Mixing
How to find the homotopy fiber of bouquet embedding $S^1⋁S^1↪S^1×S^1$
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Can somebody please explain how to find the homotopy fiber of bouquet embedding $S^1⋁S^1↪S^1×S^1$
algebraic-topology
asked Feb 1 at 17:05
jessicaajessicaa
296
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add a comment |
$begingroup$
Can somebody please explain how to find the homotopy fiber of bouquet embedding $S^1⋁S^1↪S^1×S^1$
algebraic-topology
asked Feb 1 at 17:05
jessicaajessicaa
296
$endgroup$
1
$begingroup$
(Not that in the two answers $ast$ means something completely different.)
$endgroup$
– Pedro Tamaroff♦
Feb 2 at 0:51
add a comment |
$begingroup$
Can somebody please explain how to find the homotopy fiber of bouquet embedding $S^1⋁S^1↪S^1×S^1$
algebraic-topology
asked Feb 1 at 17:05
jessicaajessicaa
296
$endgroup$
Can somebody please explain how to find the homotopy fiber of bouquet embedding $S^1⋁S^1↪S^1×S^1$
algebraic-topology
algebraic-topology
asked Feb 1 at 17:05
jessicaajessicaa
296
asked Feb 1 at 17:05
jessicaajessicaa
296
asked Feb 1 at 17:05
jessicaajessicaa
296
asked Feb 1 at 17:05
jessicaajessicaa
296
asked Feb 1 at 17:05
jessicaajessicaa
296
296
1
$begingroup$
(Not that in the two answers $ast$ means something completely different.)
$endgroup$
– Pedro Tamaroff♦
Feb 2 at 0:51
add a comment |
1
$begingroup$
(Not that in the two answers $ast$ means something completely different.)
$endgroup$
– Pedro Tamaroff♦
Feb 2 at 0:51
1
1
$begingroup$
(Not that in the two answers $ast$ means something completely different.)
$endgroup$
– Pedro Tamaroff♦
Feb 2 at 0:51
$begingroup$
(Not that in the two answers $ast$ means something completely different.)
$endgroup$
– Pedro Tamaroff♦
Feb 2 at 0:51
add a comment |
2 Answers
2
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oldest
votes
$begingroup$
I'll give a couple hints, and then a solution below.
Hint 1: Notice that $S^1vee S^1$ and $S^1 times S^1$ are Eilenberg-Maclane spaces $K(mathbb{Z}*mathbb{Z},1)$ and $K(mathbb{Z}^2, 1)$ respectively.
Hint 2: The inclusion $S^1vee S^1 to S^1 times S^1$ induces the abelianization map on $pi_1$.
Solution
Let's say $F$ is the homotopy fibre of this map (using your favourite definition), so that $Fto S^1 vee S^1 to S^1 times S^1$ is equivalent to a fibration sequence, hence we get a long exact sequence of homotopy groups. By Hints 1 and 2, the long exact sequence of homotopy groups tells use that $Fsimeq K(C, 1)$ where $C$ is the commutator subgroup of $mathbb{Z}*mathbb{Z}$. This determines the (weak?) homotopy type of $F$ since the Eilenberg-Maclane property determines the homotopy type of a CW complex (and every space is weakly equivalent to a CW complex).
Since there are many different definitions of a homotopy fibre, usually we're only concerned with determining its homotopy type. Explicit descriptions are often opaque and hard to work with, so hopefully this is what the person posing the problem had in mind.
answered Feb 1 at 18:08
WilliamWilliam
3,2111228
$endgroup$
1
$begingroup$
I like the hint and solution format.
$endgroup$
– user98602
Feb 1 at 18:34
add a comment |
$begingroup$
In general, the homotopy fiber of the inclusion $X vee Y hookrightarrow X times Y$ is homotopy equivalent to the join $Omega X * Omega Y$. To see this, note that one definition of the homotopy fiber is the pullback of the maps $X vee Y to X times Y$ and $P(X times Y) to X times Y$, which is $(Omega X times PY) cup_{Omega X times Omega Y} (PX times Omega Y) simeq Omega X * Omega Y$.
In the case $X = Y = S^1$, we have $Omega S^1 simeq mathbb{Z}$, so the homotopy fiber is homotopy equivalent to $mathbb{Z} * mathbb{Z}$, which is some infinite bouquet of circles.
answered Feb 1 at 22:04
JHFJHF
4,9711026
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2 Answers
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2 Answers
2
active
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$begingroup$
I'll give a couple hints, and then a solution below.
Hint 1: Notice that $S^1vee S^1$ and $S^1 times S^1$ are Eilenberg-Maclane spaces $K(mathbb{Z}*mathbb{Z},1)$ and $K(mathbb{Z}^2, 1)$ respectively.
