What is wrong with this application of van Kampen's Theorem?
$begingroup$
My understanding of van Kampen's Theorem (simplified to just two neighbourhoods):
Let $X$ be a topological space and let ${N_a, N_b}$ be a cover of $X$ such that $N_a cap N_b$ is path-connected (and each open set is path-connected). Then,
$$
pi_1Xcong frac{pi_1N_a*pi_1N_b}{[i_a(gamma)][i_b(gamma)]^{-1}}
$$
Where $i_a$ and $i_b$ are the inclusion maps from $N_acap N_b$ to $N_a$ and $N_b$ respectively, and $gamma$ is any loop in the intersection $N_acap N_b$ (so we quotient by the normal subgroup generated by $[i_a(gamma)][i_b(gamma)]^{-1}$).
I was trying to make sure I understand it by seeing if I can "break" it: Let $N_a = N_b = X$ (where $X$ is a path-connected topological space). Then clearly $N_a$, $N_b$, and $N_acap N_b$ are path-connected and we may apply van Kampen. But now the intersection is just $X$, and so the inclusion maps $i_a$ and $i_b$ are just the identity maps, so applying van Kampen we have ($gamma$ in $X$)
$$
pi_1Xcong frac{pi_1N_a*pi_1N_b}{[gamma][gamma]^{-1}} cong pi_1X*pi_1X
$$
Clearly this is wrong, so I was wondering where my mistake is.
algebraic-topology fundamental-groups
edited Dec 12 '18 at 0:43
Eric Wofsey
185k13213339
asked Dec 12 '18 at 0:26
FunctionalDefectFunctionalDefect
255
$endgroup$
add a comment |
$begingroup$
My understanding of van Kampen's Theorem (simplified to just two neighbourhoods):
Let $X$ be a topological space and let ${N_a, N_b}$ be a cover of $X$ such that $N_a cap N_b$ is path-connected (and each open set is path-connected). Then,
$$
pi_1Xcong frac{pi_1N_a*pi_1N_b}{[i_a(gamma)][i_b(gamma)]^{-1}}
$$
Where $i_a$ and $i_b$ are the inclusion maps from $N_acap N_b$ to $N_a$ and $N_b$ respectively, and $gamma$ is any loop in the intersection $N_acap N_b$ (so we quotient by the normal subgroup generated by $[i_a(gamma)][i_b(gamma)]^{-1}$).
I was trying to make sure I understand it by seeing if I can "break" it: Let $N_a = N_b = X$ (where $X$ is a path-connected topological space). Then clearly $N_a$, $N_b$, and $N_acap N_b$ are path-connected and we may apply van Kampen. But now the intersection is just $X$, and so the inclusion maps $i_a$ and $i_b$ are just the identity maps, so applying van Kampen we have ($gamma$ in $X$)
$$
pi_1Xcong frac{pi_1N_a*pi_1N_b}{[gamma][gamma]^{-1}} cong pi_1X*pi_1X
$$
Clearly this is wrong, so I was wondering where my mistake is.
algebraic-topology fundamental-groups
edited Dec 12 '18 at 0:43
Eric Wofsey
185k13213339
asked Dec 12 '18 at 0:26
FunctionalDefectFunctionalDefect
255
$endgroup$
add a comment |
$begingroup$
My understanding of van Kampen's Theorem (simplified to just two neighbourhoods):
Let $X$ be a topological space and let ${N_a, N_b}$ be a cover of $X$ such that $N_a cap N_b$ is path-connected (and each open set is path-connected). Then,
$$
pi_1Xcong frac{pi_1N_a*pi_1N_b}{[i_a(gamma)][i_b(gamma)]^{-1}}
$$
Where $i_a$ and $i_b$ are the inclusion maps from $N_acap N_b$ to $N_a$ and $N_b$ respectively, and $gamma$ is any loop in the intersection $N_acap N_b$ (so we quotient by the normal subgroup generated by $[i_a(gamma)][i_b(gamma)]^{-1}$).
