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Let $X$ be a connected hyperbolic 3-manifold (without boundary), $S^3$ the 3-sphere and $Map(X,S^3)$ the space of continuous maps between $X$ and $S^3$.

Question: Is the space $Map(X,S^3)$ connected ?

Thanks

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No, homotopic maps have the same degree, but it's an exercise (common to qualifying exams) to construct maps of any degree from a closed, oriented, connected $n$-manifold X to the $n$-sphere. It is less trivial, and I think due to Hopf, that two maps $f,g: X\rightarrow S^n$ are homotopic if and only if they have the same degree. Hence, the components of this space are labelled by the degree of the mapping.

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The object you are seeking is the third cohomotopy group $\pi^3(M)$ of a $3$-dimensional manifold $M.$ It is known (H. Hopf, 1953) that the $n$-th cohomotopy group of an $n$-dimensional complex is isomorphic to the $n$-th cohomology group, which is $Z$ for an orientable manifold, so the answer is NO.

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    $\begingroup$ One can see that there is an isomorphism since if $K(Z,3)$ is the appropriate Postnikov stage of $S^3$, whence there is a map $tr_3\colon S^3 \to K(Z,3)$ inducing isomorphisms on homotopy groups in dimension 3 and below, since $M$ is 3-dimensional $tr_{3\ast}\colon[M,S^3] \to [M,K(Z,3)]$ is an isomorphism. I'm trying to think of a geometric model (in case that's useful) for $tr_3$, involving $SU(2)$ and $BPU(H)$, but it's eluding me for now. $\endgroup$ – David Roberts Jul 15 '15 at 1:52
  • $\begingroup$ @David: what's wrong with the infinite symmetric product of $S^3$? $\endgroup$ – Qiaochu Yuan Jul 15 '15 at 2:14
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    $\begingroup$ @QiaochuYuan I didn't know that was a K(Z,3) :-) I was also trying to think of a geometric (smooth!) construction, rather than a topological/homotopical one. $\endgroup$ – David Roberts Jul 15 '15 at 5:17
  • $\begingroup$ @David: this is a corollary of the Dold-Thom theorem; more generally, the infinite symmetric product of $S^n$ is $B^n \mathbb{Z}$. $\endgroup$ – Qiaochu Yuan Jul 15 '15 at 5:47
  • $\begingroup$ @QiaochuYuan yes, I looked it up. And, I forgot to say above, clearly $[M,K(Z,3)] \simeq H^3(M,Z)$! $\endgroup$ – David Roberts Jul 15 '15 at 7:20

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