Homotopy of continuous map from a space with finite fundamental group
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Problem source: 2c on the UMD January, 2018 topology qualifying exam, seen here https://www-math.umd.edu/images/pdfs/quals/Topology/Topology-January-2018.pdf I have an argument for it, but I am not at all sure that it is correct.
In particular, I would like to know 1) if it is correct and 2) if there is a better way of doing it (especially if my way is incorrect).
Let X be a connected CW complex with $pi_1(X)$ finite. Let Y be a CW complex. Let $f:Xto Ytimes S^1$ be a continuous map. Show there exists a $g:Xto Y$ such that $f$ is homotopic to the map $(g,1):Xto Ytimes S^1$.
I am fairly confident in the first part of the proof, ie that $|f_*pi_1(X)|<infty$ (although my group theory is rusty). Since $f$ is continuous, it induces a group homomorphism on $pi_1(X)$, so 1) $f_*pi_1(X)simeqpi_1(X)/Ker(f_*)$ and 2) $f_*pi_1(X)leqpi_1(Ytimes S^1)simeqpi_1(Y)times mathbb{Z}$. In particular, (1) gives that $f_*pi_1(X)$ is finite, hence $simeq Htimes {0}$, where $H$ is some subgroup of $pi_1(Y)$, by (2).
The part I am much less sure of (that it is either correct or necessary) is the remainder. Take $p:Ytimesmathbb{R}to Ytimes S^1$ by $p(y,x)=(y,e^{2pi ix})$. This is a covering map, and $p_*pi_1(Ytimesmathbb{R})=pi_1(Y)times{0}$, so $f$ lifts to some $tilde{f}$. Since the image of any closed loop in $X$ is taken to a closed loop that does not wind around $S^1$ (else, we would get a generator for an infinite fundamental group), $tilde{f}$ is homotopic to some $(g,0)$, which maps to $(g,1)$ under $p$.
I have not used the fact that X and Y are CW complexes anywhere, but the only theorem that I know offhand in the context of CW complexes, subsets, and homotopy of functions would be the Homotopy Extension Theorem, which wouldn't make sense here (since $f(X)$ is not necessarily a CW complex).
algebraic-topology homotopy-theory fundamental-groups cw-complexes
asked Dec 28 '18 at 22:47
PepperPepper
508
$endgroup$
add a comment |
$begingroup$
Problem source: 2c on the UMD January, 2018 topology qualifying exam, seen here https://www-math.umd.edu/images/pdfs/quals/Topology/Topology-January-2018.pdf I have an argument for it, but I am not at all sure that it is correct.
In particular, I would like to know 1) if it is correct and 2) if there is a better way of doing it (especially if my way is incorrect).
Let X be a connected CW complex with $pi_1(X)$ finite. Let Y be a CW complex. Let $f:Xto Ytimes S^1$ be a continuous map. Show there exists a $g:Xto Y$ such that $f$ is homotopic to the map $(g,1):Xto Ytimes S^1$.
I am fairly confident in the first part of the proof, ie that $|f_*pi_1(X)|<infty$ (although my group theory is rusty). Since $f$ is continuous, it induces a group homomorphism on $pi_1(X)$, so 1) $f_*pi_1(X)simeqpi_1(X)/Ker(f_*)$ and 2) $f_*pi_1(X)leqpi_1(Ytimes S^1)simeqpi_1(Y)times mathbb{Z}$. In particular, (1) gives that $f_*pi_1(X)$ is finite, hence $simeq Htimes {0}$, where $H$ is some subgroup of $pi_1(Y)$, by (2).
The part I am much less sure of (that it is either correct or necessary) is the remainder. Take $p:Ytimesmathbb{R}to Ytimes S^1$ by $p(y,x)=(y,e^{2pi ix})$. This is a covering map, and $p_*pi_1(Ytimesmathbb{R})=pi_1(Y)times{0}$, so $f$ lifts to some $tilde{f}$. Since the image of any closed loop in $X$ is taken to a closed loop that does not wind around $S^1$ (else, we would get a generator for an infinite fundamental group), $tilde{f}$ is homotopic to some $(g,0)$, which maps to $(g,1)$ under $p$.
