How does $mathbb R^nsetminus 0$ being simply connected follow as a corollary from $mathbb S^n-1$ being a strong deformation retract?
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From Introduction to Topological Manifolds by Lee:
We know $mathbb S^n-1$ is simply connected for $nge 3$.
We know that since $mathbb S^n-1$ is a strong deformation retract of $mathbb R^nsetminus 0$, then $r circ iota = mathrmId_mathbb S^n-1$ and $iota circ r simeq mathrmId_mathbb R^nsetminus 0$, where $iota:mathbb S^n-1 hookrightarrow mathbb R^nsetminus 0$ is inclusion.
We do not know homotopy invariance of $pi_1$.
So, how does the corollary follow from the proposition?
general-topology algebraic-topology proof-explanation homotopy-theory
edited Aug 26 at 19:50
Pedro Tamaroff♦
94.4k10143292
asked Aug 2 at 17:22
Al Jebr
4,01943071
 |Â
show 20 more comments
up vote
6
down vote
favorite
From Introduction to Topological Manifolds by Lee:
We know $mathbb S^n-1$ is simply connected for $nge 3$.
We know that since $mathbb S^n-1$ is a strong deformation retract of $mathbb R^nsetminus 0$, then $r circ iota = mathrmId_mathbb S^n-1$ and $iota circ r simeq mathrmId_mathbb R^nsetminus 0$, where $iota:mathbb S^n-1 hookrightarrow mathbb R^nsetminus 0$ is inclusion.
We do not know homotopy invariance of $pi_1$.
So, how does the corollary follow from the proposition?
general-topology algebraic-topology proof-explanation homotopy-theory
edited Aug 26 at 19:50
Pedro Tamaroff♦
94.4k10143292
asked Aug 2 at 17:22
Al Jebr
4,01943071
1
If $X$ is a strong deformation retract of $Y$ then $X$ and $Y$ are homotopy equivalent.
– Lord Shark the Unknown
Aug 2 at 17:24
3
Even without any theory this is an easy result: take a loop in $mathbb R^n-0$, retract it into $S^n-1$, contract it into a point and you're done.
– Wojowu
Aug 2 at 17:31
1
@TedShifrin We know $r circ iota = mathrmid$ and so $r_* circ iota_* = mathrmid_*$. But we only know $iota circ r simeq mathrmid$. So, we don't know $iota_* circ r_* = mathrmid_*$.
– Al Jebr
Aug 2 at 17:32
5
Wojowu has given a proof that doesn't invoke homotopy invariance (the strong deformation retract provides a homotopy taking a curve in $Bbb R^n setminus 0$ to a curve in $S^n-1$, where you already know that you can homotope it to a point, so all loops may be contracted). But beyond that, I do not understand your resistance to using elementary facts that you should be building in the theory anyway: the point of $pi_1$ is that it is a homotopy invariant, and the proof (essentially Jason DeVito's comment that homotopic maps induce the same homomorphism) is very straightforward.
– Mike Miller
Aug 2 at 17:50
1
@Max I don't know the book, too. If it should introduce the fundamental group without studying induced maps and the basic features at least on the pointed homotopy category, it would be somewhat strange. But perhaps that comes later in the book. And you are right, most comments didn't consider the pointed vs. unpointed issue. So perhaps we should know how "simply connected" is defined in the book ($pi_1(X,x_0) = 0$ for some (or for all?) $x_0$, or each loop in $X$ is inessential?). But I believe it is not that important.
– Paul Frost
Aug 3 at 7:24
 |Â
show 20 more comments
up vote
6
down vote
favorite
up vote
6
down vote
favorite
From Introduction to Topological Manifolds by Lee:
We know $mathbb S^n-1$ is simply connected for $nge 3$.
We know that since $mathbb S^n-1$ is a strong deformation retract of $mathbb R^nsetminus 0$, then $r circ iota = mathrmId_mathbb S^n-1$ and $iota circ r simeq mathrmId_mathbb R^nsetminus 0$, where $iota:mathbb S^n-1 hookrightarrow mathbb R^nsetminus 0$ is inclusion.
