Timeline for When is finding an explicit inverse of an isomorphism not possible
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| May 14, 2018 at 7:18 | comment | added | Rene Recktenwald | Thank you for the answer. Is it possible to write down explicitely where an element of $H^n(H,CoInd_H^G(A))$ gets mapped to? | |
| May 11, 2018 at 17:37 | comment | added | Todd Trimble | that there exists a map between any two projective resolutions that is unique up to homotopy (that's what I was referring to in "acyclic models"), and this induces an isomorphism in homology. However, the relevant map $\mathbb{Z}[G^n] \to \mathbb{Z}[H^n]$ between projective resolutions involves choosing an $H$-basis of the free modules $\mathbb{Z}[G^n]$, and that's what I was referring to at the end. | |
| May 11, 2018 at 17:33 | comment | added | Todd Trimble | $\psi(g_1, \ldots, g_n)(g) = \phi(gg_1, \ldots, gg_n)$. [The inverse of this takes $\psi$ to the map $\phi$ defined by $\phi(g_1, \ldots, g_n) = \psi(g_1, \ldots, g_n)(1)$. If you want to check carefully that this works, remember that "homogeneous" refers to the diagonal action, and that the left $G$-module structure on $\hom_{\mathbb{Z}H}(\mathbb{Z}G, A)$ is defined by $(g \cdot f)(g') = f(g'g)$.] As for the first map $C^n(H, A) \to C^n(G, A)$: that's induced by a map between projective resolutions of the $H$-module $\mathbb{Z}$. A basic result in any text on homological algebra is (cont.) | |
| May 11, 2018 at 17:28 | comment | added | Todd Trimble | @Roundthecorner At the $n$-cocycle level, this involves a composite of $H$-module maps $Z^n(H, A) \to Z^n(G, A) \to Z^n(G, \hom_{\mathbb{Z}H}(\mathbb{Z}G, A))$, which is a restriction of a composite of $n$-cochain maps $C^n(H, A) \to C^n(G, A) \to C^n(G, \hom_{\mathbb{Z}H}(\mathbb{Z}G, A))$. The second of these is the canonical isomorphism $\hom_{\mathbb{Z}H}(\mathbb{Z}[G^n], A) \cong \hom_{\mathbb{Z}G}(\mathbb{Z}[G^n], \hom_{\mathbb{Z}H}(\mathbb{Z}G, A))$ mentioned in my answer; it takes $\phi: G^n \to A$ to the map $\psi: G^n \to \hom_{\mathbb{Z}H}(\mathbb{Z}G, A))$ defined by (continued) | |
| May 11, 2018 at 11:14 | comment | added | Rene Recktenwald | If one chooses such a splitting, how would you make this choice into an inverse for the Shapiro Lemma? | |
| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Apr 10, 2015 at 9:53 | vote | accept | user114539 | ||
| Mar 28, 2015 at 22:18 | history | answered | Todd Trimble | CC BY-SA 3.0 |