Show that there are no retractions $r: X \rightarrow A$ in the following cases:
(c) $X = S^1 \times D^2$ and $A$ the circle shown in the figure.
My attempt : I found the answer here But i didn't undertstand the answer.If $A$ is circle then why is $A $ contractible ?
My thinking: I had two things in my mind.
case I: Here $\pi_1(A) \cong \mathbb{Z}$ and $\pi_1(X) \cong \mathbb{Z} \implies $ Every subgroup of $\mathbb{Z}$ is cyclic $\implies$ the homomorphism $i_* : \mathbb{Z} \to \mathbb{Z}$ is injective
Cases II :let $i : A \hookrightarrow X$ be the inclusion , and fix $x_0 \in A$.The induced map $i_*:\pi_1(A,x_0) \to \pi_1(X,x_0)$ is defined by $i_*([f])=[if]$
we know that $[f] \in \pi_1(A,x_0)$ where $f: S^1 \to A$ is a loop based at $x_0$
similarly $[if] \in \pi_1(X,x_0) $ where $if: S^1 \to X$ is a loop based at $x_0$
Use the fact : circle is not contractible
Also, it is given that $A$ is circle $\implies \pi_1(A,x_0) \cong \mathbb{Z}$ which is non-trivial
similarly $\pi_1(S^1 \times D^2) \simeq \pi_1(S^1) \times \pi_1(D^2) \simeq \mathbb{Z}$ which is non-trivial
This implies the inclusion $i: A \hookrightarrow X$ induces a non-trivial map $i_* : \pi_1(A) \to \pi_1(X)$
Therefore $i_*$ is injective
Tell me where I'm wrong?
The space $A$ is circle. So, as a topological space, $A$ is not contractible. On the other hand, $A$ is embedded in $X$. Let $i_A : A \to X$ be the inclusion map. The answer you found shows that $A$ is contractible as a subspace of $X$. That is, there is a homotopy $H : A \times I \to X$ that begins at $i_A$ and ends at a constant map. Thus $(i_A)_* : \pi_1(A) \to \pi_1(X)$ is the trivial homomorphism.