The Whitehead theorem states that a map of connected CW complexes that induces an isomorphism in homotopy groups is a homotopy equivalence. In particular, isomorphisms in the homotopy category of pointed CW complexes can be detected by homming out of spheres $S^n$. But the equality of two morphisms cannot. The fact that this “relative Whitehead theorem” fails was the subject of a MO question. Today, I want to discuss another example along these lines. (I will assume a little more familiarity with algebraic topology than I have in previous posts.)

Recall that a common technique to show that a map is not nullhomotopic is to show that it does not induce the trivial morphism on some functor in algebraic topology. For instance, the fact that ${\pi_1(S^1) \neq 0}$ is used to show that ${S^1}$ is not contractible; this is probably the most basic example. But the basic invariants of algebraic topology can be insufficient. Here is an example which Eric Larson showed me yesterday.

There is a map

$\displaystyle (S^1)^3 \rightarrow S^3$

defined by quotienting by the 2-skeleton. One way to see this is that ${(S^1)^3}$ is a cube with opposite faces identified. ${S^3}$ is a cube with all the faces collapsed to a point. So we can define the quotient map by crushing the boundary of the cube defining ${(S^1)^3}$. Then, we compose this with the Hopf fibration

$\displaystyle S^3 \rightarrow S^2.$

The composite is a map ${f:(S^1)^3 \rightarrow S^2}$, which induces the trivial morphism on the homotopy groups. Indeed,

$\displaystyle \pi_n((S^1)^3) = \pi_n(S^1)^3$

is zero except in dimension one, since ${S^1 }$ admits a covering map from the contractible space ${\mathbb{R}}$. ${\pi_n(S^2)}$ is zero for ${n=1}$ (by a corollary of the simplicial approximation theorem). So the two spaces ${T=(S^1)^3, S^2}$ don’t have nonvanishing homotopy groups in the same dimension.

It is also true that ${f}$ induces the trivial morphism in reduced homology. The reason is that ${\widetilde{H}_*(S^2)}$ is zero except in dimension 2. But ${f}$ maps the 2-skeleton of ${T}$ into zero. Since the map ${T^2 \rightarrow T}$ induces a surjection on ${H_2}$, we see that ${f_*: H_2(T) \rightarrow H_2(S^2)}$ is zero.

Proposition 1 ${f}$ is not nullhomotopic.

The reason is that the Hopf fibration is, well, a fibration. If ${f}$ were nullhomotopic, then the map ${T \rightarrow S^3}$ would be nullhomotopic. This is because of the homotopy lifting property is precisely what defines a fibration. If there were a nullhomotopy of ${f: T \rightarrow S^3 \rightarrow S^2}$ to the constant map at ${\ast \in S^3}$, then we could lift this nullhomotopy to get a homotopy of $f$ to a map of $T$ into one of the fibers $S^1 \subset S^3$; the fibers are contractible, though.

But ${T \rightarrow S^3}$ does not induce the trivial map in homology. In terms of cellular homology in dimension 3, the map is actually just the identity. So ${T \rightarrow S^3}$ is not nullhomotopic. Another way to see this is to look at simplicial homology. The third homology class of the torus ${T}$ is represented by the cube itself. The same is true for ${S^3}$.