Timeline for Cup product with arbitrary coefficients
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 13, 2012 at 0:24 | answer | added | Ralph | timeline score: 3 | |
| Apr 12, 2012 at 21:01 | comment | added | Andreas Blass | It seems to me that you get a homomorphism $H^k(X;A)\to\text{Hom}(H_k(X;B),C)$ when (and probably only when) you have a pairing $A\otimes B\to C$. The $h$ in the question comes from the fact that an abelian group $G$ is a $\mathbb Z$-module, i.e., a pairing $G\otimes\mathbb Z\to G$. The $\tilde h$ comes from the ring multiplication $R\otimes R\to R$. | |
| Apr 12, 2012 at 19:16 | comment | added | David White | Here's some pure speculation... given $G$ there should always be a group homomorphism from $\mathbb{Z}\rightarrow G$, right? This induces a group homomorphism $Hom(H_k(X),\mathbb{Z})\righarrow Hom(H_k(X),G)$. Maybe this can help you use the $\tilde{h}$ from $R=\mathbb{Z}$ to construct the map you're after. Of course, it can only be a group homomorphism, since $Hom(H_k(X),G)$ is only a group, so there is no guarantee it'll have the nice properties that $\tilde{h}$ has. Now, here's a question: what do you need $\tilde{h}_G$ for? Why isn't $h$ good enough? | |
| Apr 12, 2012 at 17:44 | comment | added | Lee Mosher | @Ralph: ah, right. Got it. | |
| Apr 12, 2012 at 17:30 | comment | added | Ralph | @Lee: Note that homology on the right hand side of $h$ has coefficients in $\mathbb Z$ while $\tilde{h}$ has coefficients in $R$. Hence it's a bit more subtil than just arguing that a ring is a group with some additional properties. | |
| Apr 12, 2012 at 16:54 | comment | added | Lee Mosher | A ring $R$, in Hatcher's context, already has the structure of an abelian group with respect to addition, so everything in 3.1 applies. It is true that a ring has additional structure besides addition --- namely multiplication --- but your questions (a,b,c) are answered by ignoring multiplication. The whole point of cup product is to exploit the additional multiplicative structure on $R$. | |
| Apr 12, 2012 at 15:28 | history | asked | Mr-Cups | CC BY-SA 3.0 |