# Kernel of restriction and cokernel of corestriction of group cohomology

Let $$G$$ be an abelian group and $$M$$ a $$G$$-module. The basic definitions: Let $$H < G$$ be a subgroup of finite index. We have a map $$tr: H^0(H, M) \rightarrow H^0(G, M)$$ on group cohomology defined by $$m \mapsto \sum_{g \in G/H} gm$$. This can be extended to a map $$H^*(H, M) \rightarrow H^*(G, M)$$ of cohomological functors. It is explicitly written down as following and called $$\text{Cor} : H^1(H,M)\to H^1(G,M)$$.

Let $$f$$ be a cocyle for $$H$$. Take a set of representatives $$X$$ of $$G/H$$ in $$G$$. Then $$\operatorname{Cor}(f)(g) = \sum_{x \in X} y\cdot f(y^{-1}gx)$$ where $$y\in X$$ is the unique representative such that $$gxH=yH$$. Then $$\operatorname{cor}(f)$$ is a cocycle whose class is the well-defined corestriction of the class of $$f$$.

My question is,

is it true that $$\#\text{Ker}(res)=\#\text{Coker}({Cor})$$ hold?

Background of this question : I want to understand the following line of the answer by Christian Wuthrich, ''The more interesting question, and the one analogue to the kernel of restriction, is to ask what is the cokernel of corestriction.'' https://mathoverflow.net/questions/449060/ker-of-corestriction-of-galois-cohomology

No, this is not always true. Let $$K$$ be a characteristic $$0$$ local field with Galois group $$G_K$$. There is an isomorphism $$H^2(G_K, \overline{K}^\times) = \mathbb{Q}/\mathbb{Z}$$. Let $$L/K$$ be a finite nontrivial Galois field extension, so that $$G_L$$ has index $$[L : K]$$ in $$G_K$$. We have the formula $$\text{Cor} \circ \text{Res} = [G_K : G_L]$$ (multiplication by the index), so that $$\text{Cor} \circ \text{Res}: H^2(G_K, \overline{K}^\times) \to H^2(G_H, \overline{K}^\times) \to H^2(G_K, \overline{K}^\times)$$ may be identified with multiplication by $$[L : K]$$ on $$\mathbb{Q}/\mathbb{Z}$$. Since this group is divisible, this map is surjective, hence in particular $$\text{Cor}$$ has trivial cokernel.
Meanwhile, we claim that $$\text{Res}: H^2(G_K, \overline{K}^\times) \to H^2(G_H, \overline{K}^\times)$$ has nontrivial kernel. By Hilbert Theorem 90, $$H^1(G_L, \overline{K}^\times) = 0$$, so we get the Inflation-Restriction exact sequence $$0 \to H^2(\text{Gal}(L/K), L^\times) \to H^2(G_K, \overline{K}^\times) \to H^2(G_L, \overline{K}^\times).$$ Thus the kernel of restriction is the group $$H^2(\text{Gal}(L/K), L^\times)$$, and this is known to be the nontrivial cyclic group $$\mathbb{Z}/[L : K]\mathbb{Z}$$.
Let $$G$$ be $$\mathbb{Z}$$ and let $$H$$ be $$2\mathbb{Z}$$. Have $$G$$ act on $$M = \mathbb{Q}/\mathbb{Z}$$ trivially. (So every $$H^1$$ that shows up here will really just be a hom.)
If we restrict and then corestrict, $$H^1(G, M) \to H^1(H, M) \to H^1(G, M)$$, we multiply by $$2$$, which is surjective as an endomorphism of the divisible group $$H^1(\mathbb{Z}, M) \sim \mathbb{Q}/\mathbb{Z}$$, so in particular, the corestriction is surjective.
On the other hand, the restriction is not injective, since the kernel is the image of the inflation map $$H^1(\mathbb{Z}/2\mathbb{Z}, M) \to H^1(G, M)$$ and there is a non-trivial map from $$\mathbb{Z}/2\mathbb{Z}$$ to $$\mathbb{Q}/\mathbb{Z}$$ sending the generator to $$1/2$$.