_1) = p(zeta_2)$ and we conclude $zeta_1 = zeta_2$ and thus $z_1 = z_2$. This shows that $p$ is injective on $V'_pm$. Therefore $p$ maps $V'_pm$ homeomorphically onto $U'$. It remains to show that $p^-1(U') = V'_+ cup V'_-$. So let $zeta in p^-1(U')$, i.e. $p(zeta) in U' = h_z(U)$. Thus $zeta^2 = -z zeta' = w^2 zeta'$ with some $zeta' in U$. We conclude $p(-zeta/w) = (-zeta/w)^2 in U$, hence $-zeta/w in V_pm$. This means $zeta = h_w(-zeta/w) in h_w(V_pm) = V'_pm$.



        Remark:



        This proof is completely elementary. It does not use any theorems of complex analysis. The only ingredients are the continuity of complex multplication and the fact that each complex number has a root (which can be explictly calculated, see How do I get the square root of a complex number?
        ).







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 7 hours ago

























        answered 16 hours ago









        Paul FrostPaul Frost

        11.5k3934




        11.5k3934



























            draft saved

            draft discarded
















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid


            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.

            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function ()
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3139028%2felementary-clarification-about-complex-square-root-maps%23new-answer', 'question_page');

            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            How should I support this large drywall patch? Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?How do I cover large gaps in drywall?How do I keep drywall around a patch from crumbling?Can I glue a second layer of drywall?How to patch long strip on drywall?Large drywall patch: how to avoid bulging seams?Drywall Mesh Patch vs. Bulge? To remove or not to remove?How to fix this drywall job?Prep drywall before backsplashWhat's the best way to fix this horrible drywall patch job?Drywall patching using 3M Patch Plus Primer

            random experiment with two different functions on unit interval Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Random variable and probability space notionsRandom Walk with EdgesFinding functions where the increase over a random interval is Poisson distributedNumber of days until dayCan an observed event in fact be of zero probability?Unit random processmodels of coins and uniform distributionHow to get the number of successes given $n$ trials , probability $P$ and a random variable $X$Absorbing Markov chain in a computer. Is “almost every” turned into always convergence in computer executions?Stopped random walk is not uniformly integrable

            Lowndes Grove History Architecture References Navigation menu32°48′6″N 79°57′58″W / 32.80167°N 79.96611°W / 32.80167; -79.9661132°48′6″N 79°57′58″W / 32.80167°N 79.96611°W / 32.80167; -79.9661178002500"National Register Information System"Historic houses of South Carolina"Lowndes Grove""+32° 48' 6.00", −79° 57' 58.00""Lowndes Grove, Charleston County (260 St. Margaret St., Charleston)""Lowndes Grove"The Charleston ExpositionIt Happened in South Carolina"Lowndes Grove (House), Saint Margaret Street & Sixth Avenue, Charleston, Charleston County, SC(Photographs)"Plantations of the Carolina Low Countrye