This is from Hatcher's Algebraic Topology page 136.
Suppose continuous map $f: S^n \to S^n$ $(n>0)$ has the property that for some $y \in S^n$, the preimage $f^{-1}(y)$ consists only finitely many points $x_1, \ldots, x_n$.
Let $U_1 \ldots, U_m$ be disjoint neighborhoods of these points, mapped by $f$ to a neighborhood $V$ of $y$, then $f(U_i-x_i)\subset V-y$ for each $i$.
We use the following commutative diagram to define local degree of $f$ at $x_i$:
$k_i, p_i$ are induced by inclusions, and $j:H_n(S^n) \to H_n(S^n,S^n-f^{-1}(y))$ is the homomorphism in long exaxt sequence for pair $(S^n,S^n-f^{-1}(y))$.
The top homomorphism $f_*$ becomes multiplication by an integer, defined to be local degree of $f$ at $x_i$, denoted by $\deg f \mid {x_i}$.
My question:
How can I show that this diagram really commutes?
Related question: Local Degree of a map between $n$-spheres
Best Answer
The commutativity of the upper half is trivial because it is obtained from a commutative diagram of pairs of topological spaces and maps of pairs by applying the functor $H_n$.
The commutativity of the right lower square follows from the fact that the long exact sequence of a pair is natural (see the chapter "Naturality" in Hatcher Section 2.1). Simply consider the map of pairs $f : (S^n,S^n \setminus f^{-1}(y)) \to (S^n,S^n \setminus \{y \})$.
Similarly the left lower triangle commutes. Just consider the map of pairs $id : (S^n,S^n \setminus f^{-1}(y)) \to (S^n,S^n \setminus \{x_i \})$.