If $A\subset X$ is a deformation retract, then do we have $\pi_k(X,A)=0$

Question

Let $A$ be a deformation retract of the topological space $X$; the example in my mind is for example: $X=(\mathbb C^*)^n$ and $A=(S^1)^n$.

Notice that if we consider the homology instead, then we know $H_k(X,A)=0$ by using the long exact sequence in homology theory and using the fact that $H_*(A)\cong H_*(X)$.

However, do we have still have $\pi_k(X,A)=0$ for the relative homotopy groups? I am not familiar with them; do we have some similar properties?

Answer 1

Let $j:(X,x_0)\to (X,A)$ denote an inclusion of pairs, and $i:A\to X$ the obvious inclusion. There is a long exact sequence of homotopy groups: $$ \cdots \to\pi_{k+1}(X,A)\to\pi_k(A,x_0)\xrightarrow{i_*}\pi_k(X,x_0)\xrightarrow{j_*} \pi_k(X,A)\to\cdots.$$ Since $A\subseteq X$ is a deformation retract of $X$, it follows that $i:A\to X$ induces an isomorphism on homotopy groups and hence that the middle map $i_*$ is an isomorphism in all degrees. Hence, $\pi_k(X,A)=0$ for all $k$ by exactness.