Degree of a map from the surface of a tetrahedron to a triangle with identified edges This problem comes from Topology and Geometry of Bredon :

Consider the surface $S$ of a tetrahedron. Number its vertices in any order and number three vertices of a triangle $1,2,3$. Let $S'$ be the triangle with the sides collapsed to a point. Define a map $S \to S'$ by taking a triangle $T$ in $S$ and mapping its vertices to the points $1,2,3$ of the triangle in order of the numbering of the vertices of $T$. Extend that to an affine map of $T$ to the triangle and then collapse to get a map to $S'$. Find the degree of this map up to sign. Does the answer depend on the numbering?

The inverse image of the point at the center of the face $S'$ consists of 4 fours points, each of which is located on a different face of the tetrahedron. How can I compute the sign of the degree of the map? In order to do so, I think that one could say the degree is the sum of the local degree.
 A: Here is an intuitive visual argument. Picture the outward normal vector to each face of $S$ and picture the triangle, $\Delta$, flat on the $x, y$ plane in $\mathbb{R}^3$, with an upward-pointing normal vector. These vectors specify the orientations of these spaces (the faces of $S$ viewed as subspaces with induced orientation). Visualize taking each of these faces and placing them on $\Delta$ according to the numbering that you chose and see whether the normal vectors point in the same or opposite direction. If they point in the same direction, then the local degree around a point of the corresponding face is $+1$, otherwise, it is $-1$. If you draw a picture and label everything you'll see that two faces will have $+1$ and two $-1$ so the degree must be $0$.
Here is a sketch of a more rigorous approach using Corollary 7.5 in Chapter IV of Bredon.
Model $S$ as the boundary of the standard $3$-simplex:
$$ S = \partial\Delta_3 = \{(t_i)_{i=1}^4\in\mathbb{R}^4|\sum_{i=1}^4 t_i = 1, t_i = 0 \textrm{ for some }i\}.$$ Let $T_i\subset S$ be the subset of points for which $t_i = 0$. These are the faces of the tetrahedron. Let $v_i$ be the $i$-th vertex, which is also just the $i$-th standard basis vector in $\mathbb{R}^4$. Model the triangle $\Delta$ as just one of these faces, say $T_4$. We order the vertices of $S$ according to $i$ and similarly the vertices of $\Delta = T_4$. Now, the resulting map $f:S\to S'$ doesn't factor through a map into $\Delta$, but it does if we restrict to the interior of each face. Let $f_i:\textrm{int}(T_i)\to \Delta$ be the restriction. Then these maps are given by, for example, $f_1(0,t_2,t_3,t_4) = (t_2,t_3,t_4,0)$, and similarly defined for other $i$. Now, in order to compute the local degree, we need to take coordinates for the interior of each face. For int$(T_4) = \textrm{int}(\Delta)$, we can take for example
$$(0,1)\times (0,1)\xrightarrow{\phi_4}\textrm{int}(T_4).$$
$$(u,v)\mapsto (u,v,1-u-v,0) $$
Choose local coordinates for the rest of the $T_i$, keeping in mind that they should be rotations of the coordinates we chose for $T_4$. From there you can compute the Jacobian determinants of each of the maps $\phi_4^{-1}\circ f_i\circ\phi_i$ and find the same result as the visual argument above.
