Mixing
Is every map from $mathbb{R}P^2 to S^2$ nullhomotopic?
$begingroup$
I think the answer is no, but I'm not sure. Consider the CW-structure on $mathbb{R}P^2$ given by one $0$-cell $x$, one oriented $1$-cell $a$ attached to $x$ at both ends, and one $2$-cell whose boundary is glued along the loop $a^2$. Let $X^1 subset mathbb{R}P^2$ be the $1$-skeleton. My spidey senses are suggesting the map $mathbb{R}P^2 to mathbb{R}P^2/X^1 cong S^2$ given by crushing the $1$-skeleton to a point should be homotopically nontrivial. Is this true? If so, how does one prove this?
algebraic-topology homotopy-theory projective-space
asked Jan 5 at 22:50
Sameer KailasaSameer Kailasa
5,50621843
$endgroup$
add a comment |
$begingroup$
I think the answer is no, but I'm not sure. Consider the CW-structure on $mathbb{R}P^2$ given by one $0$-cell $x$, one oriented $1$-cell $a$ attached to $x$ at both ends, and one $2$-cell whose boundary is glued along the loop $a^2$. Let $X^1 subset mathbb{R}P^2$ be the $1$-skeleton. My spidey senses are suggesting the map $mathbb{R}P^2 to mathbb{R}P^2/X^1 cong S^2$ given by crushing the $1$-skeleton to a point should be homotopically nontrivial. Is this true? If so, how does one prove this?
algebraic-topology homotopy-theory projective-space
asked Jan 5 at 22:50
Sameer KailasaSameer Kailasa
5,50621843
$endgroup$
6
$begingroup$
How about considering $H_2$ with coefficients in $Bbb Z_2$?
$endgroup$
– Lord Shark the Unknown
Jan 5 at 22:56
1
$begingroup$
... Or $H^2$ with integer coefficients.
$endgroup$
– Tyrone
Jan 6 at 12:26
1
$begingroup$
One way to see the above is to use cellular chains, if you take the CW-structure on $S^2$ with one $0$-cell and one $2$-cell, you will see that your map is cellular, and maps the $2$-cell to the $2$-cell. Then, if you calculate the cellular chain complex with $mathbb Z/2$ coefficients, you clearly get the identity on $H_2$, which is not possible for a nullhomotopic map.
$endgroup$
– Justin Young
Jan 9 at 18:32
add a comment |
$begingroup$
I think the answer is no, but I'm not sure. Consider the CW-structure on $mathbb{R}P^2$ given by one $0$-cell $x$, one oriented $1$-cell $a$ attached to $x$ at both ends, and one $2$-cell whose boundary is glued along the loop $a^2$. Let $X^1 subset mathbb{R}P^2$ be the $1$-skeleton. My spidey senses are suggesting the map $mathbb{R}P^2 to mathbb{R}P^2/X^1 cong S^2$ given by crushing the $1$-skeleton to a point should be homotopically nontrivial. Is this true? If so, how does one prove this?
algebraic-topology homotopy-theory projective-space
asked Jan 5 at 22:50
Sameer KailasaSameer Kailasa
5,50621843
$endgroup$
I think the answer is no, but I'm not sure. Consider the CW-structure on $mathbb{R}P^2$ given by one $0$-cell $x$, one oriented $1$-cell $a$ attached to $x$ at both ends, and one $2$-cell whose boundary is glued along the loop $a^2$. Let $X^1 subset mathbb{R}P^2$ be the $1$-skeleton. My spidey senses are suggesting the map $mathbb{R}P^2 to mathbb{R}P^2/X^1 cong S^2$ given by crushing the $1$-skeleton to a point should be homotopically nontrivial. Is this true? If so, how does one prove this?
algebraic-topology homotopy-theory projective-space
algebraic-topology homotopy-theory projective-space
asked Jan 5 at 22:50
Sameer KailasaSameer Kailasa
5,50621843
asked Jan 5 at 22:50
Sameer KailasaSameer Kailasa
5,50621843
asked Jan 5 at 22:50
Sameer KailasaSameer Kailasa
5,50621843
asked Jan 5 at 22:50
Sameer KailasaSameer Kailasa
5,50621843
asked Jan 5 at 22:50
Sameer KailasaSameer Kailasa
5,50621843
5,50621843
6
$begingroup$
How about considering $H_2$ with coefficients in $Bbb Z_2$?