I have not used the fact that X and Y are CW complexes anywhere, but the only theorem that I know offhand in the context of CW complexes, subsets, and homotopy of functions would be the Homotopy Extension Theorem, which wouldn't make sense here (since $f(X)$ is not necessarily a CW complex).
algebraic-topology homotopy-theory fundamental-groups cw-complexes
asked Dec 28 '18 at 22:47
PepperPepper
508
$endgroup$
$begingroup$
The map to $Y$ is a red herring, since maps into a product are determined by the maps into the factors. For a fancy proof using cohomology/homology see: math.stackexchange.com/questions/380383/… .
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– Justin Young
Dec 30 '18 at 13:30
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That's a very nice proof, although I'll need to make it to homology in my notes before I can properly parse what's going on.
$endgroup$
– Pepper
Dec 31 '18 at 14:10
add a comment |
$begingroup$
Problem source: 2c on the UMD January, 2018 topology qualifying exam, seen here https://www-math.umd.edu/images/pdfs/quals/Topology/Topology-January-2018.pdf I have an argument for it, but I am not at all sure that it is correct.
In particular, I would like to know 1) if it is correct and 2) if there is a better way of doing it (especially if my way is incorrect).
Let X be a connected CW complex with $pi_1(X)$ finite. Let Y be a CW complex. Let $f:Xto Ytimes S^1$ be a continuous map. Show there exists a $g:Xto Y$ such that $f$ is homotopic to the map $(g,1):Xto Ytimes S^1$.
I am fairly confident in the first part of the proof, ie that $|f_*pi_1(X)|<infty$ (although my group theory is rusty). Since $f$ is continuous, it induces a group homomorphism on $pi_1(X)$, so 1) $f_*pi_1(X)simeqpi_1(X)/Ker(f_*)$ and 2) $f_*pi_1(X)leqpi_1(Ytimes S^1)simeqpi_1(Y)times mathbb{Z}$. In particular, (1) gives that $f_*pi_1(X)$ is finite, hence $simeq Htimes {0}$, where $H$ is some subgroup of $pi_1(Y)$, by (2).
The part I am much less sure of (that it is either correct or necessary) is the remainder. Take $p:Ytimesmathbb{R}to Ytimes S^1$ by $p(y,x)=(y,e^{2pi ix})$. This is a covering map, and $p_*pi_1(Ytimesmathbb{R})=pi_1(Y)times{0}$, so $f$ lifts to some $tilde{f}$. Since the image of any closed loop in $X$ is taken to a closed loop that does not wind around $S^1$ (else, we would get a generator for an infinite fundamental group), $tilde{f}$ is homotopic to some $(g,0)$, which maps to $(g,1)$ under $p$.
I have not used the fact that X and Y are CW complexes anywhere, but the only theorem that I know offhand in the context of CW complexes, subsets, and homotopy of functions would be the Homotopy Extension Theorem, which wouldn't make sense here (since $f(X)$ is not necessarily a CW complex).
algebraic-topology homotopy-theory fundamental-groups cw-complexes
asked Dec 28 '18 at 22:47
PepperPepper
508
$endgroup$
Problem source: 2c on the UMD January, 2018 topology qualifying exam, seen here https://www-math.umd.edu/images/pdfs/quals/Topology/Topology-January-2018.pdf I have an argument for it, but I am not at all sure that it is correct.
In particular, I would like to know 1) if it is correct and 2) if there is a better way of doing it (especially if my way is incorrect).
Let X be a connected CW complex with $pi_1(X)$ finite. Let Y be a CW complex. Let $f:Xto Ytimes S^1$ be a continuous map. Show there exists a $g:Xto Y$ such that $f$ is homotopic to the map $(g,1):Xto Ytimes S^1$.
