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4

Since $S^n$ is simply-connected for $n>1$, any such map $f$ factors through $\mathbb{R}^n$, which is the universal cover of $T^n$ and has $n$-th degree trivial homology since its contractible. It follows that the $n$-th degree induced map of $f$ on homology is trivial by functoriality of homology. The key point here is the Lifting Theorem.


3

If we take the definition of a (global) lefschetz number as in $7.1$ of these notes, then the proof goes as follows: If $f$ is homotopic to $g$, the inclusion maps $id×f$ and $id×g$ of graph $f$ and graph $g$ are homotopic. Thus the global Lefschetz number of a map is homotopy invariant.


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One can do this without covering spaces using the cup product in cohomology. Namely $H^n(T^n,\Bbb Z)=\Bbb Z\langle e_1\cup e_2\dots \cup e_n\rangle$ (this is supposed to denote integer multiples of $ e_1\cup e_2\dots \cup e_n$), where $e_i$ are the generators of $H^1(T^n,\Bbb Z)$. Now if $f:S^n \to T^n$ has nonzero topological degree then $f^*( e_1\cup e_2\...


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You have to be careful when $F$ is not compact, and I don't want to be careful today, so let $F$ be a smooth compact manifold without boundary. Then $\text{Diff}(F)$ is a Frechet manifold: it's locally modeled not on Euclidean spaces, but on topological vector spaces called Frechet spaces. Here the Frechet space that's the tangent space at the identity is ...


3

Assuming that $B$ is a submanifold of $\mathbb R^n$, $f$ is a diffeomorphism if and only if it is an immersion; i.e. if the differential is injective everywhere. (Without this first assumption you need to think about what you mean by diffeomorphism - what is the smooth structure on $B$ meant to be?) I doubt this is much help to you, though, since this ...


3

Assuming you want to compute the degree of the composition of maps, otherwise please reformulate if you spell out the definition the proof is easy. So let me give you some advices: Let $M \xrightarrow{f} N \xrightarrow{g} O$ be your composition, let $z\in O$ be a regular value for $gf$. Then set $\{y_i\}_1^n$ as the preimage of $g^{-1}(z)$ and $\{x^i_j\}...


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Eric, first of all, $\omega$ is a $1$-form on the image of $g$ and $R$ is a region contained in the image of $g$, so it certainly makes sense to integrate $\omega$ over $\partial R$. Here's what I suggest: (1) Compute $g^*\omega$ (you should get $\cos(s-t)\,ds + dt$). (2) Compute $d(g^*\omega)$ and $g^*(dx\wedge dy)$. How do they compare? (3) Using the ...


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Let $\omega$ be a closed $k$-form on the torus. Let $\theta_1,\dots,\theta_n$ be the $n$ angle $1$-forms on $T^n$. We know that these generate the cohomology of $T^n$, so we know that $\omega$ is cohomologous to a (unique!) form of the form $$\eta=\sum_{1\leq i_1<\cdots<i_k\leq n}\eta_{i_1,\dots,u_k}\theta_{i_1}\wedge\cdots\wedge\theta_{i_k}$$ with all ...


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Yes, this is true. Let $P_1, P_2:V \to V$ be projections onto $R$. I.e., $P_iP_i = P_i$ and $P_i|_R$ is the identity for $i = 1,2$. Let $c_1, c_2 \geq 0$ be such that $c_1 + c_2 = 1$. Define $P_3:= c_1 P_1 + c_2 P_2$. It is immediate that $P_3(V) \subseteq R$. We have $$P_3P_3 = (c_1P_1+c_2P_2)(c_1P_1 + c_2 P_2) = c_1^2 P_1^2 + c_2^2 P_2^2 + c_1 c_2 P_1P_2 ...



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