Doubt about Hatcher's proof of degree calculation In Hatcher's Algebraic Topology, Proposition 2.30 shows that the degree of a map $S^n \to S^n$ is the sum of its local degrees across a finite set $\{x_i\}$ that is the preimage of a point $y$. The definition of the local degree and the proof of the proposition use the commutative diagram below. 
My question: why can't we follow the outer isomorphisms to get the degree of the bottom horizontal map? All of the outer vertical maps send generators to generators because they are isomorphisms, so why is it that the degree of the bottom map is not the degree of the top map? I think I'm overlooking something silly but can't figure it out. Effectively the same question was asked here but I do not find the answer convincing because it ignores the outer maps.
 A: After Hatcher has considered his diagram he proves Proposition 2.30 which says that $\deg f = \sum_i \deg f \mid_{x_i}$. Commutativity of the "outer isomorphism diagram"
$\require{AMScd}$
\begin{CD}
H_n(U_i,U_i-x_i)  @>{(f_i)_*}>> H_n(V,V -y) \\
@V{\approx}VV @V{\approx}VV \\
H_n(S^n) @>{f_*}>> H_n(S^n) \end{CD}
would mean that $\deg f = \deg f \mid_{x_i}$ for each $i$. In that case the above degree-formula couldn't be correct. This is a priori no argument against commutativity - perhaps the degree formula is wrong? Let us come to this point later.
What Hatcher is doing with his diagram are two things:

*

*He shows that $A = H_n(U_i,U_i-x_i)$ and $B =  H_n(V,V -y)$ are infinite cyclic groups. For this purpose he wouldn't have needed the whole diagram, he only uses excision and the exact sequence of a pair to establish explicit isomorphisms $\phi : A \to H_n(S^n)$ and $\psi : B \to H_n(S^n)$. Thus $(f_i)_* : A \to B$ is a homomorphism between infinite cyclic groups. To describe this map, we have to make a choice of generators $g_A$ for $A$ and $g_B$ for $B$. These choices are in general independent since normally $A \ne B$. Only if $U_i = V$ and $x_i = y$ we have $A = B$ and one choice is sufficient. Based on our choices we get $(f_i)_*(g_A) = n g_B$ for some factor $n \in \mathbb Z$. Since also  $-g_A$ and $-g_B$ are generators, we do not get a canonical factor $n = n((f_i)_*)$ - without an explicit choice of generators it is determined only up to sign. And here $\phi, \psi$ enter: To the map $\psi (f_i)_* \phi^{-1} : H_n(S^n) \to H_n(S^n)$ we can associate a canonical factor in $\mathbb Z$ because we may choose one generator of $H_n(S^n)$. This factor is denoted by $\deg f \mid_{x_i}$.


*The full diagram is used in the proof of Proposition 2.30. Have a look at the proof!
Let us finally come to the outer isomorphism diagram. In Example 2.31  Hatcher shows how to construct a map $f : S^n \to S^n$ of any degree $k \in \mathbb Z$. However, $\deg f \mid_{x_i} = \pm 1$, thus in general $\deg f \ne \deg f \mid_{x_i}$ - the outer isomorphism diagram does not commute in that case.
Although we have found explicit examples in which the outer isomorphism diagram does not commute, an obvious heuristic argument  why we cannot expect commutativity is this:

*

*The map $p_i : H_n(S^n,S^n-f^{-1}(y)) \to H_n(S^n,S^n-x_i)$ is in general not an isomorphism. $H_n(S^n,S^n-f^{-1}(y))$ is a free abelian group with $m$ generators, where $m$ is the number of points in $f^{-1}(y)$.


*$p_i$ goes into the "wrong direction" which makes it impossible for $m > 1$ to show via standard diagram chasing that the outer isomorphism diagram commutes. It only works for $m=1$ where we have $p_i = id$.