I am fairly confident in the first part of the proof, ie that $|f_*pi_1(X)|<infty$ (although my group theory is rusty). Since $f$ is continuous, it induces a group homomorphism on $pi_1(X)$, so 1) $f_*pi_1(X)simeqpi_1(X)/Ker(f_*)$ and 2) $f_*pi_1(X)leqpi_1(Ytimes S^1)simeqpi_1(Y)times mathbb{Z}$. In particular, (1) gives that $f_*pi_1(X)$ is finite, hence $simeq Htimes {0}$, where $H$ is some subgroup of $pi_1(Y)$, by (2).
The part I am much less sure of (that it is either correct or necessary) is the remainder. Take $p:Ytimesmathbb{R}to Ytimes S^1$ by $p(y,x)=(y,e^{2pi ix})$. This is a covering map, and $p_*pi_1(Ytimesmathbb{R})=pi_1(Y)times{0}$, so $f$ lifts to some $tilde{f}$. Since the image of any closed loop in $X$ is taken to a closed loop that does not wind around $S^1$ (else, we would get a generator for an infinite fundamental group), $tilde{f}$ is homotopic to some $(g,0)$, which maps to $(g,1)$ under $p$.
I have not used the fact that X and Y are CW complexes anywhere, but the only theorem that I know offhand in the context of CW complexes, subsets, and homotopy of functions would be the Homotopy Extension Theorem, which wouldn't make sense here (since $f(X)$ is not necessarily a CW complex).
algebraic-topology homotopy-theory fundamental-groups cw-complexes
algebraic-topology homotopy-theory fundamental-groups cw-complexes
asked Dec 28 '18 at 22:47
PepperPepper
508
asked Dec 28 '18 at 22:47
PepperPepper
508
edited Dec 29 '18 at 3:47
Pepper
asked Dec 28 '18 at 22:47
PepperPepper
508
asked Dec 28 '18 at 22:47
PepperPepper
508
asked Dec 28 '18 at 22:47
PepperPepper
508
508
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The map to $Y$ is a red herring, since maps into a product are determined by the maps into the factors. For a fancy proof using cohomology/homology see: math.stackexchange.com/questions/380383/… .
$endgroup$
– Justin Young
Dec 30 '18 at 13:30
$begingroup$
That's a very nice proof, although I'll need to make it to homology in my notes before I can properly parse what's going on.
$endgroup$
– Pepper
Dec 31 '18 at 14:10
add a comment |
$begingroup$
The map to $Y$ is a red herring, since maps into a product are determined by the maps into the factors. For a fancy proof using cohomology/homology see: math.stackexchange.com/questions/380383/… .
$endgroup$
– Justin Young
Dec 30 '18 at 13:30
$begingroup$
That's a very nice proof, although I'll need to make it to homology in my notes before I can properly parse what's going on.
$endgroup$
– Pepper
Dec 31 '18 at 14:10
$begingroup$
The map to $Y$ is a red herring, since maps into a product are determined by the maps into the factors. For a fancy proof using cohomology/homology see: math.stackexchange.com/questions/380383/… .
$endgroup$
– Justin Young
Dec 30 '18 at 13:30
$begingroup$
The map to $Y$ is a red herring, since maps into a product are determined by the maps into the factors. For a fancy proof using cohomology/homology see: math.stackexchange.com/questions/380383/… .
$endgroup$
– Justin Young
Dec 30 '18 at 13:30
$begingroup$
That's a very nice proof, although I'll need to make it to homology in my notes before I can properly parse what's going on.
$endgroup$
– Pepper
Dec 31 '18 at 14:10
$begingroup$
That's a very nice proof, although I'll need to make it to homology in my notes before I can properly parse what's going on.
