Retraction of the Möbius strip to its boundary
$begingroup$
Prove that there is no retraction (i.e. continuous function constant on the codomain) $r: M rightarrow S^1 = partial M$ where $M$ is the Möbius strip.
I've tried to find a contradiction using $r_*$ homomorphism between the fundamental groups, but they are both $mathbb{Z}$ and nothing seems to go wrong...
algebraic-topology mobius-band
edited Apr 8 '15 at 18:23
Najib Idrissi
41.6k473140
$endgroup$
add a comment |
$begingroup$
Prove that there is no retraction (i.e. continuous function constant on the codomain) $r: M rightarrow S^1 = partial M$ where $M$ is the Möbius strip.
I've tried to find a contradiction using $r_*$ homomorphism between the fundamental groups, but they are both $mathbb{Z}$ and nothing seems to go wrong...
algebraic-topology mobius-band
edited Apr 8 '15 at 18:23
Najib Idrissi
41.6k473140
$endgroup$
add a comment |
$begingroup$
Prove that there is no retraction (i.e. continuous function constant on the codomain) $r: M rightarrow S^1 = partial M$ where $M$ is the Möbius strip.
I've tried to find a contradiction using $r_*$ homomorphism between the fundamental groups, but they are both $mathbb{Z}$ and nothing seems to go wrong...
algebraic-topology mobius-band
edited Apr 8 '15 at 18:23
Najib Idrissi
41.6k473140
$endgroup$
Prove that there is no retraction (i.e. continuous function constant on the codomain) $r: M rightarrow S^1 = partial M$ where $M$ is the Möbius strip.
I've tried to find a contradiction using $r_*$ homomorphism between the fundamental groups, but they are both $mathbb{Z}$ and nothing seems to go wrong...
algebraic-topology mobius-band
algebraic-topology mobius-band
edited Apr 8 '15 at 18:23
Najib Idrissi
41.6k473140
edited Apr 8 '15 at 18:23
Najib Idrissi
41.6k473140
edited Apr 8 '15 at 18:23
Najib Idrissi
41.6k473140
edited Apr 8 '15 at 18:23
Najib Idrissi
41.6k473140
edited Apr 8 '15 at 18:23
Najib Idrissi
41.6k473140
41.6k473140
asked Sep 25 '12 at 21:10
user32847
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
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If $alphainpi_1(partial M)$ is a generator, its image $i_*(alpha)inpi_1(M)$ under the inclusion $i:partial Mto M$ is the square of an element of $pi_1(M)$, so that if $r:Mtopartial M$ is a retraction, $alpha=r_*i_*(alpha)$ is also the square of an element of $pi_1(partial M)$. This is not so.
(For all this to work, one has to pick a basepoint $x_0inpartial M$ and use it to compute both $pi_1(M)$ and $pi_1(partial M)$)
answered Sep 25 '12 at 21:24
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
111k7157288
$endgroup$
1
$begingroup$
why $i_* (alpha) $ is the square of an element in $pi_1 (M)$?
$endgroup$
– user32847
Sep 26 '12 at 7:31
2
$begingroup$
Essentially because it turns around the band twice, and a generator of the fundamental group of the band is a curve which turns around it only once. Notice that $i$ is a map one knows, so one can completely compute the morphism $i_*:pi_1(partial M,x_0)topi_1(M,x_0)$.
$endgroup$
– Mariano Suárez-Álvarez
Sep 26 '12 at 17:36
2
$begingroup$
Why is the following true: "if $r:Mtopartial M$ is a retraction, $alpha=r_*i_*(alpha)$ is also the square of an element of $pi_1(partial M)$"?
$endgroup$
– fierydemon
Mar 15 '16 at 1:06
$begingroup$
@AyushKhaitan this is because $i_*(alpha) = beta^2$ and as we are dealing with homomorphism this means $r_*i_*(alpha)$ = r_*(beta)^2$
$endgroup$
– B.A
Feb 22 '18 at 5:26
$begingroup$
@MarianoSuárez-Álvarez what is the point of computing $pi_1(M)$ and $pi_1(partial M)$? even if there is retraction, it is not a homotopy equivalence. Can we find a contradiction by these computations? Thanks!
