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In Bredon's Topology and Geometry it‘s postulated that for a $k$-fold covering projection $p\colon X \to Y$, $\deg(p)=k$, where $\deg(p)=k$ The example given is of $X$ the $3$-dimensional sphere $S^3$ and $Y$ the Lens space $L(k,m)$, both of which have $H_3 = \mathbb Z$.

So I'd like to know where this comes from. In the case of $X=Y=S^1$ it‘s clear, but what about the general case?

Thanks for putting my question into TEX, I have to find it out myself, so it's going to take some time to edit. Here the short answer: For a continuous function $f\colon M \to N$ with $M$ and $N$ compact oriented $n$-dimensional manifolds, and its induced homological homomorphism $f_\#$ following is true

$f_\#([M])=\pm deg(f)[N]$, $[M]$ and $[N]$ the generators of the n-th homological groups $H_n(M)= H_n(N) = \mathbb Z$.

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What definition of degree are you working with? –  Mariano Suárez-Alvarez Aug 28 '12 at 18:54

1 Answer 1

up vote 4 down vote accepted

I think this is easiest to see in dimension $2$, so go ahead and assume $M$ and $N$ are compact oriented surfaces, and let $f\colon M\to N$ be a covering map of degree $k$. One way to visualize what's happening is as follows.

First, triangulate $N$, and do it in such a way that all the triangles are small. Let $\mathcal{T}$ denote the set of all the triangles in this triangulation. Now, because all the triangles $T\in \mathcal{T}$ are small, when you look at $f^{-1}(T)$, it should consist of $k$ disjoint triangles. In this way, you get a triangulation $\mathcal{S} := f^{-1}(\mathcal{T})$ of $M$.

The generator $[M]$ of $H_n(M)$ is represented by the $n$-chain $\sum_{S\in \mathcal{S}} S$. Thus the image of $[M]$ under $f_\sharp$ is $\sum_{S\in \mathcal{S}} f_\sharp S$. Since each $S\in \mathcal{S}$ maps to a triangle in $\mathcal{T}$, and moreover, every triangle in $\mathcal{T}$ has exactly $k$ triangles $S\in \mathcal{S}$ mapping onto it, it follows that $$f_\sharp[M]= \pm\sum_{T\in \mathcal{T}} kT = \pm k\sum_{T\in \mathcal{T}}T = \pm k[N].$$ The sign $\pm$ depends on whether or not $f$ is orientation preserving or reversing.

Of course, dimension 2 really doesn't matter here. It's just easier to visualize triangles. Hope this helps.

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Are higher-dimensional (compact, oriented) manifolds triangulable? –  Jesse Madnick Aug 29 '12 at 3:20
@JesseMadnick: they are if they are smooth. If not, $E_8$ is a counterexample. –  user641 Aug 29 '12 at 8:40
This helps for this $3$-dimensional case, thanks. –  x x Aug 29 '12 at 12:26

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