We do not know homotopy invariance of $pi_1$.
So, how does the corollary follow from the proposition?
general-topology algebraic-topology proof-explanation homotopy-theory
edited Aug 26 at 19:50
Pedro Tamaroff♦
94.4k10143292
asked Aug 2 at 17:22
Al Jebr
4,01943071
From Introduction to Topological Manifolds by Lee:
We know $mathbb S^n-1$ is simply connected for $nge 3$.
We know that since $mathbb S^n-1$ is a strong deformation retract of $mathbb R^nsetminus 0$, then $r circ iota = mathrmId_mathbb S^n-1$ and $iota circ r simeq mathrmId_mathbb R^nsetminus 0$, where $iota:mathbb S^n-1 hookrightarrow mathbb R^nsetminus 0$ is inclusion.
We do not know homotopy invariance of $pi_1$.
So, how does the corollary follow from the proposition?
general-topology algebraic-topology proof-explanation homotopy-theory
edited Aug 26 at 19:50
Pedro Tamaroff♦
94.4k10143292
asked Aug 2 at 17:22
Al Jebr
4,01943071
edited Aug 26 at 19:50
Pedro Tamaroff♦
94.4k10143292
edited Aug 26 at 19:50
Pedro Tamaroff♦
94.4k10143292
edited Aug 26 at 19:50
Pedro Tamaroff♦
94.4k10143292
94.4k10143292
asked Aug 2 at 17:22
Al Jebr
4,01943071
asked Aug 2 at 17:22
Al Jebr
4,01943071
asked Aug 2 at 17:22
Al Jebr
4,01943071
4,01943071
1
If $X$ is a strong deformation retract of $Y$ then $X$ and $Y$ are homotopy equivalent.
– Lord Shark the Unknown
Aug 2 at 17:24
3
Even without any theory this is an easy result: take a loop in $mathbb R^n-0$, retract it into $S^n-1$, contract it into a point and you're done.
– Wojowu
Aug 2 at 17:31
1
@TedShifrin We know $r circ iota = mathrmid$ and so $r_* circ iota_* = mathrmid_*$. But we only know $iota circ r simeq mathrmid$. So, we don't know $iota_* circ r_* = mathrmid_*$.
– Al Jebr
Aug 2 at 17:32
5
Wojowu has given a proof that doesn't invoke homotopy invariance (the strong deformation retract provides a homotopy taking a curve in $Bbb R^n setminus 0$ to a curve in $S^n-1$, where you already know that you can homotope it to a point, so all loops may be contracted). But beyond that, I do not understand your resistance to using elementary facts that you should be building in the theory anyway: the point of $pi_1$ is that it is a homotopy invariant, and the proof (essentially Jason DeVito's comment that homotopic maps induce the same homomorphism) is very straightforward.
– Mike Miller
Aug 2 at 17:50
1
@Max I don't know the book, too. If it should introduce the fundamental group without studying induced maps and the basic features at least on the pointed homotopy category, it would be somewhat strange. But perhaps that comes later in the book. And you are right, most comments didn't consider the pointed vs. unpointed issue. So perhaps we should know how "simply connected" is defined in the book ($pi_1(X,x_0) = 0$ for some (or for all?) $x_0$, or each loop in $X$ is inessential?). But I believe it is not that important.
– Paul Frost
Aug 3 at 7:24
 |Â
show 20 more comments
1
If $X$ is a strong deformation retract of $Y$ then $X$ and $Y$ are homotopy equivalent.
– Lord Shark the Unknown
Aug 2 at 17:24
3
Even without any theory this is an easy result: take a loop in $mathbb R^n-0$, retract it into $S^n-1$, contract it into a point and you're done.
– Wojowu
Aug 2 at 17:31
1
@TedShifrin We know $r circ iota = mathrmid$ and so $r_* circ iota_* = mathrmid_*$. But we only know $iota circ r simeq mathrmid$. So, we don't know $iota_* circ r_* = mathrmid_*$.
