Mixing
representing covering space by permutations
$begingroup$
When I read the section of representing covering space by permutations of Hatcher's Algebraic Topology,he points that:
n-sheeted covering spaces of $X$(path-connected,locallypath-connected,semilocally simply connected) are classified by equivalence classes(cojugate class) of homomorphism $pi_1(X,x_0) to Sigma_n$,where $Sigma_n$ is n-symmetric group.
Then does it imply that given a homomorphism $pi_1(X,x_0) to Sigma_n$,there is corresponding covering space?If so,how to deal with it?I can't find it in the book.
algebraic-topology
asked Jan 27 at 15:57
Daniel XuDaniel Xu
877
$endgroup$
add a comment |
$begingroup$
When I read the section of representing covering space by permutations of Hatcher's Algebraic Topology,he points that:
n-sheeted covering spaces of $X$(path-connected,locallypath-connected,semilocally simply connected) are classified by equivalence classes(cojugate class) of homomorphism $pi_1(X,x_0) to Sigma_n$,where $Sigma_n$ is n-symmetric group.
Then does it imply that given a homomorphism $pi_1(X,x_0) to Sigma_n$,there is corresponding covering space?If so,how to deal with it?I can't find it in the book.
algebraic-topology
asked Jan 27 at 15:57
Daniel XuDaniel Xu
877
$endgroup$
add a comment |
$begingroup$
When I read the section of representing covering space by permutations of Hatcher's Algebraic Topology,he points that:
n-sheeted covering spaces of $X$(path-connected,locallypath-connected,semilocally simply connected) are classified by equivalence classes(cojugate class) of homomorphism $pi_1(X,x_0) to Sigma_n$,where $Sigma_n$ is n-symmetric group.
Then does it imply that given a homomorphism $pi_1(X,x_0) to Sigma_n$,there is corresponding covering space?If so,how to deal with it?I can't find it in the book.
algebraic-topology
asked Jan 27 at 15:57
Daniel XuDaniel Xu
877
$endgroup$
When I read the section of representing covering space by permutations of Hatcher's Algebraic Topology,he points that:
n-sheeted covering spaces of $X$(path-connected,locallypath-connected,semilocally simply connected) are classified by equivalence classes(cojugate class) of homomorphism $pi_1(X,x_0) to Sigma_n$,where $Sigma_n$ is n-symmetric group.
Then does it imply that given a homomorphism $pi_1(X,x_0) to Sigma_n$,there is corresponding covering space?If so,how to deal with it?I can't find it in the book.
algebraic-topology
algebraic-topology
asked Jan 27 at 15:57
Daniel XuDaniel Xu
877
asked Jan 27 at 15:57
Daniel XuDaniel Xu
877
asked Jan 27 at 15:57
Daniel XuDaniel Xu
877
asked Jan 27 at 15:57
Daniel XuDaniel Xu
877
asked Jan 27 at 15:57
Daniel XuDaniel Xu
877
877
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Consider a representation $h:pi_1(M)rightarrow Sigma_n$ and $U_n={1,...,n}$ the following $n$-cover can be associated: the quotient of $hat Xtimes U_n$ by the diagonal action of $pi_1(M)$ where $hat X$ is the universal cover of $X$.
Conversely given a $n$-cover $p:Nrightarrow M$ you can associate to it its holonomy obtained by the action of $pi_1(M)$ on the fibre of any element of $M$, these actions are conjugate.
answered Jan 27 at 16:34
Tsemo AristideTsemo Aristide
59.8k11446
$endgroup$
$begingroup$
It's worth pointing out that this is an instance of the "Borel construction": given a group $G$ and two $G$-spaces $X$, $Y$, their Borel construction is $Xtimes Y/sim$ where $(x,y)sim(xg, gy)$ (often people require $X$ to be a right $G$-space and $Y$ a left $G$-space for notation's sake). A typical situation is when $X$ is a principal $G$-bundle and $Y$ is an arbitrary $G$-space, and the effect of the construction is "changing the fibres" of $X$ from $G$ to $Y$. In our case the principal $pi_1M$ bundle is $hat{X}$ and the new fibres are $Y = U_n$.
