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Suppose $M$ is a closed smooth n-manifold.

a)Does there always exist a smooth map $f:M\to S^n$ from $M$ into the n-sphere, such that $f$ is essential(i.e. $f$ is not homotopic to a constant map)? Justify your answer.

b) Same question, replacing $S^n$ by the n-torus $T^n$.

For the question a), I think since $M$ is a n-manifold, there is a neighbourhood $U$ of $M$ and $U$ is homeomorphic to $\mathbb R^n$,and so $U$ is homeomorphic to $S^n \backslash N$, $N$ is the north pole of $S^n$.And then we map $M\backslash U$ to $N$. Does the map constructed above satisfy the condition smooth map?

Could anybody give some hints on how to slove a) and b)? Thanks!

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    I guess (a) is true, (b) is false. The construction you gave makes sense to me.2012-06-16
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    Your idea for (a) works and the map can be made smooth. Do you know why it is not nullhomotopic?2012-06-16
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    @JuanS I'm sorry,I tried to figure out a proof for nullhomotopic but I can't. Could you give any hints? Thank you!:)2012-06-17
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    For part b), you might try thinking about the case $n=2$ and using covering space theory (consider the universal cover of $T^n$).2012-06-18
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    For part a) using homology is probably the simplest thing. It might help if you shrink $U$ slightly so that it lies inside a nice open set in $M$. Think about the long exact sequence for the pair $(M, M-U)$.2012-06-18
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    An argument using degree will suffice for a) (I believe this was a problem in the Yau college competition last year)2012-06-18
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    @DanRamras Thank you! I'm considering your advise from this morning, but I'm a little confused from which step we can get $f$ is not null-homotopy? Thank you!2012-06-19
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    @MichaelLuo Yes! It is the first problem of the 2011 Yau Competition of Geometry and Topology :) Thank you for your hint!2012-06-19
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    For (a), you can take $S^n$ to be the $(n+1)$-skeleton of $K(\mathbb{Z},n)$. So assuming $M$ is orientable, you can use cellular approximation to obtain an essential map $M \rightarrow K(\mathbb{Z},n)$ which lands in $S^n$, and then smooth approximation to demand that it be smooth. For (b), the answer is no: take $M=S^2$; then $\pi_2(T^2)=0$ since the universal cover of $T^2$ is $\mathbb{R}^2$, which is contractible.2012-06-19
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    The pair $(M, M-U)$ has a LES of the form $H_n (M-U) \rightarrow H_n (M) \rightarrow H_n (M/(M-U)) = H_n (S^n)\rightarrow \cdots$ (so long as you choose $U$ nicely enough). If the second map is zero then $H_n (M-U)$ surjects onto $H_n (M)$. But $H_n (M-U)$ is homotopy equivalent to $M-pt$ and a closed manifold with a point removed has trivial top homology. (If $M$ isn't orientable, then all this should be done with mod 2 coefficients.) As Aaron explained, you can always smooth out the map $M\rightarrow M/(M-U) = S^n$ up to homotopy.2012-06-19
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    @AaronMazel-Gee Thank you for your hints!But I need to learn more to understand.2012-06-21
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    @DanRamras Thank you!:)2012-06-21
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    @JiangnanYu: Sure thing. You should try to learn about "Eilenberg-MacLane spaces", which have rather obvious CW structures that make what I said true. (They also have functorial but bigger CW structures, but that's another story.)2012-06-26
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    @AaronMazel-Gee Thank you for your advise :)2012-07-10

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