If $A$ is a retract of $X$, then is the (reduced) suspension $SA$ a retract of $SX$?

Question

I have a compact Hausdorff space $X$ and $A$ a closed subset. Suppose $A$ is a retract of $X$, that is, there is a continuous map $r:X\to A$ that restricts to the identity on $A$.

Is the suspension $SA$ still a retract of $SX$?

Is the reduced suspension $\Sigma A$ still a retract of $\Sigma X$?

I believe only the latter is true. If it helps, the latter may also be identified with the smash product $S^1\wedge X$.

Answer 1

Both $S$ and $\Sigma$ are functors, so both preserve retracts: given maps $i: A \to X$ and $r: X \to A$ such that $r \circ i = 1_A$, then applying any functor $F$ gives $F(r) \circ F(i) = F(1_A) = 1_{F(A)}$.