$endgroup$
– Lord Shark the Unknown
Jan 5 at 22:56
1
$begingroup$
... Or $H^2$ with integer coefficients.
$endgroup$
– Tyrone
Jan 6 at 12:26
1
$begingroup$
One way to see the above is to use cellular chains, if you take the CW-structure on $S^2$ with one $0$-cell and one $2$-cell, you will see that your map is cellular, and maps the $2$-cell to the $2$-cell. Then, if you calculate the cellular chain complex with $mathbb Z/2$ coefficients, you clearly get the identity on $H_2$, which is not possible for a nullhomotopic map.
$endgroup$
– Justin Young
Jan 9 at 18:32
add a comment |
6
$begingroup$
How about considering $H_2$ with coefficients in $Bbb Z_2$?
$endgroup$
– Lord Shark the Unknown
Jan 5 at 22:56
1
$begingroup$
... Or $H^2$ with integer coefficients.
$endgroup$
– Tyrone
Jan 6 at 12:26
1
$begingroup$
One way to see the above is to use cellular chains, if you take the CW-structure on $S^2$ with one $0$-cell and one $2$-cell, you will see that your map is cellular, and maps the $2$-cell to the $2$-cell. Then, if you calculate the cellular chain complex with $mathbb Z/2$ coefficients, you clearly get the identity on $H_2$, which is not possible for a nullhomotopic map.
$endgroup$
– Justin Young
Jan 9 at 18:32
6
6
$begingroup$
How about considering $H_2$ with coefficients in $Bbb Z_2$?
$endgroup$
– Lord Shark the Unknown
Jan 5 at 22:56
$begingroup$
How about considering $H_2$ with coefficients in $Bbb Z_2$?
$endgroup$
– Lord Shark the Unknown
Jan 5 at 22:56
1
1
$begingroup$
... Or $H^2$ with integer coefficients.
$endgroup$
– Tyrone
Jan 6 at 12:26
$begingroup$
... Or $H^2$ with integer coefficients.
$endgroup$
– Tyrone
Jan 6 at 12:26
1
1
$begingroup$
One way to see the above is to use cellular chains, if you take the CW-structure on $S^2$ with one $0$-cell and one $2$-cell, you will see that your map is cellular, and maps the $2$-cell to the $2$-cell. Then, if you calculate the cellular chain complex with $mathbb Z/2$ coefficients, you clearly get the identity on $H_2$, which is not possible for a nullhomotopic map.
$endgroup$
– Justin Young
Jan 9 at 18:32
$begingroup$
One way to see the above is to use cellular chains, if you take the CW-structure on $S^2$ with one $0$-cell and one $2$-cell, you will see that your map is cellular, and maps the $2$-cell to the $2$-cell. Then, if you calculate the cellular chain complex with $mathbb Z/2$ coefficients, you clearly get the identity on $H_2$, which is not possible for a nullhomotopic map.
$endgroup$
– Justin Young
Jan 9 at 18:32
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
There is a cofibration sequence
$$S^1xrightarrow{2} S^1rightarrowmathbb{R}P^2xrightarrow{q}S^2rightarrowdots$$
where $q$ is the map pinching to the top cell. This cofibring leads to an exact Puppe sequence of homotopy sets
$$dotsleftarrowpi_1S^2leftarrow [mathbb{R}P^2,S^2]xleftarrow{q^*}pi_2S^2xleftarrow{times 2}pi_2S^2leftarrowdots$$
Here $[mathbb{R}P^2,S^2]$ is only a set with $pi_2(S^2)$-action, but the rest of these objects are (abelian) groups. Likewise the map $q^*$ is a map of sets, but the $times 2$ map is a homomorphism.
