Same winding number implies homotopic In the lecture we defined the winding number $\omega$ to be the unique integer $d$ such that for a continuous map $f:S^{n}\to S^{n}$ it holds that $f_{\ast}:H_{n}(S^{n})\to H_{n}(S^{n}), [a]\mapsto d[a]$, where $H_{n}(S^{n})$ denotes the singular homology.
Is there an easy way to prove that if $\omega(f)=\omega(g)$ for some continuous satisfying $f,g:S^{n}\to S^{n}$ such that $f(p_{0})=g(p_{0})=q_{0}$ where $p_{0},q_{0}\in S^{n}$ then $f$ and $g$ are homotopic relative to $\{p_{0}\}$?
The inverse statement is clear, by the homotopy invariance of $\omega$, which directly follows by the homotopy invarianve of induced maps. But the statement above seems to be not so trivial.
(This is not a homework)
 A: This is not such an easy theorem to prove.
The proof in Hatcher's Topology appears in Corollary 4.25, as an application of the Freudenthal Suspension Theorem stated in Corollary 4.24, which itself depends on a restricted version of the excision theorem that holds in the setting of homotopy groups (the full version of the excision theorem for homology groups does not generalize to homotopy groups).
It is also a consequence of the Hurewicz theorem, although if you're not careful that can lead you into a circular argument, given that some proofs of the Hurewicz theorem actually assume the statement you are asking to prove.
I do believe there exists a "low technology" proof, where one first homotopes an arbitrary continuous function $S^n \mapsto S^n$ to one which is a close approximation possessing better local properties (i.e. smooth; or cellular with respect to a CW decomposition). Next one uses that approximation to compute the degree $d$ of the map. Finally one uses the local structure as a guideline for constructing a homotopy from the given continuous function to some "standard" example of a continuous function of degree $d$.