$endgroup$
– Pepper
Dec 31 '18 at 14:10
add a comment |
1 Answer
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This is basically correct and is the natural argument to make but your argument at the end is unclear. The reason that $tilde{f}$ is homotopic to some $(g,0)$ is simply that $mathbb{R}$ is contractible, so you can take $tilde{f}$ and contract its second coordinate to $0$ via a homotopy. Your explanation seemed to be mixing this up with the reason that the lift $tilde{f}$ exists at all.
As for where the assumption that the spaces are CW-complexes comes in, you need something about $X$ to be able to say that the lift $tilde{f}$ exists at all (the usual lifting theorems for covering spaces only work for sufficiently nice spaces). In particular, it suffices to know that $X$ is path-connected and locally path-connected, which follows if you know that $X$ is a connected CW-complex. The assumption that $Y$ is a CW-complex is completely unnecessary.
By the way, the fact that $f_*pi_1(X)$ is finite is completely trivial and requires no group theory: it is the image of the finite set $pi_1(X)$ under a function.
answered Dec 28 '18 at 23:13
Eric WofseyEric Wofsey
193k14221352
$endgroup$
$begingroup$
So, in the lifting portion of the argument, it follows as soon as there is a lift to $Ytimesmathbb{R}$? The comments about not winding around $S^1$ were holdovers from my first attempt, and things got a little muddled along the way.Also, thank you for pointing out that I implicitly used X path connected and locally path connected - I had forgotten that that was a requirement for the theorem I used for the existence of a lift.
$endgroup$
– Pepper
Dec 29 '18 at 3:53
$begingroup$
The following seems a much simpler argument, is it incorrect? Let $p: Ytimes S^1to S^1$ be the projection, it follows from $pi_1(X)$ finite that $(pcirc f)_*$ is constant and $pcirc f$ is null-homotop. If $h_t$ is the homotopy contracting the map, then $(f_Y, h_t)$ is a homotopy between $f=(f_Y,f_{S^1})$ and $g=(f_Y,1)$.
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– s.harp
Dec 29 '18 at 18:04
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@s.harp: That's essentially the same argument when you fill in the details. In particular, to prove $pcirc f$ is nullhomotopic you need to lift it to the universal cover. So you're doing the same thing, just separating out the second coordinate of $f$ before lifting to a covering space instead of after.
$endgroup$
– Eric Wofsey
Dec 29 '18 at 18:10
$begingroup$
Ah, you are right.
$endgroup$
– s.harp
Dec 29 '18 at 18:12
add a comment |
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$begingroup$
This is basically correct and is the natural argument to make but your argument at the end is unclear. The reason that $tilde{f}$ is homotopic to some $(g,0)$ is simply that $mathbb{R}$ is contractible, so you can take $tilde{f}$ and contract its second coordinate to $0$ via a homotopy. Your explanation seemed to be mixing this up with the reason that the lift $tilde{f}$ exists at all.
As for where the assumption that the spaces are CW-complexes comes in, you need something about $X$ to be able to say that the lift $tilde{f}$ exists at all (the usual lifting theorems for covering spaces only work for sufficiently nice spaces). In particular, it suffices to know that $X$ is path-connected and locally path-connected, which follows if you know that $X$ is a connected CW-complex. The assumption that $Y$ is a CW-complex is completely unnecessary.
By the way, the fact that $f_*pi_1(X)$ is finite is completely trivial and requires no group theory: it is the image of the finite set $pi_1(X)$ under a function.
answered Dec 28 '18 at 23:13
Eric WofseyEric Wofsey
193k14221352
$endgroup$
$begingroup$
So, in the lifting portion of the argument, it follows as soon as there is a lift to $Ytimesmathbb{R}$? The comments about not winding around $S^1$ were holdovers from my first attempt, and things got a little muddled along the way.Also, thank you for pointing out that I implicitly used X path connected and locally path connected - I had forgotten that that was a requirement for the theorem I used for the existence of a lift.
