Let $X = S^1 \times D^2$, pick $x_0 \in X$. Suppose $\gamma$ is an embedded loop based at $x_0$ which represents an $n \in \mathbb{Z} = \pi_1(X)$ with $n \neq \pm1$. Prove that there is no retraction from $X$ to the image of $\gamma$.
My attempt at a proof:
Assume there is a retraction $r: X \rightarrow \gamma(t)$.
Then $r\circ i = Id_{\gamma(t)}$, where $i: \gamma(t) \rightarrow X$ is the inclusion map.
Then, looking at the homomorphisms induced on the fundamental groups, we have:
$r_{*}\circ i_{*}=(r\circ i)_{*} = Id_{\gamma(t)_*}$
So, we have $\pi_1(\gamma(t)) \cong \mathbb{Z}$,
and $i_{*}: \pi_1(\gamma(t)) \rightarrow \pi_1(X)$ where $i_{*}(1)=n$ where $1$ is a generator of $\pi_1(\gamma(t))$
Thus $r_*\circ i_*(1)=r_*(n)=r_*(1+1+....+1)=r_*(1)+....+r_*(1)=1$
Where the equality $r_*(1)+....+r_*(1)=1$ is because $r_{*}\circ i_{*}=(r\circ i)_{*} = Id_{\gamma(t)_*}$
But this is impossible, for $n$ things in $\pi_1(\gamma(t))$ to add up to 1. Thus we have a contraidction, and so there is no retraction $r: X \rightarrow \gamma(t)$