Now since $pi_1S^2=0$, we identify $[mathbb{R}P^2,S^2]congpi_2(S^2)/(2cdot pi_2(S^2))congmathbb{Z}_2$ as sets with $pi_2(S^2)$-action. Here we have used that $pi_2S^2congmathbb{Z}$. In particular there are two homotopically distinct maps $mathbb{R}P^2rightarrow S^2$. One is clearly the trivial map, and exactness of the above sequence tells us that the other is $q^*id_{S^2}=q$, since $id_{S^2}$ generates $pi_2S^2$. Hence the pinch map $q$ is homotopically non-trivial.
At this point we would like to get our hands dirty and see all this in another way. We want to use the Hopf Degree Theorem for non-orientable manifolds, which, as I understand it, states that two maps $f,g:Mrightarrow S^n$ from a boundaryless n-dimensional non-orientable manifold $M$ are homotopic only if they have the same Brouwer-Hopf degree modulo 2.
The real projective plane $mathbb{R}P^2$ is a 2-dimensional closed non-orientable manifold, which is a quotient of $S^2subseteqmathbb{R}^3$ by the equivalence relation $xsim -x$. We write its points $[x,y,z]=[-x,-y,-z]$, where $(x,y,z)in S^2$.
There are equivalent definitions of the degree of a map $f:mathbb{R}P^2rightarrow S^2$. The first takes the mod 2 fundamental class $[mathbb{R}P^2]in H_2(mathbb{R}P^2;mathbb{Z}_2)congmathbb{Z}_2$ and studies its image $f_*[mathbb{R}P^2]in H_2(S^2;mathbb{Z}_2)congmathbb{Z}_2$. From this definition it is clear that the trivial map $ast:mathbb{R}P^2rightarrow S^2$ has mod 2 degree 0.
The second definition takes a smooth representative for the homotopy class of $f:mathbb{R}P^2rightarrow S^2$, finds a regular value $pin S^2$ for it, and sets $deg_2(f)=|f^{-1}(p)|mod 2$. The preimage $f^{-1}(p)$ should be a finite set, so this makes sense.
Now we will calculate $deg_2(q)$, the degree of the pinch map. Recall that the subspace ${[x,y,0]}subseteq mathbb{R}P^2$ is exactly $mathbb{R}P^1cong S^1$. Hence $q$ sends all of this subspace to the basepoint of $S^2$. This is handy, because we have a chart $phi:Urightarrow mathbb{R}^2$ defined over $U={[x,y,z]mid zne0}$ by
$$phi[x,y,z]=(x/z,y/z).$$
Viewing $S^2$ as the 1-point compactification of $mathbb{R}^2$ by using stereographic projection, this chart gives the pinch map
$$q[x,y,z]=begin{cases}(x/z,y/z)in S^2&zneq 0\infty&z=0 end{cases}$$
This is clearly smooth, and you can verify that any point away from $infty$ is regular. In particular, the point $(0,0)in S^2cong (mathbb{R}^2)^+$ is regular. hence
$$deg_2(q)=|q^{-1}(0,0)|=|{[0,0,1]}|=1$$
It follows that $deg_2(q)=1neq0=deg_2(ast)$, and therefore that $q$ is homotopically non-trivial.
Returning to the first definition, we see that this is in line with Lord Shark's hint, for $q_*[mathbb{R}P^2]=1cdot[S^2]in H_2(S^2;mathbb{Z}_2)$.
answered Jan 6 at 13:13
TyroneTyrone
4,55011225
$endgroup$
$begingroup$
Your phrasing of the Hopf degree theorem for unoriented manifolds is correct. The proof is the same as usual (annihilation of opposite-signed zeroes in a chart) with the trick that we may push a point along an unoriented loop to change its sign. As is implicit in my comment, it is nice to think of this as the simplest case of Pontryagin-Thom. I feel that I should note for OP's sake that the cellular homology computation of $q_*$ is straightforward.