$endgroup$
– PerelMan
Feb 22 at 17:35
add a comment |
$begingroup$
For each $alphainpartial M$, let $gamma_alpha$ be the closed loop in $M$ that starts at $alpha$, goes directly across the strip to its antipode and then halfway around the boundary to its starting point in positive direction. Then $alphamapstogamma_alpha$ is a homotopy -- in particular every $gamma_alpha$ has the same homotopy class.
On the other hand, if $x$ and $y$ are antipodes, then when we form $gamma_x+gamma_y$, the "directly across" sections cancel out, and the concatenated curve is homotopic to a single turn around the entire boundary. So the homotopy class of $r(gamma_x+gamma_y)$ in $partial M$ is $1$. On the other hand, $r$ ought to induce a homomorphism between the homotopy groups, but $1$ is not twice anything in $mathbb Z$, which is a contradiction.
answered Sep 25 '12 at 21:37
Henning MakholmHenning Makholm
241k17308546
$endgroup$
$begingroup$
(Annoyingly, one has to do this with based loops :-/ )
$endgroup$
– Mariano Suárez-Álvarez
Sep 25 '12 at 21:39
$begingroup$
@Mariano: Reparameterizing all of the loops to be based at $x$ and then throwing away the $alpha$s where $gamma_alpha$ does not already pass through $x$ ought to take care of that.
$endgroup$
– Henning Makholm
Sep 25 '12 at 21:43
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@HenningMakholm is $pi_1(partial M)=1$ or I misunderstood something?
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– PerelMan
Feb 22 at 17:42
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@PerelMan: $partial M$ is a circle, so $pi_1(partial M)congmathbb Z$. When I write $1$ I mean $1inmathbb Z$.
$endgroup$
– Henning Makholm
Feb 22 at 18:42
add a comment |
$begingroup$
You can also prove this using homology, but it's somewhat more effort. The basic idea is that, if $B$ is the boundary circle and $r:Mrightarrow M$ is a retraction onto $B$ then the inclusion map also induces an injection $i_*:H_1(B)rightarrow H_1(X)$ (to see this, apply $r_*$ to $i_*alpha=i_*beta$). Thus we have an exact sequence
$$
0rightarrow H_1(B)xrightarrow{i_*}H_1(M)xrightarrow{q_*} H_1(M,B)rightarrow 0
$$
coming from the reduced long exact sequence for the pair $(M,B)$. We know that $H_1(B)$ and $H_1(M)$ are both $mathbb Z$ (because $B=S^1$ and $M$ deformation retracts onto its central circle) and, since $(M,B)$ is a good pair, $H_1(M,B)cong H_1(M/B)$. But $M/B=mathbb Rmathbb P^2$, as can be seen by their cell decompositions, where the pink indicates the boundary circle $B$:
thus $H_1(X,B)cong mathbb Z/2mathbb Z$. The fact that $r$ is a retraction means that the above sequence splits, as $r_*:H_2(M)rightarrow H_2(B)$ composed with $i_*$ is the identity. This is a contradiction.
answered Jan 10 at 4:06
ArbutusArbutus
701715
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $alphainpi_1(partial M)$ is a generator, its image $i_*(alpha)inpi_1(M)$ under the inclusion $i:partial Mto M$ is the square of an element of $pi_1(M)$, so that if $r:Mtopartial M$ is a retraction, $alpha=r_*i_*(alpha)$ is also the square of an element of $pi_1(partial M)$. This is not so.
(For all this to work, one has to pick a basepoint $x_0inpartial M$ and use it to compute both $pi_1(M)$ and $pi_1(partial M)$)
answered Sep 25 '12 at 21:24
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
111k7157288
$endgroup$
1
$begingroup$
why $i_* (alpha) $ is the square of an element in $pi_1 (M)$?
$endgroup$
– user32847
Sep 26 '12 at 7:31
2
$begingroup$
Essentially because it turns around the band twice, and a generator of the fundamental group of the band is a curve which turns around it only once. Notice that $i$ is a map one knows, so one can completely compute the morphism $i_*:pi_1(partial M,x_0)topi_1(M,x_0)$.