I was trying to make sure I understand it by seeing if I can "break" it: Let $N_a = N_b = X$ (where $X$ is a path-connected topological space). Then clearly $N_a$, $N_b$, and $N_acap N_b$ are path-connected and we may apply van Kampen. But now the intersection is just $X$, and so the inclusion maps $i_a$ and $i_b$ are just the identity maps, so applying van Kampen we have ($gamma$ in $X$)
$$
pi_1Xcong frac{pi_1N_a*pi_1N_b}{[gamma][gamma]^{-1}} cong pi_1X*pi_1X
$$
Clearly this is wrong, so I was wondering where my mistake is.
algebraic-topology fundamental-groups
edited Dec 12 '18 at 0:43
Eric Wofsey
185k13213339
asked Dec 12 '18 at 0:26
FunctionalDefectFunctionalDefect
255
$endgroup$
My understanding of van Kampen's Theorem (simplified to just two neighbourhoods):
Let $X$ be a topological space and let ${N_a, N_b}$ be a cover of $X$ such that $N_a cap N_b$ is path-connected (and each open set is path-connected). Then,
$$
pi_1Xcong frac{pi_1N_a*pi_1N_b}{[i_a(gamma)][i_b(gamma)]^{-1}}
$$
Where $i_a$ and $i_b$ are the inclusion maps from $N_acap N_b$ to $N_a$ and $N_b$ respectively, and $gamma$ is any loop in the intersection $N_acap N_b$ (so we quotient by the normal subgroup generated by $[i_a(gamma)][i_b(gamma)]^{-1}$).
I was trying to make sure I understand it by seeing if I can "break" it: Let $N_a = N_b = X$ (where $X$ is a path-connected topological space). Then clearly $N_a$, $N_b$, and $N_acap N_b$ are path-connected and we may apply van Kampen. But now the intersection is just $X$, and so the inclusion maps $i_a$ and $i_b$ are just the identity maps, so applying van Kampen we have ($gamma$ in $X$)
$$
pi_1Xcong frac{pi_1N_a*pi_1N_b}{[gamma][gamma]^{-1}} cong pi_1X*pi_1X
$$
Clearly this is wrong, so I was wondering where my mistake is.
algebraic-topology fundamental-groups
algebraic-topology fundamental-groups
edited Dec 12 '18 at 0:43
Eric Wofsey
185k13213339
asked Dec 12 '18 at 0:26
FunctionalDefectFunctionalDefect
255
edited Dec 12 '18 at 0:43
Eric Wofsey
185k13213339
asked Dec 12 '18 at 0:26
FunctionalDefectFunctionalDefect
255
edited Dec 12 '18 at 0:43
Eric Wofsey
185k13213339
edited Dec 12 '18 at 0:43
Eric Wofsey
185k13213339
edited Dec 12 '18 at 0:43
Eric Wofsey
185k13213339
185k13213339
asked Dec 12 '18 at 0:26
FunctionalDefectFunctionalDefect
255
asked Dec 12 '18 at 0:26
FunctionalDefectFunctionalDefect
255
asked Dec 12 '18 at 0:26
FunctionalDefectFunctionalDefect
255
255
add a comment |
add a comment |
1 Answer
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$begingroup$
Your mistake is in identifying the subgroup you are quotienting out. You quotient out by the normal subgroup generated by all elements of the form $[i_a(gamma)][i_b(gamma)]^{-1}$. When you write this, $[i_a(gamma)]$ is to be interpreted as an element of $pi_1(N_a)$, which is then considered as an element of $pi_1(N_a)*pi_1(N_b)$ via the canonical inclusion map $pi_1(N_a)to pi_1(N_a)*pi_1(N_b)$. Similarly, $[i_b(gamma)]$ is to be interpreted as an element of $pi_1(N_b)$, which is then considered as an element of $pi_1(N_a)*pi_1(N_b)$ via the canonical inclusion map $pi_1(N_b)to pi_1(N_a)*pi_1(N_b)$.
