I am currently enrolled in an introductory course on algebraic topology.
In a recent research seminar at my home institution, it was stated that the homotopy group functors $\pi_1, \pi_2, \pi_3, …$ are not a generalized homology theory, and I'd like to know why.
I know that $\pi_n(X)$ consists of the homotopy classes of basepoint-preserving maps from $S^n$ into $X$, and I know of the four axioms (homotopy, exactness, excision, additivity) that need to be satisfied to define a generalized homology theory. What axiom isn't satisfied when trying to define a generalized homology theory via the homotopy group functors ?
Thanks!
Best Answer
First, I'll note that homotopy groups do not even solely take values in groups since $\pi_0$ is a set. Even if we ignore $\pi_0$ they do not take values in abelian groups since $\pi_1$ may be noncommutative.
Let's suppose we ignore both of those issues. If the homotopy groups were a generalized homology theory, then they would have a suspension isomorphism. This would mean that the homotopy groups of $S^1$ would be the same as the shifted homotopy groups of $S^2$. However, $\pi_2(S^1)=0$ while $\pi_3 (S^2)=\mathbb{Z}$.
When one forces the suspension isomorphism to work for homotopy groups, one arrives at the stable homotopy groups $\pi_* ^s$ which do form a homology theory. This has the result of making the stable homotopy groups of $S^0$ very complicated, even though the homotopy groups of $S^0$ are trivial.