# Examples of failure of excision for homotopy groups ($\pi_k(X, A)$ is not $\pi_k(X/A, *)$)

Let $A$ be a subcomplex of CW-complex $X$. The excision axiom for homology implies that $H_i(X, A)\cong H_i(X/A, *)$, and it is widely known that homotopy groups don't have this property. However, they satisfy a significantly weaker Blakers-Massey theorem. One of its consequences is that when $A$ is $n$-connected and the inclusion $A\to X$ is an isomorphism on the first $s$ homotopy groups, $\pi_k(X, A) \cong \pi_k(X/A, *)$ for $k < s + n - 1$.

Thus to devise an example where $\pi_k(X, A) \not\cong \pi_k(X/A, *)$ it's necessary to take large enough $k$, and the calculation of both sides becomes hard. The original paper of Blakers and Massey claims there are simple examples, but I wasn't able to make them up myself. What are some simple examples of the pairs $(X, A)$ and $(X/A, *)$ with different homotopy groups? Preferably with both $A$ and $X$ simply connected.

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Well, I figured some examples. The easiest is probably $S^2 \hookrightarrow S^3$ as an equator. Here $\pi_4(S^3, S^2) \cong \mathbb{Z}\oplus \mathbb{Z}/2\mathbb{Z}$ by the exact sequence of a pair (the inclusion is obviously null-homotopic, so maps $\pi_k(S^2)\to\pi_k(S^3)$ are all zero). On the other hand, the quotient is a wedge of two three-spheres, and $\pi_4(S^3 \vee S^3)$ is equal to $\pi_4(S^3 \times S^3) = \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ since $\pi_4$ depends only on 5-skeleton, while $S^3 \times S^3$ is $S^3\vee S^3$ with a 6-cell attached.

It is also true for $S^1\hookrightarrow S^2$, though you need more machinery. $\pi_3(S^2, S^1)$ is $\mathbb{Z}^2$ by the exact sequence of a pair as well. On the other hand, $\pi_3(S^2 \vee S^2) \cong \mathbb{Z}^3$ using the standard trick: fiber of $S^2 \vee S^2 \to K(\mathbb{Z}, 2)\times K(\mathbb{Z}, 2)$ has $H^3 \cong \mathbb{Z}^3$ by Serre's spectral sequence.

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Well, perhaps "more machinery" is a lie: for the first part, you need to know $\pi_4(S^3)$, and that's hard to prove without spectral sequences. – Dmitry Aug 19 '13 at 9:23

Dropping the preference for $X,A$ to be simply connected, then a factor involved is the influence of the fundamental group. So suppose we have a space $X$ which is the union of two open subspaces $A,B$ and let $Z=A \cap B$ contain a base point $a$. Suppose all these spaces are path connected and that the pair $(B,Z)$ is $(n-1)$-connected, $n \geqslant 3$ . Let $\lambda:\pi_1(Z) \to \pi_1(A)$ be the morphism of groups induced by inclusion. Then the module $\pi_n(X,A)$ is isomorphic to the $\pi_1(A)$-module induced from the $\pi_1(Z)$-module $\pi_n(B,Z)$ by the morphism $\lambda$. So one can write $$\pi_n(X,A) \cong \lambda_* \pi_n(B,Z).$$ This result was first proved in

R. Brown and P.J. Higgins, "Colimit theorems for relative homotopy groups", J. Pure Appl. Algebra 22 (1981) 11-41.

It implies the Relative Hurewicz Theorem. For example, it implies that if $A \to X$ is a cofibration, and $(X,A)$ is $(n-1)$-connected, then $\pi_n(X/A)$ is obtained from the $\pi_1(A)$-module $\pi_n(X,A)$ by factoring out the action of $\pi_1(A)$.

Full details are also in the book Nonabelian Algebraic Topology.

However if you want to deal with the simply connected case, I will write on that later.

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