$endgroup$
– William
Jan 27 at 17:23
add a comment |
$begingroup$
One (maybe roundabout) way to see this is with classifying spaces and homotopy theory, at least when $X$ is a CW-space. (You will encounter all of these ideas in Hatcher at some point if you have not already.) Tsemo Aristide's answer is certainly cleaner and works for a broader class of spaces, but the concepts here might help in other situations.
"Intuitively" the idea is that since $BSigma_n$ only has non-vanishing homotopy in degree 1, homotopy classes of maps into it only depend on homotopical information up to degree 2, and in particular homotopy classes of maps to $BSigma_n$ are the same for spaces that have the same $2$-skeleton. Then you have to know that $X$ and $Bpi_1X$ have the same $2$-skeleton, and about how group homomorphisms correspond to maps between classifying spaces. I will elaborate.
An $n$-sheeted covering is the same thing as a fibre bundle whose fibre is a set of cardinality $n$ and whose structure group is $Sigma_n$; therefore they are classified by homotopy classes of maps $[X, BSigma_n]$ where $Bcolon Grp to Top$ is a classifying space functor. Since $Sigma_n$ is a discrete group it follows that $BSigma_n sim K(Sigma_n, 1)$, the "Eilenberge-Maclane space" defined up to homotopy equivalence by the properties $pi_1K(Sigma_n, 1)cong Sigma_n$ and $pi_iK(Sigma_n, 1)=0$ for other values of $i$. Since the higher homotopy groups of $BSigma_n$ all vanish, a result from obstruction theory is that
$[X,BSigma_n] cong [X^{(2)}, BSigma_n]$
where $X^{(2)}$ is the $2$-skeleton of $X$. That is, the (isomorphism class of the) covering space space over $X$ is determined by its restriction to the $2$-skeleton.
Here's where things get a bit funny: the $2$-skeleton of $X$ is also the $2$-skeleton of a model of $Bpi_1 X$. This is because since $pi_1X$ is discrete its classifying space is again an Eilenberg-Maclane space in degree 1, so we can construct a CW model from a group presentation of $pi_1X$ by taking a 1-cell for every generator and attaching 2-cell along every relation, and then adding higher-dimensional cells to kill off any higher homotopy we may have introduced. But the $2$-skeleton of $X$ determines a presentation of $pi_1X$ so it is also the $2$-skeleton of the Eilenberg-Maclane construction. Therefore we get
$[X,BSigma_n] cong [X^{(2)}, BSigma_n] = [(Bpi_1X)^{(2)}, BSigma_n]cong [Bpi_1 X, BSigma_n]$
Now the last step is to establish the correspondence between $[BG, BH]$ and conjugacy classes of homomorphisms $Gto H$. I will see if I can remember a clean way of showing this and make an edit later... Again, I believe it is also in Hatcher.
answered Jan 27 at 17:14
WilliamWilliam
2,8201224
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Consider a representation $h:pi_1(M)rightarrow Sigma_n$ and $U_n={1,...,n}$ the following $n$-cover can be associated: the quotient of $hat Xtimes U_n$ by the diagonal action of $pi_1(M)$ where $hat X$ is the universal cover of $X$.
Conversely given a $n$-cover $p:Nrightarrow M$ you can associate to it its holonomy obtained by the action of $pi_1(M)$ on the fibre of any element of $M$, these actions are conjugate.
answered Jan 27 at 16:34
Tsemo AristideTsemo Aristide
59.8k11446
$endgroup$
$begingroup$
It's worth pointing out that this is an instance of the "Borel construction": given a group $G$ and two $G$-spaces $X$, $Y$, their Borel construction is $Xtimes Y/sim$ where $(x,y)sim(xg, gy)$ (often people require $X$ to be a right $G$-space and $Y$ a left $G$-space for notation's sake). A typical situation is when $X$ is a principal $G$-bundle and $Y$ is an arbitrary $G$-space, and the effect of the construction is "changing the fibres" of $X$ from $G$ to $Y$. In our case the principal $pi_1M$ bundle is $hat{X}$ and the new fibres are $Y = U_n$.
$endgroup$
– William
Jan 27 at 17:23
add a comment |
$begingroup$
Consider a representation $h:pi_1(M)rightarrow Sigma_n$ and $U_n={1,...,n}$ the following $n$-cover can be associated: the quotient of $hat Xtimes U_n$ by the diagonal action of $pi_1(M)$ where $hat X$ is the universal cover of $X$.
