Here is my attempt: The degree of a smooth map $f:M \to N$ (where $M,N$ are manifolds of same dimension) is defined on the top form. Since the integral operator induces a natural isomorphism from the top de Rham cohomology group to $\mathbb{R}$, there exists a unique real number $k=deg(f)$ such that, for $\omega\in \Omega^n(N)$ we have, $$\int_{M}F^{\ast}\omega=deg(f)\int_{N}\omega$$ In this question we have $f:S^n\to T^n$, as we know that if two maps $g,g':S^n\to T^n$ are homotopic then they have the same cohomology groups, i.e $H(g)=H(g').$ from this we conclude that $deg(g)=deg(g')$ Now, as $S^n$ is simply connected, any map from $f:S^n\to T^n$ can be lifted(by homotopy lifting) to $\tilde{f}:S^n\to \mathbb{R}^n$.

Now as $\mathbb{R}^n$ is contractible $\tilde{f}, p$ are null homotopic and composition of null-homotopic map is nullhomotopic, so is $f$. So, the map $f$ cannot be surjective(if it were then $T^n$ has to be contractible to a point, which is absurd). So, it suffice to show that non surjective smooth maps have degree 0. As $f:S^n\to T^n$ is not surjective, we have $q\in T^n$ such that $f^{-1}(q)=\emptyset$. So, there exists neighborhood $V_q$ around q such that $V_q\cap F(S^n)=\emptyset$, then we can use a bump function to choose $\omega\in \Omega^n(T^n)$ supported in $V_q$ such that $$\int_{T^n}\omega=1$$ And we have $F^{\ast}(\omega)=0$, so by definition of degree we have $deg(f)=0$

Are the reasoning correct? I feel like something is going wrong somehow though..

Thanks for any help!


1 Answer 1


Once you know that $f$ is nullhomotopic, you know that $\deg f = 0$ as $f$ is homotopic to a constant map.

Your claim about $f$ not being surjective is false, there are degree zero surjective maps $S^n \to T^n$.


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