$endgroup$
– Mariano Suárez-Álvarez
Sep 26 '12 at 17:36
2
$begingroup$
Why is the following true: "if $r:Mtopartial M$ is a retraction, $alpha=r_*i_*(alpha)$ is also the square of an element of $pi_1(partial M)$"?
$endgroup$
– fierydemon
Mar 15 '16 at 1:06
$begingroup$
@AyushKhaitan this is because $i_*(alpha) = beta^2$ and as we are dealing with homomorphism this means $r_*i_*(alpha)$ = r_*(beta)^2$
$endgroup$
– B.A
Feb 22 '18 at 5:26
$begingroup$
@MarianoSuárez-Álvarez what is the point of computing $pi_1(M)$ and $pi_1(partial M)$? even if there is retraction, it is not a homotopy equivalence. Can we find a contradiction by these computations? Thanks!
$endgroup$
– PerelMan
Feb 22 at 17:35
add a comment |
$begingroup$
If $alphainpi_1(partial M)$ is a generator, its image $i_*(alpha)inpi_1(M)$ under the inclusion $i:partial Mto M$ is the square of an element of $pi_1(M)$, so that if $r:Mtopartial M$ is a retraction, $alpha=r_*i_*(alpha)$ is also the square of an element of $pi_1(partial M)$. This is not so.
(For all this to work, one has to pick a basepoint $x_0inpartial M$ and use it to compute both $pi_1(M)$ and $pi_1(partial M)$)
answered Sep 25 '12 at 21:24
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
111k7157288
$endgroup$
1
$begingroup$
why $i_* (alpha) $ is the square of an element in $pi_1 (M)$?
$endgroup$
– user32847
Sep 26 '12 at 7:31
2
$begingroup$
Essentially because it turns around the band twice, and a generator of the fundamental group of the band is a curve which turns around it only once. Notice that $i$ is a map one knows, so one can completely compute the morphism $i_*:pi_1(partial M,x_0)topi_1(M,x_0)$.
$endgroup$
– Mariano Suárez-Álvarez
Sep 26 '12 at 17:36
2
$begingroup$
Why is the following true: "if $r:Mtopartial M$ is a retraction, $alpha=r_*i_*(alpha)$ is also the square of an element of $pi_1(partial M)$"?
$endgroup$
– fierydemon
Mar 15 '16 at 1:06
$begingroup$
@AyushKhaitan this is because $i_*(alpha) = beta^2$ and as we are dealing with homomorphism this means $r_*i_*(alpha)$ = r_*(beta)^2$
$endgroup$
– B.A
Feb 22 '18 at 5:26
$begingroup$
@MarianoSuárez-Álvarez what is the point of computing $pi_1(M)$ and $pi_1(partial M)$? even if there is retraction, it is not a homotopy equivalence. Can we find a contradiction by these computations? Thanks!
$endgroup$
– PerelMan
Feb 22 at 17:35
add a comment |
$begingroup$
If $alphainpi_1(partial M)$ is a generator, its image $i_*(alpha)inpi_1(M)$ under the inclusion $i:partial Mto M$ is the square of an element of $pi_1(M)$, so that if $r:Mtopartial M$ is a retraction, $alpha=r_*i_*(alpha)$ is also the square of an element of $pi_1(partial M)$. This is not so.
(For all this to work, one has to pick a basepoint $x_0inpartial M$ and use it to compute both $pi_1(M)$ and $pi_1(partial M)$)
answered Sep 25 '12 at 21:24
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
111k7157288
$endgroup$
If $alphainpi_1(partial M)$ is a generator, its image $i_*(alpha)inpi_1(M)$ under the inclusion $i:partial Mto M$ is the square of an element of $pi_1(M)$, so that if $r:Mtopartial M$ is a retraction, $alpha=r_*i_*(alpha)$ is also the square of an element of $pi_1(partial M)$. This is not so.
