Let $$X=\vee_{\alpha\in A} S_{\alpha}^n$$ be a bouquet of $n$-spheres.
Q: How does one compute the homotopy groups $\pi_k(X)$?
My motivation for this question was the following observation: Let $A$ be countable set and let $n=2$. Then $X$ is homotopy equivalent to $S^3$ minus the trivial knot $K\subseteq S^3$. A basic result on $3$-manifolds tells us that $Y=S^3\backslash K$ is a $K(\mathbf{Z},1)$ and therefore $\pi_k(Y)=0$ if $k\geq 2$. Thus an infinite bouquet of $2$-spheres has the homotopy type of a point, this sounds a bit surprising isn't!