Mixing
Cover of a simply connected and lpc space is trivial
$begingroup$
I have I question about a reduction step in the proof of the statement that a cover of simply connected and locally path-connected space is trivial (source: "Fundamental Groups and Galois Groups" by Szamuely, Tamás; page 40):
There is said that it's enough to show that for a connected cover $p: Y to X$ the lemma is injective.
My question is why is it enough to show this statement to conclude already lemma 2.4.4?
Remark: I know that there are also ways to show 2.4.4 directly but the point of my interests is to understand why it suffice to show the second statement or in other words why the second statement already imply lemma 2.4.4?
general-topology covering-spaces
edited Jan 10 at 0:47
Paul Frost
10.3k3933
asked Jan 10 at 0:08
KarlPeterKarlPeter
5461315
$endgroup$
add a comment |
$begingroup$
I have I question about a reduction step in the proof of the statement that a cover of simply connected and locally path-connected space is trivial (source: "Fundamental Groups and Galois Groups" by Szamuely, Tamás; page 40):
There is said that it's enough to show that for a connected cover $p: Y to X$ the lemma is injective.
My question is why is it enough to show this statement to conclude already lemma 2.4.4?
Remark: I know that there are also ways to show 2.4.4 directly but the point of my interests is to understand why it suffice to show the second statement or in other words why the second statement already imply lemma 2.4.4?
general-topology covering-spaces
edited Jan 10 at 0:47
Paul Frost
10.3k3933
asked Jan 10 at 0:08
KarlPeterKarlPeter
5461315
$endgroup$
add a comment |
$begingroup$
I have I question about a reduction step in the proof of the statement that a cover of simply connected and locally path-connected space is trivial (source: "Fundamental Groups and Galois Groups" by Szamuely, Tamás; page 40):
There is said that it's enough to show that for a connected cover $p: Y to X$ the lemma is injective.
My question is why is it enough to show this statement to conclude already lemma 2.4.4?
Remark: I know that there are also ways to show 2.4.4 directly but the point of my interests is to understand why it suffice to show the second statement or in other words why the second statement already imply lemma 2.4.4?
general-topology covering-spaces
edited Jan 10 at 0:47
Paul Frost
10.3k3933
asked Jan 10 at 0:08
KarlPeterKarlPeter
5461315
$endgroup$
I have I question about a reduction step in the proof of the statement that a cover of simply connected and locally path-connected space is trivial (source: "Fundamental Groups and Galois Groups" by Szamuely, Tamás; page 40):
There is said that it's enough to show that for a connected cover $p: Y to X$ the lemma is injective.
My question is why is it enough to show this statement to conclude already lemma 2.4.4?
Remark: I know that there are also ways to show 2.4.4 directly but the point of my interests is to understand why it suffice to show the second statement or in other words why the second statement already imply lemma 2.4.4?
general-topology covering-spaces
general-topology covering-spaces
edited Jan 10 at 0:47
Paul Frost
10.3k3933
asked Jan 10 at 0:08
KarlPeterKarlPeter
5461315
edited Jan 10 at 0:47
Paul Frost
10.3k3933
asked Jan 10 at 0:08
KarlPeterKarlPeter
5461315
edited Jan 10 at 0:47
Paul Frost
10.3k3933
edited Jan 10 at 0:47
Paul Frost
10.3k3933
edited Jan 10 at 0:47
Paul Frost
10.3k3933
10.3k3933
asked Jan 10 at 0:08
KarlPeterKarlPeter
5461315
asked Jan 10 at 0:08
KarlPeterKarlPeter
5461315
asked Jan 10 at 0:08
KarlPeterKarlPeter
5461315
5461315
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $p : Y to X$ be a covering, where $X$ is simply connected and lpc. Then $Y$ is lpc because $p$ is a local homeomorphism. Hence all path components $Y_alpha$ of $Y$ are open subsets of $Y$.
Moreover, the restrictions $p_alpha : Y_alpha to X$ are coverings.
1) $p_alpha$ is surjective.
Let $x in X$. Choose any $y in Y_alpha$ (which is possible because $Y_alpha ne emptyset$). Let $u : [0,1] to X$ be a path from $p(y)$ to $x$. Then $u$ can be lifted to a path $v : [0,1] to Y$ such that $v(0) = y$. Clearly $p(v(1)) = x$. But we must have $v([0,1]) subset Y_alpha$ since $Y_alpha$ is the path component of $Y$ which contains $y = v(0)$.
