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Let $\pi: \tilde{X} \rightarrow X$ be an n-sheeted covering. Hatcher (section 3G), defines the transfer homomorphism, $\pi^*: H^k(\tilde{X}, Z) \rightarrow H^k(X, Z)$ on the chain level by sending the singular chain $\sigma: \Delta^n \rightarrow X$ to $\Sigma_{1 \leq i \leq n} \sigma_i : \Delta^n \rightarrow \tilde{X}$, where each $\sigma_i$ is a lift of $\sigma$, and then taking cohomology (one checks easily that this is a chain map).

A particular feature of this homomorphism is that $\tau^*\pi^*: H^k(X, Z) \rightarrow H^k(X, Z)$ is that it corresponds to multiplication by $n$. I was particularly interested in this because it solves a problem in Milnor's characteristic classes book about computing the the $Z$-cohomology of $BO(n)$ from $BSO(n)$.

My question is: is there a description of this map without referring to chains? And can I define such a map with any generalized cohomology theory? I don't think I can motivate this with anything else, but I am just purely curious.

Thanks!

Elden Elmanto
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  • Do you mean homology or cogomology? Your notation suggests cohomology your definition of $H^{k}$ looks like homology. – DBS Jul 26 '13 at 00:33
  • In cohomology there is always a pullback map $\pi^{\ast}: H^{k}(X) \rightarrow H^{k}(\tilde{X})$. There is a proper pushforward (meaning only defined for proper maps) $\pi_{\ast} : H^{k}(\tilde{X}) \rightarrow H^{k}(X)$. I think you are taking the map $\pi_{\ast}\pi^{\ast}$. The maps on homology are defined using Poincare duality. – DBS Jul 26 '13 at 00:37

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In this case (and in general when you have a proper, oriented map) you can define the wrong-way map by starting with an element in $H^*(\widetilde{X})$, taking its Poincare dual, pushing forward along homology, and dualizing again.

Here you have to be careful about what I mean when I say "homology" if I'm talking about non-compact guys and I want Poincare duality (two ways to fix this: one is to first define the transfer only for homology and then dualize or something; the other is to use something called 'Borel-Moore homology' which plays the same role that cohomology with compact supports plays when you're trying to do Poincare duality on noncompact spaces.)

In particular, if you pushforward the fundamental class $[\widetilde{X}]$ you should get $n [X]$ and so the map on cohomology you've written is multiplication by $n$.

Dylan Wilson
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