About homotopy groups of pairs

Question

Suppose $A$ is a deformation retract of $X$, for $n\ge 2$ and for any $x_0\in A$, how to show $$\pi_n(X,x_0)=\pi_n(A,x_0)\oplus\pi_n(X,A,x_0)$$

I am not clear why the homotopy groups are not isomorphic, since every mapping of $S^n$ into $X$ is homotopic to one mapped into $A$ rel $x_0$.

Answer 1

There is a long exact sequence of homotopy groups: $$ \dots \pi_n(A) \to \pi_n(X) \to \pi_n(X,A) \to \pi_{n-1}(A) \to \dots$$

$A$ is in particular a retract $r : X \to A$ of $X$, which induces a surjection in homotopy $\pi_n(X) \to \pi_n(X,A)$, therefore the LES breaks into parts: $$0 \to \pi_n(A) \to \pi_n(X) \to \pi_n(X,A) \to 0$$ Since this short exact sequence is split (again via the retraction), and the groups are abelian for $n \ge 2$, it follows that $\pi_n(X) \cong \pi_n(A) \oplus \pi_n(X,A)$.

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In particular since $A$ is a deformation retract of $X$, the inclusion induces an isomorphism in homotopy; therefore $\pi_n(X,A) = 0$.

Answer 2

I believe the answer should be: Since $A$ is a deformation retract of $X$, $X \simeq A$. But $\pi_n$ is homotopy invariant, so you get $\pi_n(X)=\pi_n(A)$.