2) $p_alpha$ is a covering.
Let $x in X$ and $U$ be an open path connected neighborhood of $x$ which is evenly covered. Then $p^{-1}(U) = bigcup_beta U_beta$ with pairwise disjoint open $U_beta subset Y$ which are mapped by $p$ homeomorphically onto $U$. Each $U_beta$ is path connected, thus contained in a unique $Y_{f(beta)}$. Hence $U_beta cap Y_alpha = U_beta$ if $f(beta) = alpha$ and $U_beta cap Y_alpha = emptyset$ if $f(beta) ne alpha$. This shows that
$$p_alpha^{-1}(U) = p^{-1}(U) cap Y_alpha = left(bigcup_beta U_beta right) cap Y_alpha = bigcup_beta (U_beta cap Y_alpha) = bigcup_{beta text{ with } f(beta) = alpha} U_beta .$$
Therefore $U$ is evenly covered by $p_alpha$. Note that the $U_beta$ with $f(beta) = alpha$ are pairwise disjoint open subsets of $Y_alpha$ which are mapped by $p_alpha = p mid_{Y_alpha}$ homeomorphically onto $U$.
If we can prove what is claimed to be enough, we know the all connected coverings of $X$ are continuous, open (as coverings!) and bijective, i.e. are homeomorphisms. Now apply this to the $p_alpha$ and deduce that $X$ itself is evenly covered.
answered Jan 10 at 1:10
Paul FrostPaul Frost
10.3k3933
$endgroup$
$begingroup$
Hi, thank you for you answer. One point is unclear for me: When you show that $p_alpha : Y_alpha to X$ are coverings how do you see that $p_alpha^{-1}(U) = bigcup_{beta text{ with } f(beta) = alpha} U_beta$ isn't empty? Therefore why there exist a $beta$ with $f(beta) = alpha$ and $#{beta vert f(beta) = alpha}$ is constant. Therefore independent of the point $x$ for which we take an open pc neighbourhood $U =U_x$?
$endgroup$
– KarlPeter
Jan 10 at 1:49
$begingroup$
You are right, my proof contains a gap. I forgot to show that the $p_alpha$ are surjective. I shall edit my answer.
$endgroup$
– Paul Frost
Jan 10 at 12:24
$begingroup$
Note that $p_alpha^{-1}(U)$ is non-empty because $p_alpha$ is surjective.
$endgroup$
– Paul Frost
Jan 10 at 13:00
add a comment |
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$begingroup$
Let $p : Y to X$ be a covering, where $X$ is simply connected and lpc. Then $Y$ is lpc because $p$ is a local homeomorphism. Hence all path components $Y_alpha$ of $Y$ are open subsets of $Y$.
Moreover, the restrictions $p_alpha : Y_alpha to X$ are coverings.
1) $p_alpha$ is surjective.
Let $x in X$. Choose any $y in Y_alpha$ (which is possible because $Y_alpha ne emptyset$). Let $u : [0,1] to X$ be a path from $p(y)$ to $x$. Then $u$ can be lifted to a path $v : [0,1] to Y$ such that $v(0) = y$. Clearly $p(v(1)) = x$. But we must have $v([0,1]) subset Y_alpha$ since $Y_alpha$ is the path component of $Y$ which contains $y = v(0)$.
2) $p_alpha$ is a covering.
Let $x in X$ and $U$ be an open path connected neighborhood of $x$ which is evenly covered. Then $p^{-1}(U) = bigcup_beta U_beta$ with pairwise disjoint open $U_beta subset Y$ which are mapped by $p$ homeomorphically onto $U$. Each $U_beta$ is path connected, thus contained in a unique $Y_{f(beta)}$. Hence $U_beta cap Y_alpha = U_beta$ if $f(beta) = alpha$ and $U_beta cap Y_alpha = emptyset$ if $f(beta) ne alpha$. This shows that
$$p_alpha^{-1}(U) = p^{-1}(U) cap Y_alpha = left(bigcup_beta U_beta right) cap Y_alpha = bigcup_beta (U_beta cap Y_alpha) = bigcup_{beta text{ with } f(beta) = alpha} U_beta .$$
Therefore $U$ is evenly covered by $p_alpha$. Note that the $U_beta$ with $f(beta) = alpha$ are pairwise disjoint open subsets of $Y_alpha$ which are mapped by $p_alpha = p mid_{Y_alpha}$ homeomorphically onto $U$.
