To let $ G $ be a group and $ H $ his subgroup ($ H $ doesn't have to be normal). Consider a chain complex $ (C _ {*} (G), partial) $ Where $ C_n (G) $ is the free Abelian group that is generated by the set $ G ^ {n + 1} $ and $ partial = sum_i (-1) ^ {i} partial_i $, Here $ partial ((g_0, g_1, ldots, g_n)) = (g_0, g_1, ldots, g_ {i-1}, g_ {i + 1}, ldots, g_n) Now we define relative homology groups $ H _ {*} (G, H; mathbb {Z}) $ to be the homology groups of the chain complex $ (D _ {*}, tilde partial) $ Where

$ D_n: = C_ {n-1} (H) oplus C_ {n} (G) $

and

$ tilde partial = begin {bmatrix} partial & 0 \ i_ * & – partial end {bmatrix}. $

This is proven in the theorem in The Volume and Chern-Simons Invariant of a Representation$ 2.1 $ that the relative homology can be described as free $ G $-Resolutions over $ K $ This is the kernel of the augmentation map $ epsilon: C_0 (G / H) to mathbb {Z}. $

To prove this, they first constructed a chain complex $ (F _ {*}, delta) $ from free $ G $Modules and proved that $ H _ {*} (F _ {*} otimes _ { mathbb {Z} (G)} otimes mathbb {Z}) = H _ {* + 1} (G, H). $

The construction of $ {F_i } _ {i = 0} ^ { infty} $ is as follows:

To take $ F_i = D_ {i + 1} otimes _ { mathbb {Z}} mathbb {Z} (G) $ to the $ i geq 1 $ and $ F_0 = ker (D_1 to D_0) otimes _ { mathbb {Z}} mathbb {Z} (G). $

To prove that $ (F _ {*}, delta) $ is the free resolution of $ K $it is enough to prove the card's kokernel $ F_1 to F_0 $ is isomorphic to $ K $, You write that picture of $ F_1 to F_0 $ is the core of the card $ F_0 cong C_1 (G) xrightarrow { partial} C_0 (G) xrightarrow { pi} C_0 (G / H) $ and **I'm stuck here**,

Below are the calculations I did to find the image of $ F_1 to F_0 $ (all throughout the discussion $ G $Modules are correct $ G $Modules):

begin {align}

& Delta big ((m (h, 1) otimes 1, m & # 39; (g_1, g_2,1) otimes 1) otimes g big) \

& = delta big ((m (h, 1) otimes 1, m & # 39; (g_1, g_2,1) otimes 1) big) otimes g \

& = big ((m – mh) x 1, m (h, 1) x 1 – m & # 39; (g_2,1) x 1 + m & # 39; (g_1,1) x 1 – m & # 39; (g_1, g_2) otimes 1) big) otimes g \

& = (0, m (h, 1) × 1 – m # (g_2,1) × 1 + m # (g_1,1) × 1 – m ((g_1g_2 ^ {- 1 }, 1) otimes 1) otimes g.

end {align}

Under the isomorphism $ F_0 cong C_1 (G) $, we have $ delta big ((m (h, 1) x 1, m & # 39; (g_1, g_2,1) x 1) x g big) = m (h, g) -m & # 39; (g_2, g) + m ((g_1, g) -m # (g_1g_2 ^ {- 1}, g). $

So the picture of the card $ F_1 to F_0 to C_1 (G) $ is generated by the crowd

$ S: = {m (h, g) – m & # 39; (g_2, g) + m & # 39; (g_1, g) – m & # 39; (g_1g_2 ^ {- 1}, g) ~ | ~ h in H, g_1, g_2, g in G, m, m & # 39; in mathbb {Z} } $,

Now I see no reason why the set $ S $ Cards too $ 0 $ under the map $ pi circ partial: C_1 (G) to C_0 (G / H) $,

To understand this, I have already asked questions 1 and 2 at math StackExchange. And in the comments of 1, a valid point is raised that relates to ingestion $ m = 1 $ and $ m & # 39; = $ 0 can we conclude that $ H = G $ that's funny.

I can't see where I'm making mistakes. Please help me find out.