MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

For space pairs $(X_1,A_1)$ and $(X_2,A_2)$ ,is there a groups M,such that $\pi_{k}(X_1\vee X_2,A_1\vee A_2)\cong \pi_k(X_1,A_1)\oplus M$?

share|cite|improve this question
$\pi_3(S_2 \vee S_2) = \mathbb Z^3\neq \mathbb Z^2 = \pi_3(S_2) \oplus \pi_2(S_2)$. I can't imagine that the relative homotopy groups would make it more well-behaved. – Will Sawin Oct 6 '12 at 3:55
up vote 6 down vote accepted

The pair $(X_1,A_1)$ is a retract of the pair $(X_1\vee X_2,A_1\vee A_2)$, so the map $\pi_k(X_1,A_1)\to \pi_k(X_1\vee X_2,A_1\vee A_2)$ is a split injection for all $k$. If $k\ge 3$ then the groups are abelian, and your result follows with $M$ the cokernel.

share|cite|improve this answer

There are some cases when one can give complete information. We know that the relative homotopy group $\pi_k(X_i,A_i)$ is a $\pi_1(A_i)$-module for $k >2$, crossed $\pi_1(A_i)$-module if $k=2$.

Suppose that the spaces are connected, well pointed, and the pairs $(X_1,A_1), (X_2,A_2)$ are $(k-1)$-connected. This result is a special case of the Higher Homotopy Seifert-van Kampen Theorem (HHSvKT) stated and proved in the book ``Nonabelian Algebraic Topology", that we have a pushout diagram

$$\matrix{(1,1)&\to&(\pi_1(A_1),\pi_k(X_1,A_1)) \cr \downarrow & & \downarrow \cr (\pi_1(A_2),\pi_k(X_2,A_2)) &\to & (\pi_1(A_1 \vee A_2),\pi_k(X_1 \vee X_2,A_1 \vee A_2))}$$ which is of modules if $k >2$, crossed modules if $k=2$. Here we write a $G$-module $M$ as a pair $(G,M)$, and we also use that notation in the crossed case though that involves also a morphism $\mu: M \to G$. Thus the result of the pushout is a kind of free product. This pushout diagram, or ``free product'', yields an explicit description of the resulting relative group. For convenience let us write the pushout as

$$\matrix{(1,1)& \to & (G_1,M_1) \cr \downarrow & & \downarrow \cr (G_2,M_2) & \to &(G_1*G_2, M_1 \vee M_2)} $$

Let $i; G_1 \to G_1 * G_2, i_2; G_2 \to G_1 * G_2$ be the canonical morphisms. We can form the induced $G_1 * G_2$-modules $i_*(M_1), j_*(M_2)$. Then for $k>2$, $M_1 \vee M_2 \cong i_*(M_1) \oplus j_*(M_2)$. For $k=2$ we have to use induced crossed modules and the coproduct of crossed $G_1 * G_2$-modules, but the relevant constructions are of course more complicated.

For the case $k>2$ there is presumably a proof using covering spaces and homology, but the case $k=2$, being nonabelian, looks inaccessible to those methods, and being a part of low dimensional topology, might also be relevant to geometric group theory. (The main theorem referred to dates from 1981, in a paper entitled ``Colimit theorems for relative homotopy groups'', JPAA. )

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.