How do I prove that there are no retractions from $S^1\times \bar{B^2}$ and $A$? I want to show that there is no retraction from $S^1\times \bar{B^2}$ to $A$. Where A is the following:

Let us assume that $A$ is a deformation retract of $X$. THen by definition there exists a retraction $$r:X\rightarrow A;\,\,\,\,r(a)=a\,\,\,\forall a\in A$$In addition $r\circ i=id_A$ where $i:A\rightarrow X$ is the inclusion. Thus we can consider the morphisms $$r_*:\Pi_1(X)\rightarrow \Pi_1(A)$$ $$i_*:\Pi_1(A)\rightarrow \Pi_1(X)$$ where $r_*$ is surjective and $i_*$ is incective. Now consider also $$id_{\Pi_1(A)}=(r\circ i)_*:\Pi_1(A)\rightarrow \Pi_1(A)$$ and by a corrollary we know that $r_*\circ i_*=id_{\Pi_1(A)}$
Now I only need to find a contradiction.
Therefore I wanted to ask if someone could help me?
Thanks for your help.
 A: Let me fix some notation. We write $X=\Bbb S^1 \times \Bbb D^2$, where $\Bbb D^2$ denotes the closed unit disk. We have embeddings $j:\Bbb S^1 \cong \Bbb S^1\times\{0\}\subseteq X$ and $i:\Bbb S^1\cong A\subseteq X$. The first of both has a retraction $q:X\rightarrow \Bbb S^1, (x,t)\mapsto (x,0)$ which turns $(j,q)$ into a deformation retract, hence homotopy equivalence.
Now assume that $i$ admits a retraction $p$. As you noted this induces a surjection
$\pi_1(X) \rightarrow \pi_1(A)$ on all homotopy groups. Precomposing with the homotopy equivalence $j$ we thus have a surjection $\pi_1(\Bbb S^1) \rightarrow \pi_1(\Bbb S^1)$. This is a surjective group homomorphism $\Bbb Z \rightarrow \Bbb Z$, which necessarily is an isomorphism.
Note that the composite $qi:\Bbb S^1 \rightarrow \Bbb S^1$ defines a nullhomotopic map, hence the zero map on homotopy groups. But at the same time it is a composite of two isomorphisms of homotopy groups, hence itself an isomorphism of homotopy groups. A contradiction.
