From the material I am learning, for a chain complex $ \cdots \mathop \to \limits^\partial {C_{n + 1}}\mathop \to \limits^\partial {C_n}\mathop \to \limits^\partial {C_{n - 1}}\mathop \to \limits^\partial \cdots $, the cochain group $C_n^ * $ if defined as $C_n^ * = Hom({C_n},G)$ for a given group $G$. My question is that in what sense the cochain group $C_n^ * = Hom({C_n},G)$ is a group, should it represent a homomorphism? or $C_n^ * $ is the image of some homomorphism to $G$ and thus is a subgroup of $G$?
A basic question of cohomology group
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algebraic-topology
differential-topology
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2If $G$ is a group, then $Hom(C_n,G)$ is a group: $(\varphi \cdot \psi)(x):=\varphi(x)\cdot \psi(x)$. – 2017-01-12
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0Thanks very much, that helps a lot ! – 2017-01-12
1 Answers
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The set $Hom(C_n,G) = \{\varphi \colon C_n \to G \mid \varphi \mbox{ is a homomorphism}\}$ is a group under point-wise addition so $\varphi_1+\varphi_2$ is defined by $(\varphi_1+\varphi_2)(x) = \varphi_1(x)+\varphi_2(x)$.
This is clearly an associative and commutative binary operation (because $G$ is an abelian group).
The identity is given by the zero-homomorphism $0$ given by $0(x) = 0_G$.
The inverse $\varphi'$ of $\varphi$ is given by $\varphi'(x) = -\varphi(x)$.