(For all this to work, one has to pick a basepoint $x_0inpartial M$ and use it to compute both $pi_1(M)$ and $pi_1(partial M)$)
answered Sep 25 '12 at 21:24
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
111k7157288
answered Sep 25 '12 at 21:24
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
111k7157288
answered Sep 25 '12 at 21:24
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
111k7157288
answered Sep 25 '12 at 21:24
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
111k7157288
111k7157288
1
$begingroup$
why $i_* (alpha) $ is the square of an element in $pi_1 (M)$?
$endgroup$
– user32847
Sep 26 '12 at 7:31
2
$begingroup$
Essentially because it turns around the band twice, and a generator of the fundamental group of the band is a curve which turns around it only once. Notice that $i$ is a map one knows, so one can completely compute the morphism $i_*:pi_1(partial M,x_0)topi_1(M,x_0)$.
$endgroup$
– Mariano Suárez-Álvarez
Sep 26 '12 at 17:36
2
$begingroup$
Why is the following true: "if $r:Mtopartial M$ is a retraction, $alpha=r_*i_*(alpha)$ is also the square of an element of $pi_1(partial M)$"?
$endgroup$
– fierydemon
Mar 15 '16 at 1:06
$begingroup$
@AyushKhaitan this is because $i_*(alpha) = beta^2$ and as we are dealing with homomorphism this means $r_*i_*(alpha)$ = r_*(beta)^2$
$endgroup$
– B.A
Feb 22 '18 at 5:26
$begingroup$
@MarianoSuárez-Álvarez what is the point of computing $pi_1(M)$ and $pi_1(partial M)$? even if there is retraction, it is not a homotopy equivalence. Can we find a contradiction by these computations? Thanks!
$endgroup$
– PerelMan
Feb 22 at 17:35
add a comment |
1
$begingroup$
why $i_* (alpha) $ is the square of an element in $pi_1 (M)$?
$endgroup$
– user32847
Sep 26 '12 at 7:31
2
$begingroup$
Essentially because it turns around the band twice, and a generator of the fundamental group of the band is a curve which turns around it only once. Notice that $i$ is a map one knows, so one can completely compute the morphism $i_*:pi_1(partial M,x_0)topi_1(M,x_0)$.
$endgroup$
– Mariano Suárez-Álvarez
Sep 26 '12 at 17:36
2
$begingroup$
Why is the following true: "if $r:Mtopartial M$ is a retraction, $alpha=r_*i_*(alpha)$ is also the square of an element of $pi_1(partial M)$"?
$endgroup$
– fierydemon
Mar 15 '16 at 1:06
$begingroup$
@AyushKhaitan this is because $i_*(alpha) = beta^2$ and as we are dealing with homomorphism this means $r_*i_*(alpha)$ = r_*(beta)^2$
$endgroup$
– B.A
Feb 22 '18 at 5:26
$begingroup$
@MarianoSuárez-Álvarez what is the point of computing $pi_1(M)$ and $pi_1(partial M)$? even if there is retraction, it is not a homotopy equivalence. Can we find a contradiction by these computations? Thanks!
$endgroup$
– PerelMan
Feb 22 at 17:35
1
1
$begingroup$
why $i_* (alpha) $ is the square of an element in $pi_1 (M)$?
$endgroup$
– user32847
Sep 26 '12 at 7:31
$begingroup$
why $i_* (alpha) $ is the square of an element in $pi_1 (M)$?
$endgroup$
– user32847
Sep 26 '12 at 7:31
2
2
$begingroup$
Essentially because it turns around the band twice, and a generator of the fundamental group of the band is a curve which turns around it only once. Notice that $i$ is a map one knows, so one can completely compute the morphism $i_*:pi_1(partial M,x_0)topi_1(M,x_0)$.
$endgroup$
– Mariano Suárez-Álvarez
Sep 26 '12 at 17:36
$begingroup$
Essentially because it turns around the band twice, and a generator of the fundamental group of the band is a curve which turns around it only once. Notice that $i$ is a map one knows, so one can completely compute the morphism $i_*:pi_1(partial M,x_0)topi_1(M,x_0)$.
