Let $$T^2=\{(z,w)\in\mathbb{C}^2:|z|=|w|=1\}$$ denote the 2-torus. I'm trying to compute the degree of the smooth map $f:T^2\longrightarrow T^2$ given by $f(z,w)=(z^2,w^2)$. I'd like to find this degree without using algebraic topology, so I'm attempting to compute the sum $$\operatorname{deg}(f)=\sum_{p\in f^{-1}(q)}\operatorname{sgn}(\det df_p)$$ where $q\in T^2$ is any regular value of $f$ (which is guaranteed to exist by Sard's theorem).
Is there any way that I can compute the above sum without resorting to finding the (real) coordinate representation of $f$? Since $f$ has a nice description in terms of $z$ and $w$, my immediate instinct is to work with complex coordinates. However, I'm unable to see a way to reconcile the use of these coordinates with the answer I'd get by computing $\operatorname{deg}(f)$ with real coordinate charts (and this seems particularly complicated).
Is there possibly another way to compute this, without using coordinates? I'm self-studying/revising my understanding of smooth manifolds, and I'd really appreciate any help. Thank you!
Choose product charts for $S^1\times S^1=T^2$ (say, using angle coordinates), and working locally, write $f(\theta,\varphi)=(g(\theta),h(\varphi))$. We have $$df_{(\theta,\varphi)}=\begin{pmatrix}dg_\theta & 0\\ 0 & dh_\varphi\end{pmatrix},$$ which is a block matrix with determinant $\det df_{(\theta,\varphi)}=(\det dg_\theta)\cdot(\det dh_\varphi).$
This gives (for $q$ regular)
$$\sum_{p\in f^{-1}(q)}\operatorname{sgn}\det df_p=\sum_{(\theta,\varphi)\in f^{-1}(q)}\operatorname{sgn}(\det dg_\theta\cdot\det dh_\varphi),$$ allowing us to relate $\deg f$ to the degrees of both $g,h:S^1\longrightarrow S^1$. It can be shown that $\deg g=\deg h=2$, and since there are $2\cdot 2=4$ elements in $f^{-1}(q)$, the above sum becomes $1+1+1+1=4$.