– Al Jebr
Aug 2 at 17:32
5
Wojowu has given a proof that doesn't invoke homotopy invariance (the strong deformation retract provides a homotopy taking a curve in $Bbb R^n setminus 0$ to a curve in $S^n-1$, where you already know that you can homotope it to a point, so all loops may be contracted). But beyond that, I do not understand your resistance to using elementary facts that you should be building in the theory anyway: the point of $pi_1$ is that it is a homotopy invariant, and the proof (essentially Jason DeVito's comment that homotopic maps induce the same homomorphism) is very straightforward.
– Mike Miller
Aug 2 at 17:50
1
@Max I don't know the book, too. If it should introduce the fundamental group without studying induced maps and the basic features at least on the pointed homotopy category, it would be somewhat strange. But perhaps that comes later in the book. And you are right, most comments didn't consider the pointed vs. unpointed issue. So perhaps we should know how "simply connected" is defined in the book ($pi_1(X,x_0) = 0$ for some (or for all?) $x_0$, or each loop in $X$ is inessential?). But I believe it is not that important.
– Paul Frost
Aug 3 at 7:24
1
1
If $X$ is a strong deformation retract of $Y$ then $X$ and $Y$ are homotopy equivalent.
– Lord Shark the Unknown
Aug 2 at 17:24
If $X$ is a strong deformation retract of $Y$ then $X$ and $Y$ are homotopy equivalent.
– Lord Shark the Unknown
Aug 2 at 17:24
3
3
Even without any theory this is an easy result: take a loop in $mathbb R^n-0$, retract it into $S^n-1$, contract it into a point and you're done.
– Wojowu
Aug 2 at 17:31
Even without any theory this is an easy result: take a loop in $mathbb R^n-0$, retract it into $S^n-1$, contract it into a point and you're done.
– Wojowu
Aug 2 at 17:31
1
1
@TedShifrin We know $r circ iota = mathrmid$ and so $r_* circ iota_* = mathrmid_*$. But we only know $iota circ r simeq mathrmid$. So, we don't know $iota_* circ r_* = mathrmid_*$.
– Al Jebr
Aug 2 at 17:32
@TedShifrin We know $r circ iota = mathrmid$ and so $r_* circ iota_* = mathrmid_*$. But we only know $iota circ r simeq mathrmid$. So, we don't know $iota_* circ r_* = mathrmid_*$.
– Al Jebr
Aug 2 at 17:32
5
5
Wojowu has given a proof that doesn't invoke homotopy invariance (the strong deformation retract provides a homotopy taking a curve in $Bbb R^n setminus 0$ to a curve in $S^n-1$, where you already know that you can homotope it to a point, so all loops may be contracted). But beyond that, I do not understand your resistance to using elementary facts that you should be building in the theory anyway: the point of $pi_1$ is that it is a homotopy invariant, and the proof (essentially Jason DeVito's comment that homotopic maps induce the same homomorphism) is very straightforward.
– Mike Miller
Aug 2 at 17:50
Wojowu has given a proof that doesn't invoke homotopy invariance (the strong deformation retract provides a homotopy taking a curve in $Bbb R^n setminus 0$ to a curve in $S^n-1$, where you already know that you can homotope it to a point, so all loops may be contracted). But beyond that, I do not understand your resistance to using elementary facts that you should be building in the theory anyway: the point of $pi_1$ is that it is a homotopy invariant, and the proof (essentially Jason DeVito's comment that homotopic maps induce the same homomorphism) is very straightforward.
– Mike Miller
Aug 2 at 17:50
1
1
@Max I don't know the book, too. If it should introduce the fundamental group without studying induced maps and the basic features at least on the pointed homotopy category, it would be somewhat strange. But perhaps that comes later in the book. And you are right, most comments didn't consider the pointed vs. unpointed issue. So perhaps we should know how "simply connected" is defined in the book ($pi_1(X,x_0) = 0$ for some (or for all?) $x_0$, or each loop in $X$ is inessential?). But I believe it is not that important.