$endgroup$
– Mariano Suárez-Álvarez
Sep 26 '12 at 17:36
2
2
$begingroup$
Why is the following true: "if $r:Mtopartial M$ is a retraction, $alpha=r_*i_*(alpha)$ is also the square of an element of $pi_1(partial M)$"?
$endgroup$
– fierydemon
Mar 15 '16 at 1:06
$begingroup$
Why is the following true: "if $r:Mtopartial M$ is a retraction, $alpha=r_*i_*(alpha)$ is also the square of an element of $pi_1(partial M)$"?
$endgroup$
– fierydemon
Mar 15 '16 at 1:06
$begingroup$
@AyushKhaitan this is because $i_*(alpha) = beta^2$ and as we are dealing with homomorphism this means $r_*i_*(alpha)$ = r_*(beta)^2$
$endgroup$
– B.A
Feb 22 '18 at 5:26
$begingroup$
@AyushKhaitan this is because $i_*(alpha) = beta^2$ and as we are dealing with homomorphism this means $r_*i_*(alpha)$ = r_*(beta)^2$
$endgroup$
– B.A
Feb 22 '18 at 5:26
$begingroup$
@MarianoSuárez-Álvarez what is the point of computing $pi_1(M)$ and $pi_1(partial M)$? even if there is retraction, it is not a homotopy equivalence. Can we find a contradiction by these computations? Thanks!
$endgroup$
– PerelMan
Feb 22 at 17:35
$begingroup$
@MarianoSuárez-Álvarez what is the point of computing $pi_1(M)$ and $pi_1(partial M)$? even if there is retraction, it is not a homotopy equivalence. Can we find a contradiction by these computations? Thanks!
$endgroup$
– PerelMan
Feb 22 at 17:35
add a comment |
$begingroup$
For each $alphainpartial M$, let $gamma_alpha$ be the closed loop in $M$ that starts at $alpha$, goes directly across the strip to its antipode and then halfway around the boundary to its starting point in positive direction. Then $alphamapstogamma_alpha$ is a homotopy -- in particular every $gamma_alpha$ has the same homotopy class.
On the other hand, if $x$ and $y$ are antipodes, then when we form $gamma_x+gamma_y$, the "directly across" sections cancel out, and the concatenated curve is homotopic to a single turn around the entire boundary. So the homotopy class of $r(gamma_x+gamma_y)$ in $partial M$ is $1$. On the other hand, $r$ ought to induce a homomorphism between the homotopy groups, but $1$ is not twice anything in $mathbb Z$, which is a contradiction.
answered Sep 25 '12 at 21:37
Henning MakholmHenning Makholm
241k17308546
$endgroup$
$begingroup$
(Annoyingly, one has to do this with based loops :-/ )
$endgroup$
– Mariano Suárez-Álvarez
Sep 25 '12 at 21:39
$begingroup$
@Mariano: Reparameterizing all of the loops to be based at $x$ and then throwing away the $alpha$s where $gamma_alpha$ does not already pass through $x$ ought to take care of that.
$endgroup$
– Henning Makholm
Sep 25 '12 at 21:43
$begingroup$
@HenningMakholm is $pi_1(partial M)=1$ or I misunderstood something?
$endgroup$
– PerelMan
Feb 22 at 17:42
$begingroup$
@PerelMan: $partial M$ is a circle, so $pi_1(partial M)congmathbb Z$. When I write $1$ I mean $1inmathbb Z$.
$endgroup$
– Henning Makholm
Feb 22 at 18:42
add a comment |
$begingroup$
For each $alphainpartial M$, let $gamma_alpha$ be the closed loop in $M$ that starts at $alpha$, goes directly across the strip to its antipode and then halfway around the boundary to its starting point in positive direction. Then $alphamapstogamma_alpha$ is a homotopy -- in particular every $gamma_alpha$ has the same homotopy class.