If we can prove what is claimed to be enough, we know the all connected coverings of $X$ are continuous, open (as coverings!) and bijective, i.e. are homeomorphisms. Now apply this to the $p_alpha$ and deduce that $X$ itself is evenly covered.
answered Jan 10 at 1:10
Paul FrostPaul Frost
10.3k3933
$endgroup$
$begingroup$
Hi, thank you for you answer. One point is unclear for me: When you show that $p_alpha : Y_alpha to X$ are coverings how do you see that $p_alpha^{-1}(U) = bigcup_{beta text{ with } f(beta) = alpha} U_beta$ isn't empty? Therefore why there exist a $beta$ with $f(beta) = alpha$ and $#{beta vert f(beta) = alpha}$ is constant. Therefore independent of the point $x$ for which we take an open pc neighbourhood $U =U_x$?
$endgroup$
– KarlPeter
Jan 10 at 1:49
$begingroup$
You are right, my proof contains a gap. I forgot to show that the $p_alpha$ are surjective. I shall edit my answer.
$endgroup$
– Paul Frost
Jan 10 at 12:24
$begingroup$
Note that $p_alpha^{-1}(U)$ is non-empty because $p_alpha$ is surjective.
$endgroup$
– Paul Frost
Jan 10 at 13:00
add a comment |
$begingroup$
Let $p : Y to X$ be a covering, where $X$ is simply connected and lpc. Then $Y$ is lpc because $p$ is a local homeomorphism. Hence all path components $Y_alpha$ of $Y$ are open subsets of $Y$.
Moreover, the restrictions $p_alpha : Y_alpha to X$ are coverings.
1) $p_alpha$ is surjective.
Let $x in X$. Choose any $y in Y_alpha$ (which is possible because $Y_alpha ne emptyset$). Let $u : [0,1] to X$ be a path from $p(y)$ to $x$. Then $u$ can be lifted to a path $v : [0,1] to Y$ such that $v(0) = y$. Clearly $p(v(1)) = x$. But we must have $v([0,1]) subset Y_alpha$ since $Y_alpha$ is the path component of $Y$ which contains $y = v(0)$.
2) $p_alpha$ is a covering.
Let $x in X$ and $U$ be an open path connected neighborhood of $x$ which is evenly covered. Then $p^{-1}(U) = bigcup_beta U_beta$ with pairwise disjoint open $U_beta subset Y$ which are mapped by $p$ homeomorphically onto $U$. Each $U_beta$ is path connected, thus contained in a unique $Y_{f(beta)}$. Hence $U_beta cap Y_alpha = U_beta$ if $f(beta) = alpha$ and $U_beta cap Y_alpha = emptyset$ if $f(beta) ne alpha$. This shows that
$$p_alpha^{-1}(U) = p^{-1}(U) cap Y_alpha = left(bigcup_beta U_beta right) cap Y_alpha = bigcup_beta (U_beta cap Y_alpha) = bigcup_{beta text{ with } f(beta) = alpha} U_beta .$$
Therefore $U$ is evenly covered by $p_alpha$. Note that the $U_beta$ with $f(beta) = alpha$ are pairwise disjoint open subsets of $Y_alpha$ which are mapped by $p_alpha = p mid_{Y_alpha}$ homeomorphically onto $U$.
If we can prove what is claimed to be enough, we know the all connected coverings of $X$ are continuous, open (as coverings!) and bijective, i.e. are homeomorphisms. Now apply this to the $p_alpha$ and deduce that $X$ itself is evenly covered.
answered Jan 10 at 1:10
Paul FrostPaul Frost
10.3k3933
$endgroup$
$begingroup$
Hi, thank you for you answer. One point is unclear for me: When you show that $p_alpha : Y_alpha to X$ are coverings how do you see that $p_alpha^{-1}(U) = bigcup_{beta text{ with } f(beta) = alpha} U_beta$ isn't empty? Therefore why there exist a $beta$ with $f(beta) = alpha$ and $#{beta vert f(beta) = alpha}$ is constant. Therefore independent of the point $x$ for which we take an open pc neighbourhood $U =U_x$?
$endgroup$
– KarlPeter
Jan 10 at 1:49
$begingroup$
You are right, my proof contains a gap. I forgot to show that the $p_alpha$ are surjective. I shall edit my answer.
$endgroup$
– Paul Frost
Jan 10 at 12:24
$begingroup$
Note that $p_alpha^{-1}(U)$ is non-empty because $p_alpha$ is surjective.