– Paul Frost
Aug 3 at 7:24
@Max I don't know the book, too. If it should introduce the fundamental group without studying induced maps and the basic features at least on the pointed homotopy category, it would be somewhat strange. But perhaps that comes later in the book. And you are right, most comments didn't consider the pointed vs. unpointed issue. So perhaps we should know how "simply connected" is defined in the book ($pi_1(X,x_0) = 0$ for some (or for all?) $x_0$, or each loop in $X$ is inessential?). But I believe it is not that important.
– Paul Frost
Aug 3 at 7:24
 |Â
show 20 more comments
1 Answer
1
active
oldest
votes
up vote
8
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accepted
Wow. This is just an oversight. Corollary 7.38 should have been delayed until after the statement of Theorem 7.40, which says that $pi_1$ is homotopy invariant. I've added a correction.
Thanks for pointing this out.
answered Aug 3 at 14:34
Jack Lee
25.5k44362
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
accepted
Wow. This is just an oversight. Corollary 7.38 should have been delayed until after the statement of Theorem 7.40, which says that $pi_1$ is homotopy invariant. I've added a correction.
Thanks for pointing this out.
answered Aug 3 at 14:34
Jack Lee
25.5k44362
add a comment |Â
up vote
8
down vote
accepted
Wow. This is just an oversight. Corollary 7.38 should have been delayed until after the statement of Theorem 7.40, which says that $pi_1$ is homotopy invariant. I've added a correction.
Thanks for pointing this out.
answered Aug 3 at 14:34
Jack Lee
25.5k44362
add a comment |Â
up vote
8
down vote
accepted
up vote
8
down vote
accepted
Wow. This is just an oversight. Corollary 7.38 should have been delayed until after the statement of Theorem 7.40, which says that $pi_1$ is homotopy invariant. I've added a correction.
Thanks for pointing this out.
answered Aug 3 at 14:34
Jack Lee
25.5k44362
Wow. This is just an oversight. Corollary 7.38 should have been delayed until after the statement of Theorem 7.40, which says that $pi_1$ is homotopy invariant. I've added a correction.
Thanks for pointing this out.
answered Aug 3 at 14:34
Jack Lee
25.5k44362
answered Aug 3 at 14:34
Jack Lee
25.5k44362
answered Aug 3 at 14:34
Jack Lee
25.5k44362
answered Aug 3 at 14:34
Jack Lee
25.5k44362
25.5k44362
add a comment |Â
add a comment |Â
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1
If $X$ is a strong deformation retract of $Y$ then $X$ and $Y$ are homotopy equivalent.
– Lord Shark the Unknown
Aug 2 at 17:24
3
Even without any theory this is an easy result: take a loop in $mathbb R^n-0$, retract it into $S^n-1$, contract it into a point and you're done.
– Wojowu
Aug 2 at 17:31
1
@TedShifrin We know $r circ iota = mathrmid$ and so $r_* circ iota_* = mathrmid_*$. But we only know $iota circ r simeq mathrmid$. So, we don't know $iota_* circ r_* = mathrmid_*$.
– Al Jebr
Aug 2 at 17:32
5
Wojowu has given a proof that doesn't invoke homotopy invariance (the strong deformation retract provides a homotopy taking a curve in $Bbb R^n setminus 0$ to a curve in $S^n-1$, where you already know that you can homotope it to a point, so all loops may be contracted). But beyond that, I do not understand your resistance to using elementary facts that you should be building in the theory anyway: the point of $pi_1$ is that it is a homotopy invariant, and the proof (essentially Jason DeVito's comment that homotopic maps induce the same homomorphism) is very straightforward.
– Mike Miller
Aug 2 at 17:50
1
@Max I don't know the book, too. If it should introduce the fundamental group without studying induced maps and the basic features at least on the pointed homotopy category, it would be somewhat strange. But perhaps that comes later in the book. And you are right, most comments didn't consider the pointed vs. unpointed issue. So perhaps we should know how "simply connected" is defined in the book ($pi_1(X,x_0) = 0$ for some (or for all?) $x_0$, or each loop in $X$ is inessential?). But I believe it is not that important.
– Paul Frost
Aug 3 at 7:24