$endgroup$
– Mike Miller
Jan 6 at 13:33
add a comment |
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1 Answer
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$begingroup$
There is a cofibration sequence
$$S^1xrightarrow{2} S^1rightarrowmathbb{R}P^2xrightarrow{q}S^2rightarrowdots$$
where $q$ is the map pinching to the top cell. This cofibring leads to an exact Puppe sequence of homotopy sets
$$dotsleftarrowpi_1S^2leftarrow [mathbb{R}P^2,S^2]xleftarrow{q^*}pi_2S^2xleftarrow{times 2}pi_2S^2leftarrowdots$$
Here $[mathbb{R}P^2,S^2]$ is only a set with $pi_2(S^2)$-action, but the rest of these objects are (abelian) groups. Likewise the map $q^*$ is a map of sets, but the $times 2$ map is a homomorphism.
Now since $pi_1S^2=0$, we identify $[mathbb{R}P^2,S^2]congpi_2(S^2)/(2cdot pi_2(S^2))congmathbb{Z}_2$ as sets with $pi_2(S^2)$-action. Here we have used that $pi_2S^2congmathbb{Z}$. In particular there are two homotopically distinct maps $mathbb{R}P^2rightarrow S^2$. One is clearly the trivial map, and exactness of the above sequence tells us that the other is $q^*id_{S^2}=q$, since $id_{S^2}$ generates $pi_2S^2$. Hence the pinch map $q$ is homotopically non-trivial.
At this point we would like to get our hands dirty and see all this in another way. We want to use the Hopf Degree Theorem for non-orientable manifolds, which, as I understand it, states that two maps $f,g:Mrightarrow S^n$ from a boundaryless n-dimensional non-orientable manifold $M$ are homotopic only if they have the same Brouwer-Hopf degree modulo 2.
The real projective plane $mathbb{R}P^2$ is a 2-dimensional closed non-orientable manifold, which is a quotient of $S^2subseteqmathbb{R}^3$ by the equivalence relation $xsim -x$. We write its points $[x,y,z]=[-x,-y,-z]$, where $(x,y,z)in S^2$.
There are equivalent definitions of the degree of a map $f:mathbb{R}P^2rightarrow S^2$. The first takes the mod 2 fundamental class $[mathbb{R}P^2]in H_2(mathbb{R}P^2;mathbb{Z}_2)congmathbb{Z}_2$ and studies its image $f_*[mathbb{R}P^2]in H_2(S^2;mathbb{Z}_2)congmathbb{Z}_2$. From this definition it is clear that the trivial map $ast:mathbb{R}P^2rightarrow S^2$ has mod 2 degree 0.
The second definition takes a smooth representative for the homotopy class of $f:mathbb{R}P^2rightarrow S^2$, finds a regular value $pin S^2$ for it, and sets $deg_2(f)=|f^{-1}(p)|mod 2$. The preimage $f^{-1}(p)$ should be a finite set, so this makes sense.
Now we will calculate $deg_2(q)$, the degree of the pinch map. Recall that the subspace ${[x,y,0]}subseteq mathbb{R}P^2$ is exactly $mathbb{R}P^1cong S^1$. Hence $q$ sends all of this subspace to the basepoint of $S^2$. This is handy, because we have a chart $phi:Urightarrow mathbb{R}^2$ defined over $U={[x,y,z]mid zne0}$ by
$$phi[x,y,z]=(x/z,y/z).$$
Viewing $S^2$ as the 1-point compactification of $mathbb{R}^2$ by using stereographic projection, this chart gives the pinch map
$$q[x,y,z]=begin{cases}(x/z,y/z)in S^2&zneq 0\infty&z=0 end{cases}$$
This is clearly smooth, and you can verify that any point away from $infty$ is regular. In particular, the point $(0,0)in S^2cong (mathbb{R}^2)^+$ is regular. hence
$$deg_2(q)=|q^{-1}(0,0)|=|{[0,0,1]}|=1$$
It follows that $deg_2(q)=1neq0=deg_2(ast)$, and therefore that $q$ is homotopically non-trivial.
