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## sphere, torus, degree, cohomology

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

$$\int_{\mathbb{S}^2}F^*\omega=(\deg F)\int_{\mathbb{T}^2}\omega$$

For omega any 2-form on T². Another definition would be

$$\deg F= \sum_{x\in F^{-1}(y)}sgn(dF_x)$$

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 / and let p:R²-->/ 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*:H²(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 $\int\omega \neq 0$, we could conclude that degF=0.

So the question is, does the smooth version of the lifting lemma used above holds???

Thanks.
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 Recognitions: Gold Member Homework Help Science Advisor If p is a local diffeomorphism, this will undoubtedly imply that the a priori only continuous G is smooth. I'm gonna look into this.
 I like your second approach. Since the 2-sphere is simply-connected, every map f:S^2 -> T^2 lifts to the universal cover (i.e., lifts to a continuous map F:S^2 -> R^2.) This lift is unique up to the choice of a point in the fiber above f(1,0,0), where I'm assuming (1,0,0) to be the chosen basepoint of S^2. Since the covering projection R^2 -> T^2 has smooth local sections, the lift F is smooth. Thus, you have the commutative diagram you're after. Quick question: I usually work with singular homology, in which case the degree of a smooth map f:M -> N between manifolds of the same dimension n can be defined to be the integer corresponding to the induced map f_*:H_n(M) -> H_n(N). Relative to this approach, the result (that deg(f) = 0) follows from the existence of the commutative diagram. Does this apply to the (co)homology theory you're using?

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 Don't homotopic maps induce the same map in De Rham cohomology? I would think that $$\pi_2 T^2 = 0$$ would do the trick.
 Quote by zhentil Don't homotopic maps induce the same map in De Rham cohomology? I would think that $$\pi_2 T^2 = 0$$ would do the trick.