On the other hand, if $x$ and $y$ are antipodes, then when we form $gamma_x+gamma_y$, the "directly across" sections cancel out, and the concatenated curve is homotopic to a single turn around the entire boundary. So the homotopy class of $r(gamma_x+gamma_y)$ in $partial M$ is $1$. On the other hand, $r$ ought to induce a homomorphism between the homotopy groups, but $1$ is not twice anything in $mathbb Z$, which is a contradiction.
answered Sep 25 '12 at 21:37
Henning MakholmHenning Makholm
241k17308546
$endgroup$
$begingroup$
(Annoyingly, one has to do this with based loops :-/ )
$endgroup$
– Mariano Suárez-Álvarez
Sep 25 '12 at 21:39
$begingroup$
@Mariano: Reparameterizing all of the loops to be based at $x$ and then throwing away the $alpha$s where $gamma_alpha$ does not already pass through $x$ ought to take care of that.
$endgroup$
– Henning Makholm
Sep 25 '12 at 21:43
$begingroup$
@HenningMakholm is $pi_1(partial M)=1$ or I misunderstood something?
$endgroup$
– PerelMan
Feb 22 at 17:42
$begingroup$
@PerelMan: $partial M$ is a circle, so $pi_1(partial M)congmathbb Z$. When I write $1$ I mean $1inmathbb Z$.
$endgroup$
– Henning Makholm
Feb 22 at 18:42
add a comment |
$begingroup$
For each $alphainpartial M$, let $gamma_alpha$ be the closed loop in $M$ that starts at $alpha$, goes directly across the strip to its antipode and then halfway around the boundary to its starting point in positive direction. Then $alphamapstogamma_alpha$ is a homotopy -- in particular every $gamma_alpha$ has the same homotopy class.
On the other hand, if $x$ and $y$ are antipodes, then when we form $gamma_x+gamma_y$, the "directly across" sections cancel out, and the concatenated curve is homotopic to a single turn around the entire boundary. So the homotopy class of $r(gamma_x+gamma_y)$ in $partial M$ is $1$. On the other hand, $r$ ought to induce a homomorphism between the homotopy groups, but $1$ is not twice anything in $mathbb Z$, which is a contradiction.
answered Sep 25 '12 at 21:37
Henning MakholmHenning Makholm
241k17308546
$endgroup$
For each $alphainpartial M$, let $gamma_alpha$ be the closed loop in $M$ that starts at $alpha$, goes directly across the strip to its antipode and then halfway around the boundary to its starting point in positive direction. Then $alphamapstogamma_alpha$ is a homotopy -- in particular every $gamma_alpha$ has the same homotopy class.
On the other hand, if $x$ and $y$ are antipodes, then when we form $gamma_x+gamma_y$, the "directly across" sections cancel out, and the concatenated curve is homotopic to a single turn around the entire boundary. So the homotopy class of $r(gamma_x+gamma_y)$ in $partial M$ is $1$. On the other hand, $r$ ought to induce a homomorphism between the homotopy groups, but $1$ is not twice anything in $mathbb Z$, which is a contradiction.
answered Sep 25 '12 at 21:37
Henning MakholmHenning Makholm
241k17308546
answered Sep 25 '12 at 21:37
Henning MakholmHenning Makholm
241k17308546
answered Sep 25 '12 at 21:37
Henning MakholmHenning Makholm
241k17308546
answered Sep 25 '12 at 21:37
Henning MakholmHenning Makholm
241k17308546
241k17308546
$begingroup$
(Annoyingly, one has to do this with based loops :-/ )
$endgroup$
– Mariano Suárez-Álvarez
Sep 25 '12 at 21:39
$begingroup$
@Mariano: Reparameterizing all of the loops to be based at $x$ and then throwing away the $alpha$s where $gamma_alpha$ does not already pass through $x$ ought to take care of that.
$endgroup$
– Henning Makholm
Sep 25 '12 at 21:43
$begingroup$
@HenningMakholm is $pi_1(partial M)=1$ or I misunderstood something?
$endgroup$
– PerelMan
Feb 22 at 17:42
$begingroup$
@PerelMan: $partial M$ is a circle, so $pi_1(partial M)congmathbb Z$. When I write $1$ I mean $1inmathbb Z$.