Returning to the first definition, we see that this is in line with Lord Shark's hint, for $q_*[mathbb{R}P^2]=1cdot[S^2]in H_2(S^2;mathbb{Z}_2)$.
answered Jan 6 at 13:13
TyroneTyrone
4,55011225
$endgroup$
$begingroup$
Your phrasing of the Hopf degree theorem for unoriented manifolds is correct. The proof is the same as usual (annihilation of opposite-signed zeroes in a chart) with the trick that we may push a point along an unoriented loop to change its sign. As is implicit in my comment, it is nice to think of this as the simplest case of Pontryagin-Thom. I feel that I should note for OP's sake that the cellular homology computation of $q_*$ is straightforward.
$endgroup$
– Mike Miller
Jan 6 at 13:33
add a comment |
$begingroup$
There is a cofibration sequence
$$S^1xrightarrow{2} S^1rightarrowmathbb{R}P^2xrightarrow{q}S^2rightarrowdots$$
where $q$ is the map pinching to the top cell. This cofibring leads to an exact Puppe sequence of homotopy sets
$$dotsleftarrowpi_1S^2leftarrow [mathbb{R}P^2,S^2]xleftarrow{q^*}pi_2S^2xleftarrow{times 2}pi_2S^2leftarrowdots$$
Here $[mathbb{R}P^2,S^2]$ is only a set with $pi_2(S^2)$-action, but the rest of these objects are (abelian) groups. Likewise the map $q^*$ is a map of sets, but the $times 2$ map is a homomorphism.
Now since $pi_1S^2=0$, we identify $[mathbb{R}P^2,S^2]congpi_2(S^2)/(2cdot pi_2(S^2))congmathbb{Z}_2$ as sets with $pi_2(S^2)$-action. Here we have used that $pi_2S^2congmathbb{Z}$. In particular there are two homotopically distinct maps $mathbb{R}P^2rightarrow S^2$. One is clearly the trivial map, and exactness of the above sequence tells us that the other is $q^*id_{S^2}=q$, since $id_{S^2}$ generates $pi_2S^2$. Hence the pinch map $q$ is homotopically non-trivial.
At this point we would like to get our hands dirty and see all this in another way. We want to use the Hopf Degree Theorem for non-orientable manifolds, which, as I understand it, states that two maps $f,g:Mrightarrow S^n$ from a boundaryless n-dimensional non-orientable manifold $M$ are homotopic only if they have the same Brouwer-Hopf degree modulo 2.
The real projective plane $mathbb{R}P^2$ is a 2-dimensional closed non-orientable manifold, which is a quotient of $S^2subseteqmathbb{R}^3$ by the equivalence relation $xsim -x$. We write its points $[x,y,z]=[-x,-y,-z]$, where $(x,y,z)in S^2$.
There are equivalent definitions of the degree of a map $f:mathbb{R}P^2rightarrow S^2$. The first takes the mod 2 fundamental class $[mathbb{R}P^2]in H_2(mathbb{R}P^2;mathbb{Z}_2)congmathbb{Z}_2$ and studies its image $f_*[mathbb{R}P^2]in H_2(S^2;mathbb{Z}_2)congmathbb{Z}_2$. From this definition it is clear that the trivial map $ast:mathbb{R}P^2rightarrow S^2$ has mod 2 degree 0.
The second definition takes a smooth representative for the homotopy class of $f:mathbb{R}P^2rightarrow S^2$, finds a regular value $pin S^2$ for it, and sets $deg_2(f)=|f^{-1}(p)|mod 2$. The preimage $f^{-1}(p)$ should be a finite set, so this makes sense.
Now we will calculate $deg_2(q)$, the degree of the pinch map. Recall that the subspace ${[x,y,0]}subseteq mathbb{R}P^2$ is exactly $mathbb{R}P^1cong S^1$. Hence $q$ sends all of this subspace to the basepoint of $S^2$. This is handy, because we have a chart $phi:Urightarrow mathbb{R}^2$ defined over $U={[x,y,z]mid zne0}$ by
$$phi[x,y,z]=(x/z,y/z).$$
Viewing $S^2$ as the 1-point compactification of $mathbb{R}^2$ by using stereographic projection, this chart gives the pinch map
$$q[x,y,z]=begin{cases}(x/z,y/z)in S^2&zneq 0\infty&z=0 end{cases}$$
This is clearly smooth, and you can verify that any point away from $infty$ is regular. In particular, the point $(0,0)in S^2cong (mathbb{R}^2)^+$ is regular. hence
$$deg_2(q)=|q^{-1}(0,0)|=|{[0,0,1]}|=1$$
It follows that $deg_2(q)=1neq0=deg_2(ast)$, and therefore that $q$ is homotopically non-trivial.
