Reverse use of Seifert-van Kampen Theorem?
$begingroup$
I am trying to use S-vK Theorem in reverse; what I know are as follows:
$U$ and $V$ satisfy the requirements (open, path-connected), $Ucup V = X$, $U cap V = N$
- $pi_1(N) = langle c,d| cd=dcrangle$
- $pi_1(U) = ??$
- $pi_1(V) = langle drangle$
$pi_1(X) = langle a,b|a^p=b^qrangle$, where $p,q in mathbb{Z}$
- When $c$ and $d$ are injected into $X$, they become identity and $b$ respectively.
In fact does it work this way? Any other help? Thank you very much.
algebraic-topology
asked Oct 4 '13 at 12:25
wilsonwwilsonw
478315
$endgroup$
add a comment |
$begingroup$
I am trying to use S-vK Theorem in reverse; what I know are as follows:
$U$ and $V$ satisfy the requirements (open, path-connected), $Ucup V = X$, $U cap V = N$
- $pi_1(N) = langle c,d| cd=dcrangle$
- $pi_1(U) = ??$
- $pi_1(V) = langle drangle$
$pi_1(X) = langle a,b|a^p=b^qrangle$, where $p,q in mathbb{Z}$
- When $c$ and $d$ are injected into $X$, they become identity and $b$ respectively.
In fact does it work this way? Any other help? Thank you very much.
algebraic-topology
asked Oct 4 '13 at 12:25
wilsonwwilsonw
478315
$endgroup$
add a comment |
$begingroup$
I am trying to use S-vK Theorem in reverse; what I know are as follows:
$U$ and $V$ satisfy the requirements (open, path-connected), $Ucup V = X$, $U cap V = N$
- $pi_1(N) = langle c,d| cd=dcrangle$
- $pi_1(U) = ??$
- $pi_1(V) = langle drangle$
$pi_1(X) = langle a,b|a^p=b^qrangle$, where $p,q in mathbb{Z}$
- When $c$ and $d$ are injected into $X$, they become identity and $b$ respectively.
In fact does it work this way? Any other help? Thank you very much.
algebraic-topology
asked Oct 4 '13 at 12:25
wilsonwwilsonw
478315
$endgroup$
I am trying to use S-vK Theorem in reverse; what I know are as follows:
$U$ and $V$ satisfy the requirements (open, path-connected), $Ucup V = X$, $U cap V = N$
- $pi_1(N) = langle c,d| cd=dcrangle$
- $pi_1(U) = ??$
- $pi_1(V) = langle drangle$
$pi_1(X) = langle a,b|a^p=b^qrangle$, where $p,q in mathbb{Z}$
- When $c$ and $d$ are injected into $X$, they become identity and $b$ respectively.
In fact does it work this way? Any other help? Thank you very much.
algebraic-topology
algebraic-topology
asked Oct 4 '13 at 12:25
wilsonwwilsonw
478315
asked Oct 4 '13 at 12:25
wilsonwwilsonw
478315
edited Jan 23 at 12:19
wilsonw
asked Oct 4 '13 at 12:25
wilsonwwilsonw
478315
asked Oct 4 '13 at 12:25
wilsonwwilsonw
478315
asked Oct 4 '13 at 12:25
wilsonwwilsonw
478315
478315
add a comment |
add a comment |
1 Answer
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$begingroup$
I know this is an old question, but here's an answer anyway:
let
$pi_1(U) = langle a,d | a^p = d^q rangle$
and the map $f: langle c,d | cd = dc rangle to pi_1(U)$ be given by $f(c) = e, f(d) = d$.
The pushout of $pi_1(U)$ and $pi_1(V)$ over $pi_1(N)$ is isomorphic to the $pi_1(X)$ above.
Alternatively $pi_1(U) = langle a,d,c | a^p = b^q, cd = dc rangle$ and $f(c) = c, f(d) = d$. This gives the same pushout but is not isomorphic to the previous $pi_1(U)$.
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I know this is an old question, but here's an answer anyway:
let
$pi_1(U) = langle a,d | a^p = d^q rangle$
and the map $f: langle c,d | cd = dc rangle to pi_1(U)$ be given by $f(c) = e, f(d) = d$.
The pushout of $pi_1(U)$ and $pi_1(V)$ over $pi_1(N)$ is isomorphic to the $pi_1(X)$ above.
Alternatively $pi_1(U) = langle a,d,c | a^p = b^q, cd = dc rangle$ and $f(c) = c, f(d) = d$. This gives the same pushout but is not isomorphic to the previous $pi_1(U)$.
$endgroup$
add a comment |
$begingroup$
I know this is an old question, but here's an answer anyway:
let
$pi_1(U) = langle a,d | a^p = d^q rangle$
and the map $f: langle c,d | cd = dc rangle to pi_1(U)$ be given by $f(c) = e, f(d) = d$.
The pushout of $pi_1(U)$ and $pi_1(V)$ over $pi_1(N)$ is isomorphic to the $pi_1(X)$ above.
Alternatively $pi_1(U) = langle a,d,c | a^p = b^q, cd = dc rangle$ and $f(c) = c, f(d) = d$. This gives the same pushout but is not isomorphic to the previous $pi_1(U)$.
$endgroup$
add a comment |
$begingroup$
I know this is an old question, but here's an answer anyway:
let
$pi_1(U) = langle a,d | a^p = d^q rangle$
and the map $f: langle c,d | cd = dc rangle to pi_1(U)$ be given by $f(c) = e, f(d) = d$.
The pushout of $pi_1(U)$ and $pi_1(V)$ over $pi_1(N)$ is isomorphic to the $pi_1(X)$ above.
Alternatively $pi_1(U) = langle a,d,c | a^p = b^q, cd = dc rangle$ and $f(c) = c, f(d) = d$. This gives the same pushout but is not isomorphic to the previous $pi_1(U)$.
$endgroup$
I know this is an old question, but here's an answer anyway:
let
$pi_1(U) = langle a,d | a^p = d^q rangle$
and the map $f: langle c,d | cd = dc rangle to pi_1(U)$ be given by $f(c) = e, f(d) = d$.
The pushout of $pi_1(U)$ and $pi_1(V)$ over $pi_1(N)$ is isomorphic to the $pi_1(X)$ above.
Alternatively $pi_1(U) = langle a,d,c | a^p = b^q, cd = dc rangle$ and $f(c) = c, f(d) = d$. This gives the same pushout but is not isomorphic to the previous $pi_1(U)$.
answered Oct 21 '14 at 15:44
user40167
add a comment |
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