Why doesn't the Homotopy group satisfy excision?

Question

I'm studying higher homotopy groups from the book Algebraic Topology by author Allen Hatcher, there he says that the sequence $A \to X \to X/A$ does not induce an exact sequence of homotopy groups. Why? Is this related to excision?

Answer 1

For both Homology and Homotopy functors we get a long exact sequence of pairs

$$\dots \to F_{i+1}(X,A) \to F_i(A) \to F_i(X) \to F_i(X,A) \to F_{i-1}(A) \to \dots $$

If the map $A\to X$ is a cofibration (or more generally if $(X, A)$ is a "good pair" in the sense of Hatcher) then $H_i(X, A) \cong \tilde{H}_i(X/A)$, and the proof uses excision (Proposition 2.22 in Hatcher). In fact by carefully looking at the proof you see that if $\pi_i$ also satisfied excision then we would have $\pi_i(X, A) \cong \pi_i(X/A)$ as well.

However, there cannot be a long exact sequence in homotopy groups for a general cofibre sequence of spaces. Take $X = D^{n+1}$ and $A=\partial X = S^n$, so that $X/A\cong S^{n+1}$. If we had a long exact sequence of the form

$$\dots \to \pi_{i+1}(S^{n+1}) \to \pi_i(S^{n}) \to \pi_i(D^{n+1}) \to \pi_i(S^{n+1}) \to \pi_{i-1}(S^{n}) \to \dots $$ then we would have $\pi_{i+1}S^{n+1} \cong \pi_i S^n$ for all $i> 0$, which is not true (for example, $\pi_2(S^1) = 0$ but $\pi_3(S^2) = \mathbb{Z}$). More generally we would be able to prove a suspension isomorphism theorem, which we know does not hold for homotopy groups.