Returning to the first definition, we see that this is in line with Lord Shark's hint, for $q_*[mathbb{R}P^2]=1cdot[S^2]in H_2(S^2;mathbb{Z}_2)$.
answered Jan 6 at 13:13
TyroneTyrone
4,55011225
$endgroup$
$begingroup$
Your phrasing of the Hopf degree theorem for unoriented manifolds is correct. The proof is the same as usual (annihilation of opposite-signed zeroes in a chart) with the trick that we may push a point along an unoriented loop to change its sign. As is implicit in my comment, it is nice to think of this as the simplest case of Pontryagin-Thom. I feel that I should note for OP's sake that the cellular homology computation of $q_*$ is straightforward.
$endgroup$
– Mike Miller
Jan 6 at 13:33
add a comment |
$begingroup$
There is a cofibration sequence
$$S^1xrightarrow{2} S^1rightarrowmathbb{R}P^2xrightarrow{q}S^2rightarrowdots$$
where $q$ is the map pinching to the top cell. This cofibring leads to an exact Puppe sequence of homotopy sets
$$dotsleftarrowpi_1S^2leftarrow [mathbb{R}P^2,S^2]xleftarrow{q^*}pi_2S^2xleftarrow{times 2}pi_2S^2leftarrowdots$$
Here $[mathbb{R}P^2,S^2]$ is only a set with $pi_2(S^2)$-action, but the rest of these objects are (abelian) groups. Likewise the map $q^*$ is a map of sets, but the $times 2$ map is a homomorphism.
Now since $pi_1S^2=0$, we identify $[mathbb{R}P^2,S^2]congpi_2(S^2)/(2cdot pi_2(S^2))congmathbb{Z}_2$ as sets with $pi_2(S^2)$-action. Here we have used that $pi_2S^2congmathbb{Z}$. In particular there are two homotopically distinct maps $mathbb{R}P^2rightarrow S^2$. One is clearly the trivial map, and exactness of the above sequence tells us that the other is $q^*id_{S^2}=q$, since $id_{S^2}$ generates $pi_2S^2$. Hence the pinch map $q$ is homotopically non-trivial.
At this point we would like to get our hands dirty and see all this in another way. We want to use the Hopf Degree Theorem for non-orientable manifolds, which, as I understand it, states that two maps $f,g:Mrightarrow S^n$ from a boundaryless n-dimensional non-orientable manifold $M$ are homotopic only if they have the same Brouwer-Hopf degree modulo 2.
The real projective plane $mathbb{R}P^2$ is a 2-dimensional closed non-orientable manifold, which is a quotient of $S^2subseteqmathbb{R}^3$ by the equivalence relation $xsim -x$. We write its points $[x,y,z]=[-x,-y,-z]$, where $(x,y,z)in S^2$.
There are equivalent definitions of the degree of a map $f:mathbb{R}P^2rightarrow S^2$. The first takes the mod 2 fundamental class $[mathbb{R}P^2]in H_2(mathbb{R}P^2;mathbb{Z}_2)congmathbb{Z}_2$ and studies its image $f_*[mathbb{R}P^2]in H_2(S^2;mathbb{Z}_2)congmathbb{Z}_2$. From this definition it is clear that the trivial map $ast:mathbb{R}P^2rightarrow S^2$ has mod 2 degree 0.
The second definition takes a smooth representative for the homotopy class of $f:mathbb{R}P^2rightarrow S^2$, finds a regular value $pin S^2$ for it, and sets $deg_2(f)=|f^{-1}(p)|mod 2$. The preimage $f^{-1}(p)$ should be a finite set, so this makes sense.
