I was reading Hatcher's algebraic topology book,there is a theorem for local degree as follows in Proposition 2.30 p136:
$$\deg(f) = \sum_i \deg_{x_i}f$$
Where $x_i \in f^{-1}(y)$ that $\deg_{x_i}$ is the local degree.
which is proved using the following diagram:
In the book we use the fact that $p_i\circ j(1) = 1$ thanks to the isomorphism in the lower triangle.the proof stated as below:
Proof: By excision, the central term $H_{n}\left(S^{n}, S^{n}-f^{-1}(y)\right)$ in the preceding diagram is the direct sum of the groups $H_{n}\left(U_{i}, U_{i}-x_{i}\right) \approx \mathbb{Z}$, with $k_{i}$ the inclusion of the $i^{t h}$ summand and $p_{i}$ the projection onto the $i^{t h}$ summand. Identifying the outer groups in the diagram with $\mathbb{Z}$ as before, commutativity of the lower triangle says that $p_{i} j(1)=1$, hence $j(1)=(1, \cdots, 1)=\sum_{i} k_{i}(1)$. Commutativity of the upper square says that the middle $f_{*}$ takes $k_{i}(1)$ to $\operatorname{deg} f \mid x_{i}$, hence the sum $\sum_{i} k_{i}(1)=j(1)$ is taken to $\sum_{i} \operatorname{deg} f \mid x_{i}$. Commutativity of the lower square then gives the formula $\operatorname{deg} f=\sum_{i} \operatorname{deg} f \mid x_{i}$
The question is the isomorphism only maps generator to generator,can we let $p_i\circ j(1) = -1$?