$endgroup$
– Pepper
Dec 29 '18 at 3:53
$begingroup$
The following seems a much simpler argument, is it incorrect? Let $p: Ytimes S^1to S^1$ be the projection, it follows from $pi_1(X)$ finite that $(pcirc f)_*$ is constant and $pcirc f$ is null-homotop. If $h_t$ is the homotopy contracting the map, then $(f_Y, h_t)$ is a homotopy between $f=(f_Y,f_{S^1})$ and $g=(f_Y,1)$.
$endgroup$
– s.harp
Dec 29 '18 at 18:04
$begingroup$
@s.harp: That's essentially the same argument when you fill in the details. In particular, to prove $pcirc f$ is nullhomotopic you need to lift it to the universal cover. So you're doing the same thing, just separating out the second coordinate of $f$ before lifting to a covering space instead of after.
$endgroup$
– Eric Wofsey
Dec 29 '18 at 18:10
$begingroup$
Ah, you are right.
$endgroup$
– s.harp
Dec 29 '18 at 18:12
add a comment |
$begingroup$
This is basically correct and is the natural argument to make but your argument at the end is unclear. The reason that $tilde{f}$ is homotopic to some $(g,0)$ is simply that $mathbb{R}$ is contractible, so you can take $tilde{f}$ and contract its second coordinate to $0$ via a homotopy. Your explanation seemed to be mixing this up with the reason that the lift $tilde{f}$ exists at all.
As for where the assumption that the spaces are CW-complexes comes in, you need something about $X$ to be able to say that the lift $tilde{f}$ exists at all (the usual lifting theorems for covering spaces only work for sufficiently nice spaces). In particular, it suffices to know that $X$ is path-connected and locally path-connected, which follows if you know that $X$ is a connected CW-complex. The assumption that $Y$ is a CW-complex is completely unnecessary.
By the way, the fact that $f_*pi_1(X)$ is finite is completely trivial and requires no group theory: it is the image of the finite set $pi_1(X)$ under a function.
answered Dec 28 '18 at 23:13
Eric WofseyEric Wofsey
193k14221352
$endgroup$
$begingroup$
So, in the lifting portion of the argument, it follows as soon as there is a lift to $Ytimesmathbb{R}$? The comments about not winding around $S^1$ were holdovers from my first attempt, and things got a little muddled along the way.Also, thank you for pointing out that I implicitly used X path connected and locally path connected - I had forgotten that that was a requirement for the theorem I used for the existence of a lift.
$endgroup$
– Pepper
Dec 29 '18 at 3:53
$begingroup$
The following seems a much simpler argument, is it incorrect? Let $p: Ytimes S^1to S^1$ be the projection, it follows from $pi_1(X)$ finite that $(pcirc f)_*$ is constant and $pcirc f$ is null-homotop. If $h_t$ is the homotopy contracting the map, then $(f_Y, h_t)$ is a homotopy between $f=(f_Y,f_{S^1})$ and $g=(f_Y,1)$.
$endgroup$
– s.harp
Dec 29 '18 at 18:04
$begingroup$
@s.harp: That's essentially the same argument when you fill in the details. In particular, to prove $pcirc f$ is nullhomotopic you need to lift it to the universal cover. So you're doing the same thing, just separating out the second coordinate of $f$ before lifting to a covering space instead of after.
$endgroup$
– Eric Wofsey
Dec 29 '18 at 18:10
$begingroup$
Ah, you are right.
$endgroup$
– s.harp
Dec 29 '18 at 18:12
add a comment |
$begingroup$
This is basically correct and is the natural argument to make but your argument at the end is unclear. The reason that $tilde{f}$ is homotopic to some $(g,0)$ is simply that $mathbb{R}$ is contractible, so you can take $tilde{f}$ and contract its second coordinate to $0$ via a homotopy. Your explanation seemed to be mixing this up with the reason that the lift $tilde{f}$ exists at all.