Now we will calculate $deg_2(q)$, the degree of the pinch map. Recall that the subspace ${[x,y,0]}subseteq mathbb{R}P^2$ is exactly $mathbb{R}P^1cong S^1$. Hence $q$ sends all of this subspace to the basepoint of $S^2$. This is handy, because we have a chart $phi:Urightarrow mathbb{R}^2$ defined over $U={[x,y,z]mid zne0}$ by
$$phi[x,y,z]=(x/z,y/z).$$
Viewing $S^2$ as the 1-point compactification of $mathbb{R}^2$ by using stereographic projection, this chart gives the pinch map
$$q[x,y,z]=begin{cases}(x/z,y/z)in S^2&zneq 0\infty&z=0 end{cases}$$
This is clearly smooth, and you can verify that any point away from $infty$ is regular. In particular, the point $(0,0)in S^2cong (mathbb{R}^2)^+$ is regular. hence
$$deg_2(q)=|q^{-1}(0,0)|=|{[0,0,1]}|=1$$
It follows that $deg_2(q)=1neq0=deg_2(ast)$, and therefore that $q$ is homotopically non-trivial.
Returning to the first definition, we see that this is in line with Lord Shark's hint, for $q_*[mathbb{R}P^2]=1cdot[S^2]in H_2(S^2;mathbb{Z}_2)$.
answered Jan 6 at 13:13
TyroneTyrone
4,55011225
$endgroup$
There is a cofibration sequence
$$S^1xrightarrow{2} S^1rightarrowmathbb{R}P^2xrightarrow{q}S^2rightarrowdots$$
where $q$ is the map pinching to the top cell. This cofibring leads to an exact Puppe sequence of homotopy sets
$$dotsleftarrowpi_1S^2leftarrow [mathbb{R}P^2,S^2]xleftarrow{q^*}pi_2S^2xleftarrow{times 2}pi_2S^2leftarrowdots$$
Here $[mathbb{R}P^2,S^2]$ is only a set with $pi_2(S^2)$-action, but the rest of these objects are (abelian) groups. Likewise the map $q^*$ is a map of sets, but the $times 2$ map is a homomorphism.
Now since $pi_1S^2=0$, we identify $[mathbb{R}P^2,S^2]congpi_2(S^2)/(2cdot pi_2(S^2))congmathbb{Z}_2$ as sets with $pi_2(S^2)$-action. Here we have used that $pi_2S^2congmathbb{Z}$. In particular there are two homotopically distinct maps $mathbb{R}P^2rightarrow S^2$. One is clearly the trivial map, and exactness of the above sequence tells us that the other is $q^*id_{S^2}=q$, since $id_{S^2}$ generates $pi_2S^2$. Hence the pinch map $q$ is homotopically non-trivial.
At this point we would like to get our hands dirty and see all this in another way. We want to use the Hopf Degree Theorem for non-orientable manifolds, which, as I understand it, states that two maps $f,g:Mrightarrow S^n$ from a boundaryless n-dimensional non-orientable manifold $M$ are homotopic only if they have the same Brouwer-Hopf degree modulo 2.
The real projective plane $mathbb{R}P^2$ is a 2-dimensional closed non-orientable manifold, which is a quotient of $S^2subseteqmathbb{R}^3$ by the equivalence relation $xsim -x$. We write its points $[x,y,z]=[-x,-y,-z]$, where $(x,y,z)in S^2$.
There are equivalent definitions of the degree of a map $f:mathbb{R}P^2rightarrow S^2$. The first takes the mod 2 fundamental class $[mathbb{R}P^2]in H_2(mathbb{R}P^2;mathbb{Z}_2)congmathbb{Z}_2$ and studies its image $f_*[mathbb{R}P^2]in H_2(S^2;mathbb{Z}_2)congmathbb{Z}_2$. From this definition it is clear that the trivial map $ast:mathbb{R}P^2rightarrow S^2$ has mod 2 degree 0.
The second definition takes a smooth representative for the homotopy class of $f:mathbb{R}P^2rightarrow S^2$, finds a regular value $pin S^2$ for it, and sets $deg_2(f)=|f^{-1}(p)|mod 2$. The preimage $f^{-1}(p)$ should be a finite set, so this makes sense.
