I'm currently studying "Fibre Bundles" by Husemoller and I'm trying to understand in Ch.$8$ Theorem $10.4$ which states:
$10.4$ Theorem. Let $\xi[F] = (E_F, p_F,S^n)$ be a fibre bundle with structure group $G$ and fibre $F$. Let $c_\xi : S^{n-1} \longmapsto G$ the characteristic map of $\xi$, and for $y_0 \in F$, let $c_F : S^{n-1} \longmapsto F$ be the map defined by $c_F(y_0) = c_\xi(x)y_0$. Then $\text{Im}\partial = \text{ker}\alpha$ is generated by $[c_F]$ in the exact sequence
$$\mathbb{Z} = \pi_n(S^n) \overset{\partial}{\longmapsto}\pi_{n-1}(F,y_0) \overset{\alpha}{\longmapsto}\pi_{n-1}(E_F,x_0) \longmapsto 0$$
The proof goes as follows : We only show that $\partial[f] = [c_F],$ where $\text{deg}f = \pm 1$.
I don't understand the latter. Why is it sufficient to show that $\partial[f] = [c_F],$ where $\text{deg}f = \pm 1$?
Also: Is this related to the isomorphism between $\mathbb{Z} = \pi_n(S^n)?$ The isomorphism I've had in mind was the one given by Freudenthal, i.e the suspension; Since the degree of a map (definition I have at least) is defined by the map induced in homology, I'm quite hesitant with my thoughts since the degree doesn't induce an isomorphism (between $\mathbb{Z} = \pi_n(S^n)$) because it shoudn't be an homomorphism either if I'm not mistaken, so I don't know how to relationate the concept of the degree (which I'm locking in Homology) with this isomorphism between $\mathbb{Z} = \pi_n(S^n)$ which should be the "sufficient" part the theorem to show.
The degree map from $\pi_n(S^n) \rightarrow \mathbb{Z}$ is defined as the Hurewicz map $\pi_n(S^n) \rightarrow H_n(S^n) \cong \mathbb{Z}$, among many equivalent way. For all $n>0$ it is an isomorphism, meaning the degree of a map uniquely specifies the element in $\pi_n(S^n)$ it comes from. The isomorphism $H_n(S^n) \cong \mathbb{Z}$ is one among two possible isomorphisms, often authors are nonexplicit about this isomorphism, and so it is often only described up to negation.
For this reason, it can be easier to only describe the degree of a map up to sign. In this theorem, the sign does not matter because you are asking about the subgroup the image of this element generates: the subgroup generated by $f$ is equal to the subgroup generated by $-f$.