$p:Xto Y$ covering space if $qcirc p:Xto Z$ and $q:Yto Z$ covering spaces and $Z$ locally path-connected.
$begingroup$
I have done the problem, but I'm confused about why the local path-connectedness of $Z$ is necessary.
My solution: Let $p:Xto Y$ and $q:Yto Z$ be such that $q$ and $qcirc p$ are covering spaces. Let $yin Y$ and let $W$ be a path-connected neighborhood of $z=q(y)in Z$ which is evenly covered by $q$ and $qcirc p.$ Let $V$ be the path component of $q^{-1}(W)$ which contains $y$ (observe that $V$ is then a sheet of $W$ by $q$). Any sheet $U$ of $p^{-1}(V)$ is mapped homeomorphically to $W$ by $qcirc p.$ Since $q:Vto W$ is a homeomorphism, we conclude that $p:Uto V$ is a homeomorphism, so $p$ is a covering space.
It seems like I used the local path-connectedness of $Z$ in my solution, letting $W$ be a path-connected neighborhood of $q(y),$ but this use seems artificial. It isn't clear to me why $W$ must be path-connected, other than when one tries to isolate an evenly covered neighborhood $V$ of $y.$ However, it seems like local connectedness is enough to do this.
So my question is this: is local path-connectedness of $Z$ necessary (and if so, is there some instructive counterexample when $Z$ is not locally path-connected) or can this hypothesis be weakened?
general-topology algebraic-topology covering-spaces
$endgroup$
add a comment |
$begingroup$
I have done the problem, but I'm confused about why the local path-connectedness of $Z$ is necessary.
My solution: Let $p:Xto Y$ and $q:Yto Z$ be such that $q$ and $qcirc p$ are covering spaces. Let $yin Y$ and let $W$ be a path-connected neighborhood of $z=q(y)in Z$ which is evenly covered by $q$ and $qcirc p.$ Let $V$ be the path component of $q^{-1}(W)$ which contains $y$ (observe that $V$ is then a sheet of $W$ by $q$). Any sheet $U$ of $p^{-1}(V)$ is mapped homeomorphically to $W$ by $qcirc p.$ Since $q:Vto W$ is a homeomorphism, we conclude that $p:Uto V$ is a homeomorphism, so $p$ is a covering space.
It seems like I used the local path-connectedness of $Z$ in my solution, letting $W$ be a path-connected neighborhood of $q(y),$ but this use seems artificial. It isn't clear to me why $W$ must be path-connected, other than when one tries to isolate an evenly covered neighborhood $V$ of $y.$ However, it seems like local connectedness is enough to do this.
So my question is this: is local path-connectedness of $Z$ necessary (and if so, is there some instructive counterexample when $Z$ is not locally path-connected) or can this hypothesis be weakened?
general-topology algebraic-topology covering-spaces
$endgroup$
$begingroup$
You need some additional assumptions. Probably the best setting is to require that $X,Y,Z$ are connected. Otherwise you get trivial counterexamples like $X = Z$ and $Y= Z times {0,1 }$ with $p(z) = (z,0)$ and $q(z,i) = z$.
$endgroup$
– Paul Frost
Dec 29 '18 at 14:56
add a comment |
$begingroup$
I have done the problem, but I'm confused about why the local path-connectedness of $Z$ is necessary.
My solution: Let $p:Xto Y$ and $q:Yto Z$ be such that $q$ and $qcirc p$ are covering spaces. Let $yin Y$ and let $W$ be a path-connected neighborhood of $z=q(y)in Z$ which is evenly covered by $q$ and $qcirc p.$ Let $V$ be the path component of $q^{-1}(W)$ which contains $y$ (observe that $V$ is then a sheet of $W$ by $q$). Any sheet $U$ of $p^{-1}(V)$ is mapped homeomorphically to $W$ by $qcirc p.$ Since $q:Vto W$ is a homeomorphism, we conclude that $p:Uto V$ is a homeomorphism, so $p$ is a covering space.
It seems like I used the local path-connectedness of $Z$ in my solution, letting $W$ be a path-connected neighborhood of $q(y),$ but this use seems artificial. It isn't clear to me why $W$ must be path-connected, other than when one tries to isolate an evenly covered neighborhood $V$ of $y.$ However, it seems like local connectedness is enough to do this.