Now we will calculate $deg_2(q)$, the degree of the pinch map. Recall that the subspace ${[x,y,0]}subseteq mathbb{R}P^2$ is exactly $mathbb{R}P^1cong S^1$. Hence $q$ sends all of this subspace to the basepoint of $S^2$. This is handy, because we have a chart $phi:Urightarrow mathbb{R}^2$ defined over $U={[x,y,z]mid zne0}$ by
$$phi[x,y,z]=(x/z,y/z).$$
Viewing $S^2$ as the 1-point compactification of $mathbb{R}^2$ by using stereographic projection, this chart gives the pinch map
$$q[x,y,z]=begin{cases}(x/z,y/z)in S^2&zneq 0\infty&z=0 end{cases}$$
This is clearly smooth, and you can verify that any point away from $infty$ is regular. In particular, the point $(0,0)in S^2cong (mathbb{R}^2)^+$ is regular. hence
$$deg_2(q)=|q^{-1}(0,0)|=|{[0,0,1]}|=1$$
It follows that $deg_2(q)=1neq0=deg_2(ast)$, and therefore that $q$ is homotopically non-trivial.
Returning to the first definition, we see that this is in line with Lord Shark's hint, for $q_*[mathbb{R}P^2]=1cdot[S^2]in H_2(S^2;mathbb{Z}_2)$.
answered Jan 6 at 13:13
TyroneTyrone
4,55011225
answered Jan 6 at 13:13
TyroneTyrone
4,55011225
answered Jan 6 at 13:13
TyroneTyrone
4,55011225
answered Jan 6 at 13:13
TyroneTyrone
4,55011225
4,55011225
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Your phrasing of the Hopf degree theorem for unoriented manifolds is correct. The proof is the same as usual (annihilation of opposite-signed zeroes in a chart) with the trick that we may push a point along an unoriented loop to change its sign. As is implicit in my comment, it is nice to think of this as the simplest case of Pontryagin-Thom. I feel that I should note for OP's sake that the cellular homology computation of $q_*$ is straightforward.
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– Mike Miller
Jan 6 at 13:33
add a comment |
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Your phrasing of the Hopf degree theorem for unoriented manifolds is correct. The proof is the same as usual (annihilation of opposite-signed zeroes in a chart) with the trick that we may push a point along an unoriented loop to change its sign. As is implicit in my comment, it is nice to think of this as the simplest case of Pontryagin-Thom. I feel that I should note for OP's sake that the cellular homology computation of $q_*$ is straightforward.
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– Mike Miller
Jan 6 at 13:33
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Your phrasing of the Hopf degree theorem for unoriented manifolds is correct. The proof is the same as usual (annihilation of opposite-signed zeroes in a chart) with the trick that we may push a point along an unoriented loop to change its sign. As is implicit in my comment, it is nice to think of this as the simplest case of Pontryagin-Thom. I feel that I should note for OP's sake that the cellular homology computation of $q_*$ is straightforward.
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– Mike Miller
Jan 6 at 13:33
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Your phrasing of the Hopf degree theorem for unoriented manifolds is correct. The proof is the same as usual (annihilation of opposite-signed zeroes in a chart) with the trick that we may push a point along an unoriented loop to change its sign. As is implicit in my comment, it is nice to think of this as the simplest case of Pontryagin-Thom. I feel that I should note for OP's sake that the cellular homology computation of $q_*$ is straightforward.
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– Mike Miller
Jan 6 at 13:33
add a comment |
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How about considering $H_2$ with coefficients in $Bbb Z_2$?
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– Lord Shark the Unknown
Jan 5 at 22:56
1
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... Or $H^2$ with integer coefficients.
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– Tyrone
Jan 6 at 12:26
1
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One way to see the above is to use cellular chains, if you take the CW-structure on $S^2$ with one $0$-cell and one $2$-cell, you will see that your map is cellular, and maps the $2$-cell to the $2$-cell. Then, if you calculate the cellular chain complex with $mathbb Z/2$ coefficients, you clearly get the identity on $H_2$, which is not possible for a nullhomotopic map.
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– Justin Young
Jan 9 at 18:32