How to prove $S^n \vee S^1$ is not retract of $S^n \times S^1$? How to prove $S^n \vee S^1$ is not retract of $S^n \times S^1$?
If $n=1$, it has been solved here.
For greater $n$, the fundamental groups seem not to work.
 A: If $S^n\vee S^1$ were a retract of $S^n\times S^1$ then all the homotopy groups of $S^n\vee S^1$ would retract off of those of $S^n\times S^1$ also. This, however, cannot occur, and to show this it will suffice to calculate $\pi_n$ of each space.
We calculate $\pi_n(S^1\vee S^n)$ using its universal cover $X$, following Hatcher's construction of universal covers in section 1.3 of "Algebraic Topology". The space $X$ consists of a copy of $\mathbb{R}$ with an n-sphere $S^n$ attached at each integer and is thus homotopy equivalent to the countable wedge sum $\bigvee_\mathbb{Z} S^n$. From this we get that $\pi_n(S^n\vee S^1)\cong\pi_n(X)\cong \pi_n(\bigvee_\mathbb{Z} S^n)\cong \mathbb{Z}[t,t^{-1}]$ is the integral Laurent polynomials in variables $t$, $t^{-1}$.
On the other hand we have $\pi_n(S^n\times S^1)\cong \pi_n(S^1)\oplus\pi_n(S^n)\cong 0\oplus\mathbb{Z}=\mathbb{Z}$.
It follows that $\pi_n(S^n\vee S^1)\cong \mathbb{Z}[t,t^{-1}]$ cannot retract off of $\pi_n(S^n\times S^1)\cong \mathbb{Z}$ and the lack of such an algebraic retraction rules out the possibility of the existence a topological retraction. Therefore $S^n\vee S^1$ is not a retract of $S^n\times S^1$.
