# Is there a definition of the transfer homomorphism (between cohomology of cover and base) without referring to chains?

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!

• Do you mean homology or cogomology? Your notation suggests cohomology your definition of $H^{k}$ looks like homology. – DBS Jul 26 '13 at 0:33
• 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 0:37

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.
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 $n$.