Let $$X=\vee_{\alpha\in A} S_{\alpha}^n$$ be a bouquet of $n$-spheres.
Q: How does one compute the homotopy groups $\pi_k(X)$?
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asked Mar 14, 2012 at 19:01
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11$\begingroup$ This is called the Hilton-Milnor theorem. Rationally the homotopy groups are a free lie algebra with respect to the whitehead product. Google provides you with the appropriate references. The result you state about the countable wedge of 2-spheres related to knot complements is false. $\endgroup$– Ryan BudneyCommented Mar 14, 2012 at 19:05
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3$\begingroup$ To be precise, an unknot complement in $S^3$ is homeo/diffeomorphic to $S^1 \times \mathbb R^2$. A countable wedge of 2-spheres has $\pi_2 X$ isomorphic to a direct sum of countable-many copies of $\mathbb Z$. Knot complements in the 3-sphere are all $K(\pi,1)$ type spaces. Wedges of spheres have infinitely many non-trivial homotopy groups. $\endgroup$– Ryan BudneyCommented Mar 14, 2012 at 19:10
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$\begingroup$ You are right completely right Ryan, I forgot the point at $\infty$, it is the unknot complement in $\mathbf{R}^3$ which is homotopy equivalent to a countable wedge of $S^2$. I'll remove my motivation. $\endgroup$– Hugo ChapdelaineCommented Mar 14, 2012 at 20:45
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3$\begingroup$ The complement of the unknot in $\mathbb R^3$ is homotopy-equivalent to $S^1 \vee S^2$. I suspect you're thinking of the universal cover of this space. The universal cover of $S^1 \vee S^2$ is homotopy-equivalent to $\vee_\infty S^2$. $\endgroup$– Ryan BudneyCommented Mar 14, 2012 at 20:53
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$\begingroup$ @Ryan Do you mean unlink in your lower comment? $\endgroup$– Scott CarterCommented Mar 14, 2012 at 22:36
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