Conversely given a $n$-cover $p:Nrightarrow M$ you can associate to it its holonomy obtained by the action of $pi_1(M)$ on the fibre of any element of $M$, these actions are conjugate.
answered Jan 27 at 16:34
Tsemo AristideTsemo Aristide
59.8k11446
$endgroup$
$begingroup$
It's worth pointing out that this is an instance of the "Borel construction": given a group $G$ and two $G$-spaces $X$, $Y$, their Borel construction is $Xtimes Y/sim$ where $(x,y)sim(xg, gy)$ (often people require $X$ to be a right $G$-space and $Y$ a left $G$-space for notation's sake). A typical situation is when $X$ is a principal $G$-bundle and $Y$ is an arbitrary $G$-space, and the effect of the construction is "changing the fibres" of $X$ from $G$ to $Y$. In our case the principal $pi_1M$ bundle is $hat{X}$ and the new fibres are $Y = U_n$.
$endgroup$
– William
Jan 27 at 17:23
add a comment |
$begingroup$
Consider a representation $h:pi_1(M)rightarrow Sigma_n$ and $U_n={1,...,n}$ the following $n$-cover can be associated: the quotient of $hat Xtimes U_n$ by the diagonal action of $pi_1(M)$ where $hat X$ is the universal cover of $X$.
Conversely given a $n$-cover $p:Nrightarrow M$ you can associate to it its holonomy obtained by the action of $pi_1(M)$ on the fibre of any element of $M$, these actions are conjugate.
answered Jan 27 at 16:34
Tsemo AristideTsemo Aristide
59.8k11446
$endgroup$
Consider a representation $h:pi_1(M)rightarrow Sigma_n$ and $U_n={1,...,n}$ the following $n$-cover can be associated: the quotient of $hat Xtimes U_n$ by the diagonal action of $pi_1(M)$ where $hat X$ is the universal cover of $X$.
Conversely given a $n$-cover $p:Nrightarrow M$ you can associate to it its holonomy obtained by the action of $pi_1(M)$ on the fibre of any element of $M$, these actions are conjugate.
answered Jan 27 at 16:34
Tsemo AristideTsemo Aristide
59.8k11446
answered Jan 27 at 16:34
Tsemo AristideTsemo Aristide
59.8k11446
answered Jan 27 at 16:34
Tsemo AristideTsemo Aristide
59.8k11446
answered Jan 27 at 16:34
Tsemo AristideTsemo Aristide
59.8k11446
59.8k11446
$begingroup$
It's worth pointing out that this is an instance of the "Borel construction": given a group $G$ and two $G$-spaces $X$, $Y$, their Borel construction is $Xtimes Y/sim$ where $(x,y)sim(xg, gy)$ (often people require $X$ to be a right $G$-space and $Y$ a left $G$-space for notation's sake). A typical situation is when $X$ is a principal $G$-bundle and $Y$ is an arbitrary $G$-space, and the effect of the construction is "changing the fibres" of $X$ from $G$ to $Y$. In our case the principal $pi_1M$ bundle is $hat{X}$ and the new fibres are $Y = U_n$.
$endgroup$
– William
Jan 27 at 17:23
add a comment |
$begingroup$
It's worth pointing out that this is an instance of the "Borel construction": given a group $G$ and two $G$-spaces $X$, $Y$, their Borel construction is $Xtimes Y/sim$ where $(x,y)sim(xg, gy)$ (often people require $X$ to be a right $G$-space and $Y$ a left $G$-space for notation's sake). A typical situation is when $X$ is a principal $G$-bundle and $Y$ is an arbitrary $G$-space, and the effect of the construction is "changing the fibres" of $X$ from $G$ to $Y$. In our case the principal $pi_1M$ bundle is $hat{X}$ and the new fibres are $Y = U_n$.