$endgroup$
– Paul Frost
Jan 10 at 13:00
add a comment |
$begingroup$
Let $p : Y to X$ be a covering, where $X$ is simply connected and lpc. Then $Y$ is lpc because $p$ is a local homeomorphism. Hence all path components $Y_alpha$ of $Y$ are open subsets of $Y$.
Moreover, the restrictions $p_alpha : Y_alpha to X$ are coverings.
1) $p_alpha$ is surjective.
Let $x in X$. Choose any $y in Y_alpha$ (which is possible because $Y_alpha ne emptyset$). Let $u : [0,1] to X$ be a path from $p(y)$ to $x$. Then $u$ can be lifted to a path $v : [0,1] to Y$ such that $v(0) = y$. Clearly $p(v(1)) = x$. But we must have $v([0,1]) subset Y_alpha$ since $Y_alpha$ is the path component of $Y$ which contains $y = v(0)$.
2) $p_alpha$ is a covering.
Let $x in X$ and $U$ be an open path connected neighborhood of $x$ which is evenly covered. Then $p^{-1}(U) = bigcup_beta U_beta$ with pairwise disjoint open $U_beta subset Y$ which are mapped by $p$ homeomorphically onto $U$. Each $U_beta$ is path connected, thus contained in a unique $Y_{f(beta)}$. Hence $U_beta cap Y_alpha = U_beta$ if $f(beta) = alpha$ and $U_beta cap Y_alpha = emptyset$ if $f(beta) ne alpha$. This shows that
$$p_alpha^{-1}(U) = p^{-1}(U) cap Y_alpha = left(bigcup_beta U_beta right) cap Y_alpha = bigcup_beta (U_beta cap Y_alpha) = bigcup_{beta text{ with } f(beta) = alpha} U_beta .$$
Therefore $U$ is evenly covered by $p_alpha$. Note that the $U_beta$ with $f(beta) = alpha$ are pairwise disjoint open subsets of $Y_alpha$ which are mapped by $p_alpha = p mid_{Y_alpha}$ homeomorphically onto $U$.
If we can prove what is claimed to be enough, we know the all connected coverings of $X$ are continuous, open (as coverings!) and bijective, i.e. are homeomorphisms. Now apply this to the $p_alpha$ and deduce that $X$ itself is evenly covered.
answered Jan 10 at 1:10
Paul FrostPaul Frost
10.3k3933
$endgroup$
Let $p : Y to X$ be a covering, where $X$ is simply connected and lpc. Then $Y$ is lpc because $p$ is a local homeomorphism. Hence all path components $Y_alpha$ of $Y$ are open subsets of $Y$.
Moreover, the restrictions $p_alpha : Y_alpha to X$ are coverings.
1) $p_alpha$ is surjective.
Let $x in X$. Choose any $y in Y_alpha$ (which is possible because $Y_alpha ne emptyset$). Let $u : [0,1] to X$ be a path from $p(y)$ to $x$. Then $u$ can be lifted to a path $v : [0,1] to Y$ such that $v(0) = y$. Clearly $p(v(1)) = x$. But we must have $v([0,1]) subset Y_alpha$ since $Y_alpha$ is the path component of $Y$ which contains $y = v(0)$.
2) $p_alpha$ is a covering.
Let $x in X$ and $U$ be an open path connected neighborhood of $x$ which is evenly covered. Then $p^{-1}(U) = bigcup_beta U_beta$ with pairwise disjoint open $U_beta subset Y$ which are mapped by $p$ homeomorphically onto $U$. Each $U_beta$ is path connected, thus contained in a unique $Y_{f(beta)}$. Hence $U_beta cap Y_alpha = U_beta$ if $f(beta) = alpha$ and $U_beta cap Y_alpha = emptyset$ if $f(beta) ne alpha$. This shows that
$$p_alpha^{-1}(U) = p^{-1}(U) cap Y_alpha = left(bigcup_beta U_beta right) cap Y_alpha = bigcup_beta (U_beta cap Y_alpha) = bigcup_{beta text{ with } f(beta) = alpha} U_beta .$$
Therefore $U$ is evenly covered by $p_alpha$. Note that the $U_beta$ with $f(beta) = alpha$ are pairwise disjoint open subsets of $Y_alpha$ which are mapped by $p_alpha = p mid_{Y_alpha}$ homeomorphically onto $U$.
