Suppose $X$ is a contractible space and $A \subset X$ with $A$ being homotopy equivalent to $S^2$. Prove that there is no retraction $X \rightarrow A$
Suppose there is a retraction $r: X \rightarrow A$. Then $ri = Id_A$.
Let's look at the homomorphism induced on the second homology.
$H_2(A)\stackrel{i_*}{\rightarrow}H_2(X) \stackrel{r_*}{\rightarrow} H_2(A)$
Since $A \cong S^2$ we have $H_2(A) = \mathbb{Z}$. Furthermore, $ri_*=Id_{A_*}$.
But $X$ is contractible so $H_2(X)=0$ and thus $ri_*$ factors through zero and thus cannot be the identity on $\mathbb{Z}=H_2(A)$ and thus there is no retraction $r: X \rightarrow A$
Spot on. More generally, if $X$ is contractible and $A \subset X$ is not acyclic then $A$ cannot be a retract of $X$. The argument is the same as yours: pick $q \in \mathbb{N}$ such that $0 \neq H_q(A)$. If we were to have a retraction $r : X \to A$, then
$$ 1_{H_q(A)} = H_q(1_A) = H_q(ri) = H_q(r)H_q(i) \tag{1} $$
Since $X$ is contractible, $H_qr$ is the zero map (as it has domain zero) and this together with $(1)$ implies a contradiction: since $H_q(A) \neq 0$, it can't be that $1_{H_n(A)} = 0$. Hence no such retraction exists.