In Hatcher AT, we he defined the local degree of a map, as a remark he gave the case when the given map is a (local) homeomorphism. Let $f:S^n\to S^n,n>0$ and $y\in S^n$ such that $f^{-1}(y)=\{x_1,…,x_m\}$. Let $U_1,…,U_m$ be disjoint nbds of these points mapped by $f$ into a nbd $V$ of $y$.
For example, if $f$ is a homeomorphism, then $y$ can be any point and there is only one corresponding $x_i$, so all the maps in the diagram are isomorphisms and $\deg f|x_i = \deg f= \pm 1$. More generally, if $f$ maps each $U_i$ homeomorphically onto $V$, then $\deg f|x_i = \pm 1 $ for each $i$.
In this statement, the reason $\deg f|x_i =\pm 1$ is because of the (local) homeomorphism $f|_{U_i}:U_i\to V$ so $H_n(U_i,U_i-x_i)\simeq H_n(V,V-y)$ not necessary because of the above diagram. Especially, for the case of local homeomorphism case, we don't know $f_*:H_n(S^n)\to H_n(S^n)$ is an isomorphism. Am I correct?
Best Answer
You are right, that $f_* : H_n(U_i,U_i-x_i) \to H_n(V,V-y)$ is an isomorphism is due to the fact that $f \mid_{U_i}$ is a homeomorphism. And this does definitely not imply that any other of the three maps $(p_i)_* : H_n(S^n,S^n-f^{-1}(y)) \to H_n(S^n,S^n-x_i)$, $f_* : H_n(S^n,S^n-f^{-1}(y)) \to H_n(S^n,S^n-y)$ and $f_* : H_n(S^n) \to H_n(S^n)$ is an isomorphism. See Doubt about Hatcher's proof of degree calculation.
What may be confusing in Hatcher's diagram is that one is tempted to believe that 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}
commutes. But this is not true in general.
The local degree of $f$ at $x_i$ is defined as the degree of the map $$\phi_{(f,x_i)} : H_n(S^n) \stackrel{\approx}{\longrightarrow} H_n(U_i,U_i-x_i) \stackrel{(f_i)_*}{\longrightarrow} H_n(V,V -y) \stackrel{\approx}{\longrightarrow} H_n(S^n) .$$ And in fact here you see the real purpose of the specific isomorphisms $H_n(U_i,U_i -x_i) \stackrel{\approx}{\longrightarrow} H_n(S^n)$ and $H_n(V,V -y) \stackrel{\approx}{\longrightarrow} H_n(S^n)$: We need them to get the map $\phi_{(f,x_i)}$ which has the same domain and range $H_n(S^n)$ and thus allows to speak of its degree.