Degree of maps of product CW complex If we treat the points of $S^3$ as quaternions i.e. $S^3=\{y_0+y_1i+y_2j+y_3k \mid y_0^2+y_1^2+y_2^2+y_3^2=1\}$, let $X=\mathbb{R}P^3$ by identifying antipodal points of $S^3$. Then quaternion multiplication gives rise to a group structure $f: X × X →X$ and $f$ is a cellular map.
There is a CW structure on $X$, with cells in dimensions $0, 1, 2, 3$,
the interior of $e^k$ being quotients of the $\{y\in S^3\mid y_k\neq0, y_{k+1}=...=y_3=0\}$. Therefore, $f$ induces a CW chain map $f_*:C_i^{CW}(X\times X)\to C_i^{CW}(X)$.
I want to compute this map. If I consider $e_1\otimes e_2\in C_3^{CW}(X\times X)$, I should have $f_*(e_1\otimes e_2)=me_3^X$, where $e_3^X$ is the $3$-dimensional cell in $X$ and $m$ is degree of composition. I wonder how to compute the degree.

The points in $e_1$ are of the form $(t:1:0:0)$ and those in $e_2$ are of the form $(u:v:1:0)$, and their product is $$(t:1:0:0)\cdot(u:v:1:0)=(tu-v:tv+u:t:1).$$
Now pick a point $(a:b:c:1)$ in $e_3$ and look for points in $e_1$ and $e_2$ whose product is $(a:b:c:1)$. There is one choice, which has $t=c$, $u=(ac+b)/(1+c^2)$ and $v=(cb-a)/(1+c^2)$.
This tells you that the degree is $\pm 1$. I'll let you find the sign (which is that of the Jacobian of the map $e_1\times e_2\to e_3$ at those points)
