I'm trying to show that $S^n$ is not a deformation retract of $\bigvee^k S^n$ as a generalization of a proof (or an attempt) I made showing that $S^1$ is not a deformation retract of $S^1\vee S^1$, but for all $n$ and all $k$ I couldn't prove that.
In the case when $k=2$ and $n=1$, if $S^1$ was a deformation retract of $S^1\vee S^1$, then $S^1$ should be homotopy equivalent to $S^1\vee S^1$; moreover, if $S^1$ was homotopy equivalent to $S^1 \vee S^1$, its fundamental groups should be isomorphic. But $\pi_1(S^1)\cong \mathbb{Z}$ and $\pi_1(S^1\vee S^1)\cong \pi_1(S^1)\ast \pi_1(S^1)\cong \mathbb{Z}\ast \mathbb{Z}$, which are not isomorphic, then $S^1$ cannot be homotopy equivalent to $S^1\vee S^1$ and, therefore, $S^1$ cannot be a deformation retract of $S^1\vee S^1$ (by the contrapositive, I think).
For solving the problem for all $k$ and all $n$, I tried to compute the higher homotopy groups of both spaces but I don't know how to compute $\pi_n(\bigvee^k S^n)$; I know that the fundamental group of the wedge sum is the free product (with certain conditions) but I don't know if this can be extended to all homotopy groups (but I don't believe that).
How can I show that $S^n$ is not a deformation retract of $\bigvee^k S^n$? give me some hints please.
Best Answer
Calculation of homotopy groups is difficult. I recommend to use homology groups instead. It is well-known (see for example Homology of wedge sum is the direct sum of homologies) that $$H_n(\bigvee^k S^n) \approx \mathbb Z^k$$ which gives you the desired result.