# Show that every smooth map from $S^n\to T^n$ has degree 0.

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!

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 $$S^n \to T^n$$.