Let $X=S^1\times S^1$ and $A=S^1\times\{x_0\}$ where $x_0\in S^1$. Show that there is a retraction $r:X\to A$ but that $A$ is not a deformation re

Question

Let $X=S^1\times S^1$ and $A=S^1\times\{x_0\}$ where $x_0\in S^1$. Show that there is a retraction $r:X\to A$ but that $A$ is not a deformation retraction of $X$.

I really do not know how to solve this problem but I have several ideas, I do not know how to build a continuous function $r:X\to A$ such that $r(a)=a$ for any $a\in A$, could someone please help me with this? If we reason for the absurd and assume that $A$ is a deformation retraction of $X$ then $i_*: \mathbb{Z}\simeq\pi_1(A, x_0)\to \pi_1(X,x_0)\simeq\mathbb{Z}^2$ would be an isomorphism, But could there be an isomorphism between $\pi_1(A, x_0)\simeq\mathbb{Z}$ and $\pi_1(X,x_0)\simeq\mathbb{Z}^2$?, could someone help me please? Thank you.

Answer 1

$r:S^1\times S^1\rightarrow S^1\times S^1$ defined by $r(x,y)=(x,x_0)$.

Answer 2

For the second part, $\mathbb Z$ is cyclic; $\mathbb Z^2$ is not. So there cannot be an isomorphism.