As for where the assumption that the spaces are CW-complexes comes in, you need something about $X$ to be able to say that the lift $tilde{f}$ exists at all (the usual lifting theorems for covering spaces only work for sufficiently nice spaces). In particular, it suffices to know that $X$ is path-connected and locally path-connected, which follows if you know that $X$ is a connected CW-complex. The assumption that $Y$ is a CW-complex is completely unnecessary.
By the way, the fact that $f_*pi_1(X)$ is finite is completely trivial and requires no group theory: it is the image of the finite set $pi_1(X)$ under a function.
answered Dec 28 '18 at 23:13
Eric WofseyEric Wofsey
193k14221352
$endgroup$
This is basically correct and is the natural argument to make but your argument at the end is unclear. The reason that $tilde{f}$ is homotopic to some $(g,0)$ is simply that $mathbb{R}$ is contractible, so you can take $tilde{f}$ and contract its second coordinate to $0$ via a homotopy. Your explanation seemed to be mixing this up with the reason that the lift $tilde{f}$ exists at all.
As for where the assumption that the spaces are CW-complexes comes in, you need something about $X$ to be able to say that the lift $tilde{f}$ exists at all (the usual lifting theorems for covering spaces only work for sufficiently nice spaces). In particular, it suffices to know that $X$ is path-connected and locally path-connected, which follows if you know that $X$ is a connected CW-complex. The assumption that $Y$ is a CW-complex is completely unnecessary.
By the way, the fact that $f_*pi_1(X)$ is finite is completely trivial and requires no group theory: it is the image of the finite set $pi_1(X)$ under a function.
answered Dec 28 '18 at 23:13
Eric WofseyEric Wofsey
193k14221352
answered Dec 28 '18 at 23:13
Eric WofseyEric Wofsey
193k14221352
answered Dec 28 '18 at 23:13
Eric WofseyEric Wofsey
193k14221352
answered Dec 28 '18 at 23:13
Eric WofseyEric Wofsey
193k14221352
193k14221352
$begingroup$
So, in the lifting portion of the argument, it follows as soon as there is a lift to $Ytimesmathbb{R}$? The comments about not winding around $S^1$ were holdovers from my first attempt, and things got a little muddled along the way.Also, thank you for pointing out that I implicitly used X path connected and locally path connected - I had forgotten that that was a requirement for the theorem I used for the existence of a lift.
$endgroup$
– Pepper
Dec 29 '18 at 3:53
$begingroup$
The following seems a much simpler argument, is it incorrect? Let $p: Ytimes S^1to S^1$ be the projection, it follows from $pi_1(X)$ finite that $(pcirc f)_*$ is constant and $pcirc f$ is null-homotop. If $h_t$ is the homotopy contracting the map, then $(f_Y, h_t)$ is a homotopy between $f=(f_Y,f_{S^1})$ and $g=(f_Y,1)$.
$endgroup$
– s.harp
Dec 29 '18 at 18:04
$begingroup$
@s.harp: That's essentially the same argument when you fill in the details. In particular, to prove $pcirc f$ is nullhomotopic you need to lift it to the universal cover. So you're doing the same thing, just separating out the second coordinate of $f$ before lifting to a covering space instead of after.
$endgroup$
– Eric Wofsey
Dec 29 '18 at 18:10
$begingroup$
Ah, you are right.
$endgroup$
– s.harp
Dec 29 '18 at 18:12
add a comment |
$begingroup$
So, in the lifting portion of the argument, it follows as soon as there is a lift to $Ytimesmathbb{R}$? The comments about not winding around $S^1$ were holdovers from my first attempt, and things got a little muddled along the way.Also, thank you for pointing out that I implicitly used X path connected and locally path connected - I had forgotten that that was a requirement for the theorem I used for the existence of a lift.
