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I am trying to show that any smooth map F from the 2-sphere to the 2-torus has degree zero.
The definition of the degree of F can be taken to be the (integer) degF such that
[tex]\int_{\mathbb{S}^2}F^*\omega=(\deg F)\int_{\mathbb{T}^2}\omega[/tex]
For omega any 2-form on T². Another definition would be
[tex]\deg F= \sum_{x\in F^{-1}(y)}sgn(dF_x)[/tex]
where y is any regular value of F (meaning dF_x is invertible at every x is in F^{-1}(y)) and where the sign of det(dF_x) is +1 if the isomorphism dF_x preserves the orientation and -1 if it reverses it.So! Of course, if it were to happen for some reason that no smooth map from S² to T² can be surjective, then the result would follow because we would only have to select in the second definition a y in T² that has an empty preimage, in which case degF=0 by convention. But I don't see how could argue in this direction.
Another trail is this. Consider T² as R²/Z² and let p:R²-->R²/Z² be the projection map.
R²
|
|p
|
v
T²<---F---S²
Topologically, there exists a lift G:S²-->R² making the above diagram commutative. If such a smooth lift exists, then noting that H²(S²)=R and H²(R²)=0 and passing to the cohomology in the preceeding diagram, we would end up with the following commutative diagram:
H²(R²)=0
^
|
|p*
|
H²(T²)---F*--->H²(S²)=R
(and G*²(R²)=0-->H²(S²)=R). We conclude that the pullback F* of F in cohomology is identically 0. That is to say, the usual pullback pulls every 2-form to an exact 2-form (since every 2-forms on the 2-torus is closed). But on the n-sphere, the integral of an n-form vanishes if and only if the form is exact. In particular, taking omega to be an orientation (aka volume) form so that [itex]\int\omega \neq 0[/itex], we could conclude that degF=0.
So the question is, does the smooth version of the lifting lemma used above holds?
Thanks.
The definition of the degree of F can be taken to be the (integer) degF such that
[tex]\int_{\mathbb{S}^2}F^*\omega=(\deg F)\int_{\mathbb{T}^2}\omega[/tex]
For omega any 2-form on T². Another definition would be
[tex]\deg F= \sum_{x\in F^{-1}(y)}sgn(dF_x)[/tex]
where y is any regular value of F (meaning dF_x is invertible at every x is in F^{-1}(y)) and where the sign of det(dF_x) is +1 if the isomorphism dF_x preserves the orientation and -1 if it reverses it.So! Of course, if it were to happen for some reason that no smooth map from S² to T² can be surjective, then the result would follow because we would only have to select in the second definition a y in T² that has an empty preimage, in which case degF=0 by convention. But I don't see how could argue in this direction.
Another trail is this. Consider T² as R²/Z² and let p:R²-->R²/Z² be the projection map.
R²
|
|p
|
v
T²<---F---S²
Topologically, there exists a lift G:S²-->R² making the above diagram commutative. If such a smooth lift exists, then noting that H²(S²)=R and H²(R²)=0 and passing to the cohomology in the preceeding diagram, we would end up with the following commutative diagram:
H²(R²)=0
^
|
|p*
|
H²(T²)---F*--->H²(S²)=R
(and G*²(R²)=0-->H²(S²)=R). We conclude that the pullback F* of F in cohomology is identically 0. That is to say, the usual pullback pulls every 2-form to an exact 2-form (since every 2-forms on the 2-torus is closed). But on the n-sphere, the integral of an n-form vanishes if and only if the form is exact. In particular, taking omega to be an orientation (aka volume) form so that [itex]\int\omega \neq 0[/itex], we could conclude that degF=0.
So the question is, does the smooth version of the lifting lemma used above holds?
Thanks.