$endgroup$
– William
Jan 27 at 17:23
$begingroup$
It's worth pointing out that this is an instance of the "Borel construction": given a group $G$ and two $G$-spaces $X$, $Y$, their Borel construction is $Xtimes Y/sim$ where $(x,y)sim(xg, gy)$ (often people require $X$ to be a right $G$-space and $Y$ a left $G$-space for notation's sake). A typical situation is when $X$ is a principal $G$-bundle and $Y$ is an arbitrary $G$-space, and the effect of the construction is "changing the fibres" of $X$ from $G$ to $Y$. In our case the principal $pi_1M$ bundle is $hat{X}$ and the new fibres are $Y = U_n$.
$endgroup$
– William
Jan 27 at 17:23
$begingroup$
It's worth pointing out that this is an instance of the "Borel construction": given a group $G$ and two $G$-spaces $X$, $Y$, their Borel construction is $Xtimes Y/sim$ where $(x,y)sim(xg, gy)$ (often people require $X$ to be a right $G$-space and $Y$ a left $G$-space for notation's sake). A typical situation is when $X$ is a principal $G$-bundle and $Y$ is an arbitrary $G$-space, and the effect of the construction is "changing the fibres" of $X$ from $G$ to $Y$. In our case the principal $pi_1M$ bundle is $hat{X}$ and the new fibres are $Y = U_n$.
$endgroup$
– William
Jan 27 at 17:23
add a comment |
$begingroup$
One (maybe roundabout) way to see this is with classifying spaces and homotopy theory, at least when $X$ is a CW-space. (You will encounter all of these ideas in Hatcher at some point if you have not already.) Tsemo Aristide's answer is certainly cleaner and works for a broader class of spaces, but the concepts here might help in other situations.
"Intuitively" the idea is that since $BSigma_n$ only has non-vanishing homotopy in degree 1, homotopy classes of maps into it only depend on homotopical information up to degree 2, and in particular homotopy classes of maps to $BSigma_n$ are the same for spaces that have the same $2$-skeleton. Then you have to know that $X$ and $Bpi_1X$ have the same $2$-skeleton, and about how group homomorphisms correspond to maps between classifying spaces. I will elaborate.
An $n$-sheeted covering is the same thing as a fibre bundle whose fibre is a set of cardinality $n$ and whose structure group is $Sigma_n$; therefore they are classified by homotopy classes of maps $[X, BSigma_n]$ where $Bcolon Grp to Top$ is a classifying space functor. Since $Sigma_n$ is a discrete group it follows that $BSigma_n sim K(Sigma_n, 1)$, the "Eilenberge-Maclane space" defined up to homotopy equivalence by the properties $pi_1K(Sigma_n, 1)cong Sigma_n$ and $pi_iK(Sigma_n, 1)=0$ for other values of $i$. Since the higher homotopy groups of $BSigma_n$ all vanish, a result from obstruction theory is that
$[X,BSigma_n] cong [X^{(2)}, BSigma_n]$
where $X^{(2)}$ is the $2$-skeleton of $X$. That is, the (isomorphism class of the) covering space space over $X$ is determined by its restriction to the $2$-skeleton.
Here's where things get a bit funny: the $2$-skeleton of $X$ is also the $2$-skeleton of a model of $Bpi_1 X$. This is because since $pi_1X$ is discrete its classifying space is again an Eilenberg-Maclane space in degree 1, so we can construct a CW model from a group presentation of $pi_1X$ by taking a 1-cell for every generator and attaching 2-cell along every relation, and then adding higher-dimensional cells to kill off any higher homotopy we may have introduced. But the $2$-skeleton of $X$ determines a presentation of $pi_1X$ so it is also the $2$-skeleton of the Eilenberg-Maclane construction. Therefore we get
$[X,BSigma_n] cong [X^{(2)}, BSigma_n] = [(Bpi_1X)^{(2)}, BSigma_n]cong [Bpi_1 X, BSigma_n]$
Now the last step is to establish the correspondence between $[BG, BH]$ and conjugacy classes of homomorphisms $Gto H$. I will see if I can remember a clean way of showing this and make an edit later... Again, I believe it is also in Hatcher.
answered Jan 27 at 17:14
WilliamWilliam
2,8201224
$endgroup$
add a comment |
$begingroup$
One (maybe roundabout) way to see this is with classifying spaces and homotopy theory, at least when $X$ is a CW-space. (You will encounter all of these ideas in Hatcher at some point if you have not already.) Tsemo Aristide's answer is certainly cleaner and works for a broader class of spaces, but the concepts here might help in other situations.