$endgroup$
– Henning Makholm
Feb 22 at 18:42
add a comment |
$begingroup$
(Annoyingly, one has to do this with based loops :-/ )
$endgroup$
– Mariano Suárez-Álvarez
Sep 25 '12 at 21:39
$begingroup$
@Mariano: Reparameterizing all of the loops to be based at $x$ and then throwing away the $alpha$s where $gamma_alpha$ does not already pass through $x$ ought to take care of that.
$endgroup$
– Henning Makholm
Sep 25 '12 at 21:43
$begingroup$
@HenningMakholm is $pi_1(partial M)=1$ or I misunderstood something?
$endgroup$
– PerelMan
Feb 22 at 17:42
$begingroup$
@PerelMan: $partial M$ is a circle, so $pi_1(partial M)congmathbb Z$. When I write $1$ I mean $1inmathbb Z$.
$endgroup$
– Henning Makholm
Feb 22 at 18:42
$begingroup$
(Annoyingly, one has to do this with based loops :-/ )
$endgroup$
– Mariano Suárez-Álvarez
Sep 25 '12 at 21:39
$begingroup$
(Annoyingly, one has to do this with based loops :-/ )
$endgroup$
– Mariano Suárez-Álvarez
Sep 25 '12 at 21:39
$begingroup$
@Mariano: Reparameterizing all of the loops to be based at $x$ and then throwing away the $alpha$s where $gamma_alpha$ does not already pass through $x$ ought to take care of that.
$endgroup$
– Henning Makholm
Sep 25 '12 at 21:43
$begingroup$
@Mariano: Reparameterizing all of the loops to be based at $x$ and then throwing away the $alpha$s where $gamma_alpha$ does not already pass through $x$ ought to take care of that.
$endgroup$
– Henning Makholm
Sep 25 '12 at 21:43
$begingroup$
@HenningMakholm is $pi_1(partial M)=1$ or I misunderstood something?
$endgroup$
– PerelMan
Feb 22 at 17:42
$begingroup$
@HenningMakholm is $pi_1(partial M)=1$ or I misunderstood something?
$endgroup$
– PerelMan
Feb 22 at 17:42
$begingroup$
@PerelMan: $partial M$ is a circle, so $pi_1(partial M)congmathbb Z$. When I write $1$ I mean $1inmathbb Z$.
$endgroup$
– Henning Makholm
Feb 22 at 18:42
$begingroup$
@PerelMan: $partial M$ is a circle, so $pi_1(partial M)congmathbb Z$. When I write $1$ I mean $1inmathbb Z$.
$endgroup$
– Henning Makholm
Feb 22 at 18:42
add a comment |
$begingroup$
You can also prove this using homology, but it's somewhat more effort. The basic idea is that, if $B$ is the boundary circle and $r:Mrightarrow M$ is a retraction onto $B$ then the inclusion map also induces an injection $i_*:H_1(B)rightarrow H_1(X)$ (to see this, apply $r_*$ to $i_*alpha=i_*beta$). Thus we have an exact sequence
$$
0rightarrow H_1(B)xrightarrow{i_*}H_1(M)xrightarrow{q_*} H_1(M,B)rightarrow 0
$$
coming from the reduced long exact sequence for the pair $(M,B)$. We know that $H_1(B)$ and $H_1(M)$ are both $mathbb Z$ (because $B=S^1$ and $M$ deformation retracts onto its central circle) and, since $(M,B)$ is a good pair, $H_1(M,B)cong H_1(M/B)$. But $M/B=mathbb Rmathbb P^2$, as can be seen by their cell decompositions, where the pink indicates the boundary circle $B$:
thus $H_1(X,B)cong mathbb Z/2mathbb Z$. The fact that $r$ is a retraction means that the above sequence splits, as $r_*:H_2(M)rightarrow H_2(B)$ composed with $i_*$ is the identity. This is a contradiction.
