Consider the following Proposition from p. 138, Chapter III of Bredon's Topology and Geometry:

Let $f:S^1\to X$ (continuous). Then $[f]=1\in\pi_1(X) \iff f$ extends to $D^2$.

$[f]$ is the homotopy class (rel base point) of $f$.

Does an analogous characterization extend to higher homotopy groups? I.e., is the following also true for any $n>1$:

Let $f:S^n\to X$ (continuous). Then $[f]=1\in\pi_n(X) \iff f$ extends to $D^{n+1}$?

A reference will suffice for an answer.


A null homotopy of $f$ is the same as an extension of $f$ to $D^{n+1}$ (after noting that $S^n\times I$ with $S^n\times\{1\}$ identified to a single point is homeomorphic to $D^{n+1}$), so yes.

  • $\begingroup$ Does $D^n$ mean the disk of which $S^n$ is the boundary? I always mix up the numbering conventions for spheres and disks and want to make sure I correctly understand your answer before accepting it. $\endgroup$ – Chill2Macht Nov 24 '16 at 15:04
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    $\begingroup$ @William No, you're right, $D^{n+1}$ is the disc with $S^n$ as boundary. That makes both $D^n$ and $S^n$ into $n$-dimensional spaces. That was a typo, and I should've caught it. $\endgroup$ – Arthur Nov 24 '16 at 15:33

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