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enter image description here

The above picture is from Hatcher, Algebraic Topology, pp.135-136.

Why is my argument wrong?

Consider the diagram in the picture above. We know that the four red-boxed groups are all $\Bbb Z$ and that the whole diagram is commutative. Now we only consider the four maps, as indicated by red arrows. Since the vertical two red arrows are isomorphisms, we have a commutative diagram of the form:

$\require{AMScd}$ \begin{CD} \Bbb Z @>{}>> \Bbb Z \\ @V{\cong}VV @V{\cong}VV \\ \Bbb Z @>{}>> \Bbb Z \end{CD}

Now we conclude that the local degree at $x_i$ equals the degree of $f$ up to sign.

user302934
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1 Answers1

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Your diagram does not commute. In short, the diagram in Hatcher has the form $\require{AMScd}$ \begin{CD} &&\mathbb{Z}\\ &@V{\alpha}VV\\ \mathbb{Z}@<{\pi}<<\mathbb{Z}^m@>\gamma>>\mathbb{Z}\\ &@A{\beta}AA\\ &&\mathbb{Z} \end{CD}

in which $\alpha:1\mapsto(0,\cdots,1,\cdots,0)$ ($1$ in the $i$th place) and $\beta:1\mapsto(1,\cdots ,1)$, and $\pi$ is the projection to the $i$ th coordinate. For sure, you have $\pi\circ\alpha=\pi\circ\beta$, but there is no reason to expect that $\gamma \circ \alpha=\gamma\circ\beta$. Take $m=2,\gamma(m,n)=10m+20n$, for example.

Ken
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