If we can prove what is claimed to be enough, we know the all connected coverings of $X$ are continuous, open (as coverings!) and bijective, i.e. are homeomorphisms. Now apply this to the $p_alpha$ and deduce that $X$ itself is evenly covered.
answered Jan 10 at 1:10
Paul FrostPaul Frost
10.3k3933
edited Jan 10 at 17:07
answered Jan 10 at 1:10
Paul FrostPaul Frost
10.3k3933
answered Jan 10 at 1:10
Paul FrostPaul Frost
10.3k3933
answered Jan 10 at 1:10
Paul FrostPaul Frost
10.3k3933
10.3k3933
$begingroup$
Hi, thank you for you answer. One point is unclear for me: When you show that $p_alpha : Y_alpha to X$ are coverings how do you see that $p_alpha^{-1}(U) = bigcup_{beta text{ with } f(beta) = alpha} U_beta$ isn't empty? Therefore why there exist a $beta$ with $f(beta) = alpha$ and $#{beta vert f(beta) = alpha}$ is constant. Therefore independent of the point $x$ for which we take an open pc neighbourhood $U =U_x$?
$endgroup$
– KarlPeter
Jan 10 at 1:49
$begingroup$
You are right, my proof contains a gap. I forgot to show that the $p_alpha$ are surjective. I shall edit my answer.
$endgroup$
– Paul Frost
Jan 10 at 12:24
$begingroup$
Note that $p_alpha^{-1}(U)$ is non-empty because $p_alpha$ is surjective.
$endgroup$
– Paul Frost
Jan 10 at 13:00
add a comment |
$begingroup$
Hi, thank you for you answer. One point is unclear for me: When you show that $p_alpha : Y_alpha to X$ are coverings how do you see that $p_alpha^{-1}(U) = bigcup_{beta text{ with } f(beta) = alpha} U_beta$ isn't empty? Therefore why there exist a $beta$ with $f(beta) = alpha$ and $#{beta vert f(beta) = alpha}$ is constant. Therefore independent of the point $x$ for which we take an open pc neighbourhood $U =U_x$?
$endgroup$
– KarlPeter
Jan 10 at 1:49
$begingroup$
You are right, my proof contains a gap. I forgot to show that the $p_alpha$ are surjective. I shall edit my answer.
$endgroup$
– Paul Frost
Jan 10 at 12:24
$begingroup$
Note that $p_alpha^{-1}(U)$ is non-empty because $p_alpha$ is surjective.
$endgroup$
– Paul Frost
Jan 10 at 13:00
$begingroup$
Hi, thank you for you answer. One point is unclear for me: When you show that $p_alpha : Y_alpha to X$ are coverings how do you see that $p_alpha^{-1}(U) = bigcup_{beta text{ with } f(beta) = alpha} U_beta$ isn't empty? Therefore why there exist a $beta$ with $f(beta) = alpha$ and $#{beta vert f(beta) = alpha}$ is constant. Therefore independent of the point $x$ for which we take an open pc neighbourhood $U =U_x$?
$endgroup$
– KarlPeter
Jan 10 at 1:49
$begingroup$
Hi, thank you for you answer. One point is unclear for me: When you show that $p_alpha : Y_alpha to X$ are coverings how do you see that $p_alpha^{-1}(U) = bigcup_{beta text{ with } f(beta) = alpha} U_beta$ isn't empty? Therefore why there exist a $beta$ with $f(beta) = alpha$ and $#{beta vert f(beta) = alpha}$ is constant. Therefore independent of the point $x$ for which we take an open pc neighbourhood $U =U_x$?
$endgroup$
– KarlPeter
Jan 10 at 1:49
$begingroup$
You are right, my proof contains a gap. I forgot to show that the $p_alpha$ are surjective. I shall edit my answer.
$endgroup$
– Paul Frost
Jan 10 at 12:24
$begingroup$
You are right, my proof contains a gap. I forgot to show that the $p_alpha$ are surjective. I shall edit my answer.
$endgroup$
– Paul Frost
Jan 10 at 12:24
$begingroup$
Note that $p_alpha^{-1}(U)$ is non-empty because $p_alpha$ is surjective.
$endgroup$
– Paul Frost
Jan 10 at 13:00
$begingroup$
Note that $p_alpha^{-1}(U)$ is non-empty because $p_alpha$ is surjective.
$endgroup$
– Paul Frost
Jan 10 at 13:00
add a comment |
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