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 ?



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.


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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