Hint 2: The inclusion $S^1vee S^1 to S^1 times S^1$ induces the abelianization map on $pi_1$.
Solution
Let's say $F$ is the homotopy fibre of this map (using your favourite definition), so that $Fto S^1 vee S^1 to S^1 times S^1$ is equivalent to a fibration sequence, hence we get a long exact sequence of homotopy groups. By Hints 1 and 2, the long exact sequence of homotopy groups tells use that $Fsimeq K(C, 1)$ where $C$ is the commutator subgroup of $mathbb{Z}*mathbb{Z}$. This determines the (weak?) homotopy type of $F$ since the Eilenberg-Maclane property determines the homotopy type of a CW complex (and every space is weakly equivalent to a CW complex).
Since there are many different definitions of a homotopy fibre, usually we're only concerned with determining its homotopy type. Explicit descriptions are often opaque and hard to work with, so hopefully this is what the person posing the problem had in mind.
answered Feb 1 at 18:08
WilliamWilliam
3,2111228
$endgroup$
1
$begingroup$
I like the hint and solution format.
$endgroup$
– user98602
Feb 1 at 18:34
add a comment |
$begingroup$
I'll give a couple hints, and then a solution below.
Hint 1: Notice that $S^1vee S^1$ and $S^1 times S^1$ are Eilenberg-Maclane spaces $K(mathbb{Z}*mathbb{Z},1)$ and $K(mathbb{Z}^2, 1)$ respectively.
Hint 2: The inclusion $S^1vee S^1 to S^1 times S^1$ induces the abelianization map on $pi_1$.
Solution
Let's say $F$ is the homotopy fibre of this map (using your favourite definition), so that $Fto S^1 vee S^1 to S^1 times S^1$ is equivalent to a fibration sequence, hence we get a long exact sequence of homotopy groups. By Hints 1 and 2, the long exact sequence of homotopy groups tells use that $Fsimeq K(C, 1)$ where $C$ is the commutator subgroup of $mathbb{Z}*mathbb{Z}$. This determines the (weak?) homotopy type of $F$ since the Eilenberg-Maclane property determines the homotopy type of a CW complex (and every space is weakly equivalent to a CW complex).
Since there are many different definitions of a homotopy fibre, usually we're only concerned with determining its homotopy type. Explicit descriptions are often opaque and hard to work with, so hopefully this is what the person posing the problem had in mind.
answered Feb 1 at 18:08
WilliamWilliam
3,2111228
$endgroup$
1
$begingroup$
I like the hint and solution format.
$endgroup$
– user98602
Feb 1 at 18:34
add a comment |
$begingroup$
I'll give a couple hints, and then a solution below.
Hint 1: Notice that $S^1vee S^1$ and $S^1 times S^1$ are Eilenberg-Maclane spaces $K(mathbb{Z}*mathbb{Z},1)$ and $K(mathbb{Z}^2, 1)$ respectively.
Hint 2: The inclusion $S^1vee S^1 to S^1 times S^1$ induces the abelianization map on $pi_1$.
Solution
Let's say $F$ is the homotopy fibre of this map (using your favourite definition), so that $Fto S^1 vee S^1 to S^1 times S^1$ is equivalent to a fibration sequence, hence we get a long exact sequence of homotopy groups. By Hints 1 and 2, the long exact sequence of homotopy groups tells use that $Fsimeq K(C, 1)$ where $C$ is the commutator subgroup of $mathbb{Z}*mathbb{Z}$. This determines the (weak?) homotopy type of $F$ since the Eilenberg-Maclane property determines the homotopy type of a CW complex (and every space is weakly equivalent to a CW complex).
Since there are many different definitions of a homotopy fibre, usually we're only concerned with determining its homotopy type. Explicit descriptions are often opaque and hard to work with, so hopefully this is what the person posing the problem had in mind.
answered Feb 1 at 18:08
WilliamWilliam
3,2111228
$endgroup$
I'll give a couple hints, and then a solution below.
Hint 1: Notice that $S^1vee S^1$ and $S^1 times S^1$ are Eilenberg-Maclane spaces $K(mathbb{Z}*mathbb{Z},1)$ and $K(mathbb{Z}^2, 1)$ respectively.
Hint 2: The inclusion $S^1vee S^1 to S^1 times S^1$ induces the abelianization map on $pi_1$.
Solution
Let's say $F$ is the homotopy fibre of this map (using your favourite definition), so that $Fto S^1 vee S^1 to S^1 times S^1$ is equivalent to a fibration sequence, hence we get a long exact sequence of homotopy groups. By Hints 1 and 2, the long exact sequence of homotopy groups tells use that $Fsimeq K(C, 1)$ where $C$ is the commutator subgroup of $mathbb{Z}*mathbb{Z}$. This determines the (weak?) homotopy type of $F$ since the Eilenberg-Maclane property determines the homotopy type of a CW complex (and every space is weakly equivalent to a CW complex).
