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I'm helping to teach an undergraduate algebraic topology course (out of Hatcher's textbook). We have recently defined the degree of a map of spheres using homology, and the professor and I thought it would be nice if we could give some kind of argument that such a map is determined up to homotopy by its degree. I know two proofs of this: one using the Freudenthal suspension theorem, and the other using the Pontryagin correspondence between homotopy classes of [smooth] maps and framed cobordism classes of framed submanifolds (see Milnor, Topology from a Differential Viewpoint). Unfortunately, neither of these arguments would be accessible to our students, who have only seen the fundamental group and homology (no higher homotopy theory) and who are not necessarily expected to know any differential topology.

Thus, I ask the following question:

Is there an elementary argument (i.e., that can be understood by someone who only knows about homology and the fundamental group) that the degree of a map of spheres determines its homotopy type?

More precisely, what we have (or will have) available is most of the material in the first two chapters of Hatcher, not including the "additional topics."

If necessary, I'm willing to make plausible assumptions that the students may not know how to prove, such as
-Replacing $f \colon S^n \to S^n$ by a homotopic map if necessary, we may assume that there exists points with only finitely many preimages, such that $f$ is a homeomorphism locally about each preimage. (i.e., regular values)
-Every map of CW complexes is homotopic to a cellular map.
-The degree map $\pi_n(S^n) \to \mathbb{Z}$ is a group homomorphism. [This reduces us to showing that a degree-0 map is nullhomotopic.]

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1 Answer 1

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Take a look at Exercise 15 in Section 4.1, page 359 of the book you're referring to. This outlines an argument that should be the sort of thing you're looking for. The main step is to deform a given map to be linear in a neighborhood of the preimage of a point, using either simplicial approximation or the argument that proves the cellular approximation theorem. Once this is done, the rest is essentially the Pontryagin-Thom argument (in a very simple setting), plus the fact that $GL(n,\mathbb R)$ has just two path-components.

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