So my question is this: is local path-connectedness of $Z$ necessary (and if so, is there some instructive counterexample when $Z$ is not locally path-connected) or can this hypothesis be weakened?
general-topology algebraic-topology covering-spaces
$endgroup$
I have done the problem, but I'm confused about why the local path-connectedness of $Z$ is necessary.
My solution: Let $p:Xto Y$ and $q:Yto Z$ be such that $q$ and $qcirc p$ are covering spaces. Let $yin Y$ and let $W$ be a path-connected neighborhood of $z=q(y)in Z$ which is evenly covered by $q$ and $qcirc p.$ Let $V$ be the path component of $q^{-1}(W)$ which contains $y$ (observe that $V$ is then a sheet of $W$ by $q$). Any sheet $U$ of $p^{-1}(V)$ is mapped homeomorphically to $W$ by $qcirc p.$ Since $q:Vto W$ is a homeomorphism, we conclude that $p:Uto V$ is a homeomorphism, so $p$ is a covering space.
It seems like I used the local path-connectedness of $Z$ in my solution, letting $W$ be a path-connected neighborhood of $q(y),$ but this use seems artificial. It isn't clear to me why $W$ must be path-connected, other than when one tries to isolate an evenly covered neighborhood $V$ of $y.$ However, it seems like local connectedness is enough to do this.
So my question is this: is local path-connectedness of $Z$ necessary (and if so, is there some instructive counterexample when $Z$ is not locally path-connected) or can this hypothesis be weakened?
general-topology algebraic-topology covering-spaces
general-topology algebraic-topology covering-spaces
edited Dec 28 '18 at 21:22
D. Brogan
asked Dec 28 '18 at 19:06
D. BroganD. Brogan
643513
643513
$begingroup$
You need some additional assumptions. Probably the best setting is to require that $X,Y,Z$ are connected. Otherwise you get trivial counterexamples like $X = Z$ and $Y= Z times {0,1 }$ with $p(z) = (z,0)$ and $q(z,i) = z$.
$endgroup$
– Paul Frost
Dec 29 '18 at 14:56
add a comment |
$begingroup$
You need some additional assumptions. Probably the best setting is to require that $X,Y,Z$ are connected. Otherwise you get trivial counterexamples like $X = Z$ and $Y= Z times {0,1 }$ with $p(z) = (z,0)$ and $q(z,i) = z$.
$endgroup$
– Paul Frost
Dec 29 '18 at 14:56
$begingroup$
You need some additional assumptions. Probably the best setting is to require that $X,Y,Z$ are connected. Otherwise you get trivial counterexamples like $X = Z$ and $Y= Z times {0,1 }$ with $p(z) = (z,0)$ and $q(z,i) = z$.
$endgroup$
– Paul Frost
Dec 29 '18 at 14:56
$begingroup$
You need some additional assumptions. Probably the best setting is to require that $X,Y,Z$ are connected. Otherwise you get trivial counterexamples like $X = Z$ and $Y= Z times {0,1 }$ with $p(z) = (z,0)$ and $q(z,i) = z$.
$endgroup$
– Paul Frost
Dec 29 '18 at 14:56
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It can be weakened. We shall use the following well-known fact about covering projections:
Given $p,q$ as in your question (i.e. $q circ p$ and $q$ are assumed to be covering projections). If $Z$ is locally connected and $p$ is a surjection, then $p$ is a covering projection.
In this result no further connectedness assumptions on $X,Y,Z$ are needed.
Hence we have to look for assumptions assuring that $p$ is surjective. To avoid trivial counterexamples, it seems that we should have the minimal requirement that $Y$ is connected (otherwise we may take $X = Z$, $Y = Z times F$ with a discrete $F$ having more than one point and $p(z) = (z,f_0)$, $q(z,f) = z$). This implies that also $Z$ is connected. For the sake of homogeneity we may moreover asume that $X$ is connected, but this is not really needed.
