I want to prove the map , $deg$ $:$ $\pi$$_n$$($ $S$$^n$ $)$ $\to$ $\mathbb Z$ $,$ is a homomorphism.
My proof : I used these facts to prove this map is well defined :
1- If $f$ $\simeq$ $g$ $\Rightarrow$ $deg$ $($ $f$ $)$ $=$ $deg$ $($ $g$ $)$
2- Every map $f$ $:$ $S$ $^n$ $\to$ $S$ $^n$ is homotopic to a multiple of identity map.
Proof of second fact :
$f$ is homotopic to a map that is simplicial with respect to some iterated barycentric subdivision of $S$$^n$. Hence there exists a point $q$ $\in$ $S$$^n$ such that $f$ $^{-1}$ $($$q$$)$$=$ { $p$$_1$ , ... , $p$$_k$ } and $f$ is a piecewise linear map around each $p$$_i$ and hence is locally invertible. For each $i$, we can intersect the images of the neighborhoods around $p$$_i$ on which $f$ is invertible to get a small ball around $q$. Let $g$ be the map which collapses the complement of this small ball to the basepoint. For each $i$, we then get a map $f$$_i$ $:$ $S$$^n$ $\to$ $S$$^n$ such that $f$$_i$$^{-1}$$($$q$$)$$=${$p$$_i$} by identifying the neighborhood around $p$$_i$ with $S$$^n$ (by collapsing its boundary to a point) and letting $f$$_i$ be the restriction of $gf$. Then $f$ is homotopic to the sum of the $f$$_i$, so we can reduce to the case $k$ $=$ 1. Thinking of $p$ and $q$ as points at innity, and using the fact that $f$ is linear, we can think of $S$$^n$ $\setminus$ $p$ $\to$ $S$$^n$ $\setminus$ $q$ as an invertible $n$ $\times$ $n$ matrix. Using Gaussian elimination, we can find a piecewise linear path from such a matrix either to the identity matrix, or the matrix of a reflection, depending on the sign of its determinant. Such a path gives a homotopy of $f$ either to the identity map or the reflection, which is $-$1 times the identity map.
How can i prove this map is a homomorphism ?
Why didn't you use the the fundamental theorem of Hopf (Cor. 4.25 in Hatcher) to proof 2?
It sayes: If $\deg f=\deg g$, then $f\simeq g$. The converse is true, since $H_n(f)=H_n(g)$, if $f\simeq g$. Since the induced map of $f\colon S^n\to S^n$ ($n>0$) must be of the form $H_n(f)(\alpha)=d\alpha$, the degree of $f$ is $d\in \Bbb Z$. But this is the same as the degree of $d$ times the identity. So the maps must be homotopic with respect to the fundamental theorem of Hopf.
How can you proof homomorphism, i.e. $\deg fg=\deg f\deg g$ (and therefore $\deg \mbox{id }=1$)? This is easy, since $H_n(fg)=H_n(f)H_n(g)$ because of functoriality. As a consequence of the definition of a homomorphism $\deg \mbox{id }=1$. Alternatively just consider $H_n(\mbox{id})=\mbox{id}$.
The degree map at hand is actually an isomorphism. Injectivity is obvious by the fundamental Hopf theorem. Surjectivity follows by considering $m$-times the antipodal map (with degree of $-1$) or the identity map (use homomorphism property).
I hope we are speaking about the same $\deg$-map.