Since there are many different definitions of a homotopy fibre, usually we're only concerned with determining its homotopy type. Explicit descriptions are often opaque and hard to work with, so hopefully this is what the person posing the problem had in mind.
answered Feb 1 at 18:08
WilliamWilliam
3,2111228
edited Feb 1 at 18:25
answered Feb 1 at 18:08
WilliamWilliam
3,2111228
answered Feb 1 at 18:08
WilliamWilliam
3,2111228
answered Feb 1 at 18:08
WilliamWilliam
3,2111228
3,2111228
1
$begingroup$
I like the hint and solution format.
$endgroup$
– user98602
Feb 1 at 18:34
add a comment |
1
$begingroup$
I like the hint and solution format.
$endgroup$
– user98602
Feb 1 at 18:34
1
1
$begingroup$
I like the hint and solution format.
$endgroup$
– user98602
Feb 1 at 18:34
$begingroup$
I like the hint and solution format.
$endgroup$
– user98602
Feb 1 at 18:34
add a comment |
$begingroup$
In general, the homotopy fiber of the inclusion $X vee Y hookrightarrow X times Y$ is homotopy equivalent to the join $Omega X * Omega Y$. To see this, note that one definition of the homotopy fiber is the pullback of the maps $X vee Y to X times Y$ and $P(X times Y) to X times Y$, which is $(Omega X times PY) cup_{Omega X times Omega Y} (PX times Omega Y) simeq Omega X * Omega Y$.
In the case $X = Y = S^1$, we have $Omega S^1 simeq mathbb{Z}$, so the homotopy fiber is homotopy equivalent to $mathbb{Z} * mathbb{Z}$, which is some infinite bouquet of circles.
answered Feb 1 at 22:04
JHFJHF
4,9711026
$endgroup$
add a comment |
$begingroup$
In general, the homotopy fiber of the inclusion $X vee Y hookrightarrow X times Y$ is homotopy equivalent to the join $Omega X * Omega Y$. To see this, note that one definition of the homotopy fiber is the pullback of the maps $X vee Y to X times Y$ and $P(X times Y) to X times Y$, which is $(Omega X times PY) cup_{Omega X times Omega Y} (PX times Omega Y) simeq Omega X * Omega Y$.
In the case $X = Y = S^1$, we have $Omega S^1 simeq mathbb{Z}$, so the homotopy fiber is homotopy equivalent to $mathbb{Z} * mathbb{Z}$, which is some infinite bouquet of circles.
answered Feb 1 at 22:04
JHFJHF
4,9711026
$endgroup$
add a comment |
$begingroup$
In general, the homotopy fiber of the inclusion $X vee Y hookrightarrow X times Y$ is homotopy equivalent to the join $Omega X * Omega Y$. To see this, note that one definition of the homotopy fiber is the pullback of the maps $X vee Y to X times Y$ and $P(X times Y) to X times Y$, which is $(Omega X times PY) cup_{Omega X times Omega Y} (PX times Omega Y) simeq Omega X * Omega Y$.
In the case $X = Y = S^1$, we have $Omega S^1 simeq mathbb{Z}$, so the homotopy fiber is homotopy equivalent to $mathbb{Z} * mathbb{Z}$, which is some infinite bouquet of circles.
answered Feb 1 at 22:04
JHFJHF
4,9711026
$endgroup$
In general, the homotopy fiber of the inclusion $X vee Y hookrightarrow X times Y$ is homotopy equivalent to the join $Omega X * Omega Y$. To see this, note that one definition of the homotopy fiber is the pullback of the maps $X vee Y to X times Y$ and $P(X times Y) to X times Y$, which is $(Omega X times PY) cup_{Omega X times Omega Y} (PX times Omega Y) simeq Omega X * Omega Y$.
In the case $X = Y = S^1$, we have $Omega S^1 simeq mathbb{Z}$, so the homotopy fiber is homotopy equivalent to $mathbb{Z} * mathbb{Z}$, which is some infinite bouquet of circles.
answered Feb 1 at 22:04
JHFJHF
4,9711026
answered Feb 1 at 22:04
JHFJHF
4,9711026
answered Feb 1 at 22:04
JHFJHF
4,9711026
answered Feb 1 at 22:04
JHFJHF
4,9711026
4,9711026
add a comment |
add a comment |
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$begingroup$
(Not that in the two answers $ast$ means something completely different.)
$endgroup$
– Pedro Tamaroff♦
Feb 2 at 0:51