Therefore, let us assume that $X,Y,Z$ are connected and locally connected (it suffices to assume one of these spaces to be locally connected because they are locally homeomorphic). We shall prove
If $Y$ is path connected, then $p$ is surjective, i.e. a covering projection.
This is somewhat weaker than requiring $Z$ locally path connected. Note that if the latter is satisfied, then also $Y$ is locally path connected, hence $Y$ is path connected.
Let $y in Y$. Choose any $x in X$ and any path $v$ in $Y$ from $p(x)$ to $y$. Then $w = q circ v$ is a path in $Z$ beginning at $(q circ p)(x)$. It can be lifted to path $u$ in $X$ beginning at $x$. Both paths $v$ and $p circ u$ are lifts of $w$ beginning at $p(x)$. By unique path lifting we see that $v = p circ u$, hence $y = v(1) = p(u(1)) in p(X)$.
$endgroup$
add a comment |
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
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3055199%2fpx-to-y-covering-space-if-q-circ-px-to-z-and-qy-to-z-covering-spaces-an%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It can be weakened. We shall use the following well-known fact about covering projections:
Given $p,q$ as in your question (i.e. $q circ p$ and $q$ are assumed to be covering projections). If $Z$ is locally connected and $p$ is a surjection, then $p$ is a covering projection.
In this result no further connectedness assumptions on $X,Y,Z$ are needed.
Hence we have to look for assumptions assuring that $p$ is surjective. To avoid trivial counterexamples, it seems that we should have the minimal requirement that $Y$ is connected (otherwise we may take $X = Z$, $Y = Z times F$ with a discrete $F$ having more than one point and $p(z) = (z,f_0)$, $q(z,f) = z$). This implies that also $Z$ is connected. For the sake of homogeneity we may moreover asume that $X$ is connected, but this is not really needed.
Therefore, let us assume that $X,Y,Z$ are connected and locally connected (it suffices to assume one of these spaces to be locally connected because they are locally homeomorphic). We shall prove
If $Y$ is path connected, then $p$ is surjective, i.e. a covering projection.
This is somewhat weaker than requiring $Z$ locally path connected. Note that if the latter is satisfied, then also $Y$ is locally path connected, hence $Y$ is path connected.
Let $y in Y$. Choose any $x in X$ and any path $v$ in $Y$ from $p(x)$ to $y$. Then $w = q circ v$ is a path in $Z$ beginning at $(q circ p)(x)$. It can be lifted to path $u$ in $X$ beginning at $x$. Both paths $v$ and $p circ u$ are lifts of $w$ beginning at $p(x)$. By unique path lifting we see that $v = p circ u$, hence $y = v(1) = p(u(1)) in p(X)$.
$endgroup$
add a comment |
$begingroup$
It can be weakened. We shall use the following well-known fact about covering projections:
Given $p,q$ as in your question (i.e. $q circ p$ and $q$ are assumed to be covering projections). If $Z$ is locally connected and $p$ is a surjection, then $p$ is a covering projection.
In this result no further connectedness assumptions on $X,Y,Z$ are needed.
Hence we have to look for assumptions assuring that $p$ is surjective. To avoid trivial counterexamples, it seems that we should have the minimal requirement that $Y$ is connected (otherwise we may take $X = Z$, $Y = Z times F$ with a discrete $F$ having more than one point and $p(z) = (z,f_0)$, $q(z,f) = z$). This implies that also $Z$ is connected. For the sake of homogeneity we may moreover asume that $X$ is connected, but this is not really needed.
Therefore, let us assume that $X,Y,Z$ are connected and locally connected (it suffices to assume one of these spaces to be locally connected because they are locally homeomorphic). We shall prove
If $Y$ is path connected, then $p$ is surjective, i.e. a covering projection.
This is somewhat weaker than requiring $Z$ locally path connected. Note that if the latter is satisfied, then also $Y$ is locally path connected, hence $Y$ is path connected.
Let $y in Y$. Choose any $x in X$ and any path $v$ in $Y$ from $p(x)$ to $y$. Then $w = q circ v$ is a path in $Z$ beginning at $(q circ p)(x)$. It can be lifted to path $u$ in $X$ beginning at $x$. Both paths $v$ and $p circ u$ are lifts of $w$ beginning at $p(x)$. By unique path lifting we see that $v = p circ u$, hence $y = v(1) = p(u(1)) in p(X)$.
