If Y is closed subset of X, do anyone know what is the mapping from $H_{c}^k(X,\mathbb{Z})$ to $H_{c}^k(Y,\mathbb{Z})$ ? Is it differ if X is an open subset of Y.
$H_{c}^k$ is a cohomology with compact support.
Asked
Active
Viewed 59 times
0
-
1related: Long exact sequence for cohomology with compact supports – Grigory M May 07 '14 at 14:14
-
But I don't understand the difference with open and closed case.. – Svechnikov_Timofei May 07 '14 at 14:24
-
You mean that $Y$ is an open subset of $X$, right? I don't see a reason a mapping should exist otherwise. – May 07 '14 at 15:05
-
Yes, it is. Open and close subset – Svechnikov_Timofei May 07 '14 at 17:14
-
What cohomology theory are you considering? Sheaf cohomology? – Moishe Kohan May 08 '14 at 00:59
-
Singular homology – Svechnikov_Timofei May 08 '14 at 04:44
1 Answers
1
The inclusion $Y\to X$ induces a morphism of chain complexes $h: C_*(Y)\to C_*(X)$ and, hence, a morphism of cochain complexes $h^*: C^*(X)\to C^*(Y)$. Now, the inclusion $Y\to X$ is a proper map (here we use that $Y$ is closed: a closed subset of a compact is compact!). Thus, $h^*$ sends compactly supported cochains to compactly supported cochains; hence, by restriction, we obtain a morphism $$ h': C_c^*(X)\to C^*_c(Y). $$ As any morphism of chain complexes, $h'$ will induce homomorphisms of homology groups $$ H^*_c(X):=H_*(C_c^*(X))\to H_*(C^*_c(Y))=:H^*_c(Y). $$ That's all.
Moishe Kohan
- 97,719