"Intuitively" the idea is that since $BSigma_n$ only has non-vanishing homotopy in degree 1, homotopy classes of maps into it only depend on homotopical information up to degree 2, and in particular homotopy classes of maps to $BSigma_n$ are the same for spaces that have the same $2$-skeleton. Then you have to know that $X$ and $Bpi_1X$ have the same $2$-skeleton, and about how group homomorphisms correspond to maps between classifying spaces. I will elaborate.
An $n$-sheeted covering is the same thing as a fibre bundle whose fibre is a set of cardinality $n$ and whose structure group is $Sigma_n$; therefore they are classified by homotopy classes of maps $[X, BSigma_n]$ where $Bcolon Grp to Top$ is a classifying space functor. Since $Sigma_n$ is a discrete group it follows that $BSigma_n sim K(Sigma_n, 1)$, the "Eilenberge-Maclane space" defined up to homotopy equivalence by the properties $pi_1K(Sigma_n, 1)cong Sigma_n$ and $pi_iK(Sigma_n, 1)=0$ for other values of $i$. Since the higher homotopy groups of $BSigma_n$ all vanish, a result from obstruction theory is that
$[X,BSigma_n] cong [X^{(2)}, BSigma_n]$
where $X^{(2)}$ is the $2$-skeleton of $X$. That is, the (isomorphism class of the) covering space space over $X$ is determined by its restriction to the $2$-skeleton.
Here's where things get a bit funny: the $2$-skeleton of $X$ is also the $2$-skeleton of a model of $Bpi_1 X$. This is because since $pi_1X$ is discrete its classifying space is again an Eilenberg-Maclane space in degree 1, so we can construct a CW model from a group presentation of $pi_1X$ by taking a 1-cell for every generator and attaching 2-cell along every relation, and then adding higher-dimensional cells to kill off any higher homotopy we may have introduced. But the $2$-skeleton of $X$ determines a presentation of $pi_1X$ so it is also the $2$-skeleton of the Eilenberg-Maclane construction. Therefore we get
$[X,BSigma_n] cong [X^{(2)}, BSigma_n] = [(Bpi_1X)^{(2)}, BSigma_n]cong [Bpi_1 X, BSigma_n]$
Now the last step is to establish the correspondence between $[BG, BH]$ and conjugacy classes of homomorphisms $Gto H$. I will see if I can remember a clean way of showing this and make an edit later... Again, I believe it is also in Hatcher.
answered Jan 27 at 17:14
WilliamWilliam
2,8201224
$endgroup$
add a comment |
$begingroup$
One (maybe roundabout) way to see this is with classifying spaces and homotopy theory, at least when $X$ is a CW-space. (You will encounter all of these ideas in Hatcher at some point if you have not already.) Tsemo Aristide's answer is certainly cleaner and works for a broader class of spaces, but the concepts here might help in other situations.
"Intuitively" the idea is that since $BSigma_n$ only has non-vanishing homotopy in degree 1, homotopy classes of maps into it only depend on homotopical information up to degree 2, and in particular homotopy classes of maps to $BSigma_n$ are the same for spaces that have the same $2$-skeleton. Then you have to know that $X$ and $Bpi_1X$ have the same $2$-skeleton, and about how group homomorphisms correspond to maps between classifying spaces. I will elaborate.
An $n$-sheeted covering is the same thing as a fibre bundle whose fibre is a set of cardinality $n$ and whose structure group is $Sigma_n$; therefore they are classified by homotopy classes of maps $[X, BSigma_n]$ where $Bcolon Grp to Top$ is a classifying space functor. Since $Sigma_n$ is a discrete group it follows that $BSigma_n sim K(Sigma_n, 1)$, the "Eilenberge-Maclane space" defined up to homotopy equivalence by the properties $pi_1K(Sigma_n, 1)cong Sigma_n$ and $pi_iK(Sigma_n, 1)=0$ for other values of $i$. Since the higher homotopy groups of $BSigma_n$ all vanish, a result from obstruction theory is that
$[X,BSigma_n] cong [X^{(2)}, BSigma_n]$
where $X^{(2)}$ is the $2$-skeleton of $X$. That is, the (isomorphism class of the) covering space space over $X$ is determined by its restriction to the $2$-skeleton.