$endgroup$
add a comment |
$begingroup$
It can be weakened. We shall use the following well-known fact about covering projections:
Given $p,q$ as in your question (i.e. $q circ p$ and $q$ are assumed to be covering projections). If $Z$ is locally connected and $p$ is a surjection, then $p$ is a covering projection.
In this result no further connectedness assumptions on $X,Y,Z$ are needed.
Hence we have to look for assumptions assuring that $p$ is surjective. To avoid trivial counterexamples, it seems that we should have the minimal requirement that $Y$ is connected (otherwise we may take $X = Z$, $Y = Z times F$ with a discrete $F$ having more than one point and $p(z) = (z,f_0)$, $q(z,f) = z$). This implies that also $Z$ is connected. For the sake of homogeneity we may moreover asume that $X$ is connected, but this is not really needed.
Therefore, let us assume that $X,Y,Z$ are connected and locally connected (it suffices to assume one of these spaces to be locally connected because they are locally homeomorphic). We shall prove
If $Y$ is path connected, then $p$ is surjective, i.e. a covering projection.
This is somewhat weaker than requiring $Z$ locally path connected. Note that if the latter is satisfied, then also $Y$ is locally path connected, hence $Y$ is path connected.
Let $y in Y$. Choose any $x in X$ and any path $v$ in $Y$ from $p(x)$ to $y$. Then $w = q circ v$ is a path in $Z$ beginning at $(q circ p)(x)$. It can be lifted to path $u$ in $X$ beginning at $x$. Both paths $v$ and $p circ u$ are lifts of $w$ beginning at $p(x)$. By unique path lifting we see that $v = p circ u$, hence $y = v(1) = p(u(1)) in p(X)$.
$endgroup$
It can be weakened. We shall use the following well-known fact about covering projections:
Given $p,q$ as in your question (i.e. $q circ p$ and $q$ are assumed to be covering projections). If $Z$ is locally connected and $p$ is a surjection, then $p$ is a covering projection.
In this result no further connectedness assumptions on $X,Y,Z$ are needed.
Hence we have to look for assumptions assuring that $p$ is surjective. To avoid trivial counterexamples, it seems that we should have the minimal requirement that $Y$ is connected (otherwise we may take $X = Z$, $Y = Z times F$ with a discrete $F$ having more than one point and $p(z) = (z,f_0)$, $q(z,f) = z$). This implies that also $Z$ is connected. For the sake of homogeneity we may moreover asume that $X$ is connected, but this is not really needed.
Therefore, let us assume that $X,Y,Z$ are connected and locally connected (it suffices to assume one of these spaces to be locally connected because they are locally homeomorphic). We shall prove
If $Y$ is path connected, then $p$ is surjective, i.e. a covering projection.
This is somewhat weaker than requiring $Z$ locally path connected. Note that if the latter is satisfied, then also $Y$ is locally path connected, hence $Y$ is path connected.
Let $y in Y$. Choose any $x in X$ and any path $v$ in $Y$ from $p(x)$ to $y$. Then $w = q circ v$ is a path in $Z$ beginning at $(q circ p)(x)$. It can be lifted to path $u$ in $X$ beginning at $x$. Both paths $v$ and $p circ u$ are lifts of $w$ beginning at $p(x)$. By unique path lifting we see that $v = p circ u$, hence $y = v(1) = p(u(1)) in p(X)$.
edited Dec 29 '18 at 15:58
answered Dec 29 '18 at 15:41
Paul FrostPaul Frost
11.2k3934
11.2k3934
add a comment |
add a comment |
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.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3055199%2fpx-to-y-covering-space-if-q-circ-px-to-z-and-qy-to-z-covering-spaces-an%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
$begingroup$
You need some additional assumptions. Probably the best setting is to require that $X,Y,Z$ are connected. Otherwise you get trivial counterexamples like $X = Z$ and $Y= Z times {0,1 }$ with $p(z) = (z,0)$ and $q(z,i) = z$.
$endgroup$
– Paul Frost
Dec 29 '18 at 14:56