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}.

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