About the definition of local degree

[quasar987]

In Hatcher's book, http://en.wikipedia.org/wiki/Talk:Coherent_topology, see the very end of page 135 and the beginning of page136 for the definition of the local degree of a map S^n-->S^n.

I don't get how the local degree degf|x_i could be anything else than ±degf: In the diagram of page 136 used to define the local degree, identify all the outer groups with Z. Start with 1 in the lower left [itex]H_n(S^n)[/itex] and follow the isomorphism that leads to [itex]H_n(U_i,U_i-x_i)[/itex].Since an isomorphism brings a generator to a generator, we end up with ±1. Then take this ±1 to [itex]H_n(V,V-y)[/itex] through f_*. We end up with ±degf|x_i by definition of the local degree. Now take that same initial 1 to [itex]H_n(V,V-y)[/itex] through the other outer route. We get [tex]1\stackrel{f}{\mapsto} \deg f\stackrel{\cong}{\mapsto}\pm \deg f[/tex], and so by commutativity of the diagram, \degf|x_i=±degf.

Where am I mistaken??

 

[yyat]

Commutativity of a diagram means that you get the same result if you follow the directions of the arrows. In the case where all maps are isomorphisms, the directions can be changed, but in this case the [tex]k_i[/tex] are usually not isomorphisms. If [tex]k_i[/tex] where an isomorphism for an [tex]i[/tex] you could reverse the directions of the downward arrows in the top left triangle and your argument would be valid.

Consider for example the map [tex]S^1\to S^1,z\mapsto z^2[/tex] which has degree two. Each point has two preimages and the local degree is one. So we have another proof that 1+1=2.

 

[quasar987]

Oh wow! Thanks for clearing that up.. you just saved me from a lot of confusion.