$endgroup$
– Pepper
Dec 29 '18 at 3:53
$begingroup$
The following seems a much simpler argument, is it incorrect? Let $p: Ytimes S^1to S^1$ be the projection, it follows from $pi_1(X)$ finite that $(pcirc f)_*$ is constant and $pcirc f$ is null-homotop. If $h_t$ is the homotopy contracting the map, then $(f_Y, h_t)$ is a homotopy between $f=(f_Y,f_{S^1})$ and $g=(f_Y,1)$.
$endgroup$
– s.harp
Dec 29 '18 at 18:04
$begingroup$
@s.harp: That's essentially the same argument when you fill in the details. In particular, to prove $pcirc f$ is nullhomotopic you need to lift it to the universal cover. So you're doing the same thing, just separating out the second coordinate of $f$ before lifting to a covering space instead of after.
$endgroup$
– Eric Wofsey
Dec 29 '18 at 18:10
$begingroup$
Ah, you are right.
$endgroup$
– s.harp
Dec 29 '18 at 18:12
$begingroup$
So, in the lifting portion of the argument, it follows as soon as there is a lift to $Ytimesmathbb{R}$? The comments about not winding around $S^1$ were holdovers from my first attempt, and things got a little muddled along the way.Also, thank you for pointing out that I implicitly used X path connected and locally path connected - I had forgotten that that was a requirement for the theorem I used for the existence of a lift.
$endgroup$
– Pepper
Dec 29 '18 at 3:53
$begingroup$
So, in the lifting portion of the argument, it follows as soon as there is a lift to $Ytimesmathbb{R}$? The comments about not winding around $S^1$ were holdovers from my first attempt, and things got a little muddled along the way.Also, thank you for pointing out that I implicitly used X path connected and locally path connected - I had forgotten that that was a requirement for the theorem I used for the existence of a lift.
$endgroup$
– Pepper
Dec 29 '18 at 3:53
$begingroup$
The following seems a much simpler argument, is it incorrect? Let $p: Ytimes S^1to S^1$ be the projection, it follows from $pi_1(X)$ finite that $(pcirc f)_*$ is constant and $pcirc f$ is null-homotop. If $h_t$ is the homotopy contracting the map, then $(f_Y, h_t)$ is a homotopy between $f=(f_Y,f_{S^1})$ and $g=(f_Y,1)$.
$endgroup$
– s.harp
Dec 29 '18 at 18:04
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The following seems a much simpler argument, is it incorrect? Let $p: Ytimes S^1to S^1$ be the projection, it follows from $pi_1(X)$ finite that $(pcirc f)_*$ is constant and $pcirc f$ is null-homotop. If $h_t$ is the homotopy contracting the map, then $(f_Y, h_t)$ is a homotopy between $f=(f_Y,f_{S^1})$ and $g=(f_Y,1)$.
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– s.harp
Dec 29 '18 at 18:04
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@s.harp: That's essentially the same argument when you fill in the details. In particular, to prove $pcirc f$ is nullhomotopic you need to lift it to the universal cover. So you're doing the same thing, just separating out the second coordinate of $f$ before lifting to a covering space instead of after.
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– Eric Wofsey
Dec 29 '18 at 18:10
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@s.harp: That's essentially the same argument when you fill in the details. In particular, to prove $pcirc f$ is nullhomotopic you need to lift it to the universal cover. So you're doing the same thing, just separating out the second coordinate of $f$ before lifting to a covering space instead of after.
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– Eric Wofsey
Dec 29 '18 at 18:10
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Ah, you are right.
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– s.harp
Dec 29 '18 at 18:12
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Ah, you are right.
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– s.harp
Dec 29 '18 at 18:12
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The map to $Y$ is a red herring, since maps into a product are determined by the maps into the factors. For a fancy proof using cohomology/homology see: math.stackexchange.com/questions/380383/… .
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– Justin Young
Dec 30 '18 at 13:30
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That's a very nice proof, although I'll need to make it to homology in my notes before I can properly parse what's going on.
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– Pepper
Dec 31 '18 at 14:10