Here's where things get a bit funny: the $2$-skeleton of $X$ is also the $2$-skeleton of a model of $Bpi_1 X$. This is because since $pi_1X$ is discrete its classifying space is again an Eilenberg-Maclane space in degree 1, so we can construct a CW model from a group presentation of $pi_1X$ by taking a 1-cell for every generator and attaching 2-cell along every relation, and then adding higher-dimensional cells to kill off any higher homotopy we may have introduced. But the $2$-skeleton of $X$ determines a presentation of $pi_1X$ so it is also the $2$-skeleton of the Eilenberg-Maclane construction. Therefore we get
$[X,BSigma_n] cong [X^{(2)}, BSigma_n] = [(Bpi_1X)^{(2)}, BSigma_n]cong [Bpi_1 X, BSigma_n]$
Now the last step is to establish the correspondence between $[BG, BH]$ and conjugacy classes of homomorphisms $Gto H$. I will see if I can remember a clean way of showing this and make an edit later... Again, I believe it is also in Hatcher.
answered Jan 27 at 17:14
WilliamWilliam
2,8201224
$endgroup$
One (maybe roundabout) way to see this is with classifying spaces and homotopy theory, at least when $X$ is a CW-space. (You will encounter all of these ideas in Hatcher at some point if you have not already.) Tsemo Aristide's answer is certainly cleaner and works for a broader class of spaces, but the concepts here might help in other situations.
"Intuitively" the idea is that since $BSigma_n$ only has non-vanishing homotopy in degree 1, homotopy classes of maps into it only depend on homotopical information up to degree 2, and in particular homotopy classes of maps to $BSigma_n$ are the same for spaces that have the same $2$-skeleton. Then you have to know that $X$ and $Bpi_1X$ have the same $2$-skeleton, and about how group homomorphisms correspond to maps between classifying spaces. I will elaborate.
An $n$-sheeted covering is the same thing as a fibre bundle whose fibre is a set of cardinality $n$ and whose structure group is $Sigma_n$; therefore they are classified by homotopy classes of maps $[X, BSigma_n]$ where $Bcolon Grp to Top$ is a classifying space functor. Since $Sigma_n$ is a discrete group it follows that $BSigma_n sim K(Sigma_n, 1)$, the "Eilenberge-Maclane space" defined up to homotopy equivalence by the properties $pi_1K(Sigma_n, 1)cong Sigma_n$ and $pi_iK(Sigma_n, 1)=0$ for other values of $i$. Since the higher homotopy groups of $BSigma_n$ all vanish, a result from obstruction theory is that
$[X,BSigma_n] cong [X^{(2)}, BSigma_n]$
where $X^{(2)}$ is the $2$-skeleton of $X$. That is, the (isomorphism class of the) covering space space over $X$ is determined by its restriction to the $2$-skeleton.
Here's where things get a bit funny: the $2$-skeleton of $X$ is also the $2$-skeleton of a model of $Bpi_1 X$. This is because since $pi_1X$ is discrete its classifying space is again an Eilenberg-Maclane space in degree 1, so we can construct a CW model from a group presentation of $pi_1X$ by taking a 1-cell for every generator and attaching 2-cell along every relation, and then adding higher-dimensional cells to kill off any higher homotopy we may have introduced. But the $2$-skeleton of $X$ determines a presentation of $pi_1X$ so it is also the $2$-skeleton of the Eilenberg-Maclane construction. Therefore we get
$[X,BSigma_n] cong [X^{(2)}, BSigma_n] = [(Bpi_1X)^{(2)}, BSigma_n]cong [Bpi_1 X, BSigma_n]$
Now the last step is to establish the correspondence between $[BG, BH]$ and conjugacy classes of homomorphisms $Gto H$. I will see if I can remember a clean way of showing this and make an edit later... Again, I believe it is also in Hatcher.
answered Jan 27 at 17:14
WilliamWilliam
2,8201224
answered Jan 27 at 17:14
WilliamWilliam
2,8201224
answered Jan 27 at 17:14
WilliamWilliam
2,8201224
answered Jan 27 at 17:14
WilliamWilliam
2,8201224
2,8201224
add a comment |
add a comment |
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StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
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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