What this means is that even in the case where $N_a=N_b=X$, $[i_a(gamma)]$ and $[i_b(gamma)]$ are not the same element of $pi_1(N_a)*pi_1(N_b)=pi_1(X)*pi_1(X)$. The first is the copy of $[gamma]$ in the first factor of $pi_1(X)*pi_1(X)$, and the second is the copy of $[gamma]$ in the second factor of $pi_1(X)*pi_1(X)$. So, modding out a relation that says these are equal amounts to identifying the two copies of $pi_1(X)$ inside $pi_1(X)*pi_1(X)$ with each other. When you do this, the result you get is just a single $pi_1(X)$ since the two copies have been identified, just like you want. (Of course, this is not a rigorous proof that the quotient is $pi_1(X)$, but it is the intuition from which you can build a proof.)
answered Dec 12 '18 at 0:43
Eric WofseyEric Wofsey
185k13213339
$endgroup$
add a comment |
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$begingroup$
Your mistake is in identifying the subgroup you are quotienting out. You quotient out by the normal subgroup generated by all elements of the form $[i_a(gamma)][i_b(gamma)]^{-1}$. When you write this, $[i_a(gamma)]$ is to be interpreted as an element of $pi_1(N_a)$, which is then considered as an element of $pi_1(N_a)*pi_1(N_b)$ via the canonical inclusion map $pi_1(N_a)to pi_1(N_a)*pi_1(N_b)$. Similarly, $[i_b(gamma)]$ is to be interpreted as an element of $pi_1(N_b)$, which is then considered as an element of $pi_1(N_a)*pi_1(N_b)$ via the canonical inclusion map $pi_1(N_b)to pi_1(N_a)*pi_1(N_b)$.
What this means is that even in the case where $N_a=N_b=X$, $[i_a(gamma)]$ and $[i_b(gamma)]$ are not the same element of $pi_1(N_a)*pi_1(N_b)=pi_1(X)*pi_1(X)$. The first is the copy of $[gamma]$ in the first factor of $pi_1(X)*pi_1(X)$, and the second is the copy of $[gamma]$ in the second factor of $pi_1(X)*pi_1(X)$. So, modding out a relation that says these are equal amounts to identifying the two copies of $pi_1(X)$ inside $pi_1(X)*pi_1(X)$ with each other. When you do this, the result you get is just a single $pi_1(X)$ since the two copies have been identified, just like you want. (Of course, this is not a rigorous proof that the quotient is $pi_1(X)$, but it is the intuition from which you can build a proof.)
answered Dec 12 '18 at 0:43
Eric WofseyEric Wofsey
185k13213339
$endgroup$
add a comment |
$begingroup$
Your mistake is in identifying the subgroup you are quotienting out. You quotient out by the normal subgroup generated by all elements of the form $[i_a(gamma)][i_b(gamma)]^{-1}$. When you write this, $[i_a(gamma)]$ is to be interpreted as an element of $pi_1(N_a)$, which is then considered as an element of $pi_1(N_a)*pi_1(N_b)$ via the canonical inclusion map $pi_1(N_a)to pi_1(N_a)*pi_1(N_b)$. Similarly, $[i_b(gamma)]$ is to be interpreted as an element of $pi_1(N_b)$, which is then considered as an element of $pi_1(N_a)*pi_1(N_b)$ via the canonical inclusion map $pi_1(N_b)to pi_1(N_a)*pi_1(N_b)$.
What this means is that even in the case where $N_a=N_b=X$, $[i_a(gamma)]$ and $[i_b(gamma)]$ are not the same element of $pi_1(N_a)*pi_1(N_b)=pi_1(X)*pi_1(X)$. The first is the copy of $[gamma]$ in the first factor of $pi_1(X)*pi_1(X)$, and the second is the copy of $[gamma]$ in the second factor of $pi_1(X)*pi_1(X)$. So, modding out a relation that says these are equal amounts to identifying the two copies of $pi_1(X)$ inside $pi_1(X)*pi_1(X)$ with each other. When you do this, the result you get is just a single $pi_1(X)$ since the two copies have been identified, just like you want. (Of course, this is not a rigorous proof that the quotient is $pi_1(X)$, but it is the intuition from which you can build a proof.)