answered Jan 10 at 4:06
ArbutusArbutus
701715
$endgroup$
add a comment |
$begingroup$
You can also prove this using homology, but it's somewhat more effort. The basic idea is that, if $B$ is the boundary circle and $r:Mrightarrow M$ is a retraction onto $B$ then the inclusion map also induces an injection $i_*:H_1(B)rightarrow H_1(X)$ (to see this, apply $r_*$ to $i_*alpha=i_*beta$). Thus we have an exact sequence
$$
0rightarrow H_1(B)xrightarrow{i_*}H_1(M)xrightarrow{q_*} H_1(M,B)rightarrow 0
$$
coming from the reduced long exact sequence for the pair $(M,B)$. We know that $H_1(B)$ and $H_1(M)$ are both $mathbb Z$ (because $B=S^1$ and $M$ deformation retracts onto its central circle) and, since $(M,B)$ is a good pair, $H_1(M,B)cong H_1(M/B)$. But $M/B=mathbb Rmathbb P^2$, as can be seen by their cell decompositions, where the pink indicates the boundary circle $B$:
thus $H_1(X,B)cong mathbb Z/2mathbb Z$. The fact that $r$ is a retraction means that the above sequence splits, as $r_*:H_2(M)rightarrow H_2(B)$ composed with $i_*$ is the identity. This is a contradiction.
answered Jan 10 at 4:06
ArbutusArbutus
701715
$endgroup$
add a comment |
$begingroup$
You can also prove this using homology, but it's somewhat more effort. The basic idea is that, if $B$ is the boundary circle and $r:Mrightarrow M$ is a retraction onto $B$ then the inclusion map also induces an injection $i_*:H_1(B)rightarrow H_1(X)$ (to see this, apply $r_*$ to $i_*alpha=i_*beta$). Thus we have an exact sequence
$$
0rightarrow H_1(B)xrightarrow{i_*}H_1(M)xrightarrow{q_*} H_1(M,B)rightarrow 0
$$
coming from the reduced long exact sequence for the pair $(M,B)$. We know that $H_1(B)$ and $H_1(M)$ are both $mathbb Z$ (because $B=S^1$ and $M$ deformation retracts onto its central circle) and, since $(M,B)$ is a good pair, $H_1(M,B)cong H_1(M/B)$. But $M/B=mathbb Rmathbb P^2$, as can be seen by their cell decompositions, where the pink indicates the boundary circle $B$:
thus $H_1(X,B)cong mathbb Z/2mathbb Z$. The fact that $r$ is a retraction means that the above sequence splits, as $r_*:H_2(M)rightarrow H_2(B)$ composed with $i_*$ is the identity. This is a contradiction.
answered Jan 10 at 4:06
ArbutusArbutus
701715
$endgroup$
You can also prove this using homology, but it's somewhat more effort. The basic idea is that, if $B$ is the boundary circle and $r:Mrightarrow M$ is a retraction onto $B$ then the inclusion map also induces an injection $i_*:H_1(B)rightarrow H_1(X)$ (to see this, apply $r_*$ to $i_*alpha=i_*beta$). Thus we have an exact sequence
$$
0rightarrow H_1(B)xrightarrow{i_*}H_1(M)xrightarrow{q_*} H_1(M,B)rightarrow 0
$$
coming from the reduced long exact sequence for the pair $(M,B)$. We know that $H_1(B)$ and $H_1(M)$ are both $mathbb Z$ (because $B=S^1$ and $M$ deformation retracts onto its central circle) and, since $(M,B)$ is a good pair, $H_1(M,B)cong H_1(M/B)$. But $M/B=mathbb Rmathbb P^2$, as can be seen by their cell decompositions, where the pink indicates the boundary circle $B$:
thus $H_1(X,B)cong mathbb Z/2mathbb Z$. The fact that $r$ is a retraction means that the above sequence splits, as $r_*:H_2(M)rightarrow H_2(B)$ composed with $i_*$ is the identity. This is a contradiction.
answered Jan 10 at 4:06
ArbutusArbutus
701715
edited Jan 10 at 15:54
answered Jan 10 at 4:06
ArbutusArbutus
701715
answered Jan 10 at 4:06
ArbutusArbutus
701715
answered Jan 10 at 4:06
ArbutusArbutus
701715
701715
add a comment |
add a comment |
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