answered Dec 12 '18 at 0:43
Eric WofseyEric Wofsey
185k13213339
$endgroup$
add a comment |
$begingroup$
Your mistake is in identifying the subgroup you are quotienting out. You quotient out by the normal subgroup generated by all elements of the form $[i_a(gamma)][i_b(gamma)]^{-1}$. When you write this, $[i_a(gamma)]$ is to be interpreted as an element of $pi_1(N_a)$, which is then considered as an element of $pi_1(N_a)*pi_1(N_b)$ via the canonical inclusion map $pi_1(N_a)to pi_1(N_a)*pi_1(N_b)$. Similarly, $[i_b(gamma)]$ is to be interpreted as an element of $pi_1(N_b)$, which is then considered as an element of $pi_1(N_a)*pi_1(N_b)$ via the canonical inclusion map $pi_1(N_b)to pi_1(N_a)*pi_1(N_b)$.
What this means is that even in the case where $N_a=N_b=X$, $[i_a(gamma)]$ and $[i_b(gamma)]$ are not the same element of $pi_1(N_a)*pi_1(N_b)=pi_1(X)*pi_1(X)$. The first is the copy of $[gamma]$ in the first factor of $pi_1(X)*pi_1(X)$, and the second is the copy of $[gamma]$ in the second factor of $pi_1(X)*pi_1(X)$. So, modding out a relation that says these are equal amounts to identifying the two copies of $pi_1(X)$ inside $pi_1(X)*pi_1(X)$ with each other. When you do this, the result you get is just a single $pi_1(X)$ since the two copies have been identified, just like you want. (Of course, this is not a rigorous proof that the quotient is $pi_1(X)$, but it is the intuition from which you can build a proof.)
answered Dec 12 '18 at 0:43
Eric WofseyEric Wofsey
185k13213339
$endgroup$
Your mistake is in identifying the subgroup you are quotienting out. You quotient out by the normal subgroup generated by all elements of the form $[i_a(gamma)][i_b(gamma)]^{-1}$. When you write this, $[i_a(gamma)]$ is to be interpreted as an element of $pi_1(N_a)$, which is then considered as an element of $pi_1(N_a)*pi_1(N_b)$ via the canonical inclusion map $pi_1(N_a)to pi_1(N_a)*pi_1(N_b)$. Similarly, $[i_b(gamma)]$ is to be interpreted as an element of $pi_1(N_b)$, which is then considered as an element of $pi_1(N_a)*pi_1(N_b)$ via the canonical inclusion map $pi_1(N_b)to pi_1(N_a)*pi_1(N_b)$.
What this means is that even in the case where $N_a=N_b=X$, $[i_a(gamma)]$ and $[i_b(gamma)]$ are not the same element of $pi_1(N_a)*pi_1(N_b)=pi_1(X)*pi_1(X)$. The first is the copy of $[gamma]$ in the first factor of $pi_1(X)*pi_1(X)$, and the second is the copy of $[gamma]$ in the second factor of $pi_1(X)*pi_1(X)$. So, modding out a relation that says these are equal amounts to identifying the two copies of $pi_1(X)$ inside $pi_1(X)*pi_1(X)$ with each other. When you do this, the result you get is just a single $pi_1(X)$ since the two copies have been identified, just like you want. (Of course, this is not a rigorous proof that the quotient is $pi_1(X)$, but it is the intuition from which you can build a proof.)
answered Dec 12 '18 at 0:43
Eric WofseyEric Wofsey
185k13213339
answered Dec 12 '18 at 0:43
Eric WofseyEric Wofsey
185k13213339
answered Dec 12 '18 at 0:43
Eric WofseyEric Wofsey
185k13213339
answered Dec 12 '18 at 0:43
Eric WofseyEric Wofsey
185k13213339
185k13213339
add a comment |
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