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Revision difference : tensor product of chain complexes
Version 5 Version 4
\PMlinkescapeword{tensor product} \PMlinkescapeword{tensor product}
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\newcommand{\papp}[1]{\partial_{#1} ''} \newcommand{\papp}[1]{\partial_{#1} ''}
Let $C'=\gbra{C_n',\pap n}$ and $C''=\gbra{C_n'',\papp n}$ be two chain complex, their \emph{tensor product} $C'\otimes C''=\gbra{(C'\otimes C'')_n,\pa n}$ is the chain complex defined by Let $C'=\gbra{C_n',\pap n}$ and $C''=\gbra{C_n'',\papp n}$ be two chain complex, their \emph{tensor product} $C'\otimes C''=\gbra{(C'\otimes C'')_n,\pa n}$ is the chain complex defined by
$$ (C'\otimes C'')_n = \bigoplus_{i+j=n}(C_i'\otimes C_j''), $$ $$ (C'\otimes C'')_n = \bigoplus_{i+j=n}(C_i'\otimes C_j''), $$
$$ \pa n(t'_i\otimes s''_j) = \pap i(t'_i)\otimes s''_j + (-1)^i t'_i\otimes \papp j(s''_j),\ \ \ \forall t'_i\in C_i',\ s''_j\in C_j'',\ (i+j=n),$$ $$ \pa n(t'_i\otimes s''_j) = \pap i(t'_i)\otimes s''_j + (-1)^i t'_i\otimes \papp j(s''_j),\ \ \ \forall t'_i\in C_i',\ s''_j\in C_j'',\ (i+j=n),$$
where $C_i'\otimes C_j''$ denotes the \PMlinkname{tensor product}{TensorProduct} of abelian groups $C_i'$ and $C_j''$. where $C_i'\otimes C_j''$ denotes the tensor product of abelian groups $C_i'$ and $C_j''$.
This is a good definition, in fact for each $t'_i\otimes s''_j\in C_i'\otimes C_j''\subseteq (C'\otimes C'')_{i+j}$ we have This is a good definition, in fact for each $t'_i\otimes s''_j\in C_i'\otimes C_j''\subseteq (C'\otimes C'')_{i+j}$ we have
$$\pa{i+j-1} \pa {i+j}(t'_i\otimes s''_j)=\pa{i+j-1}\cbra{ \pap i(t'_i)\otimes s''_j + (-1)^i t'_i\otimes \papp j(s''_j) }= (-1)^{i-1} \pap i(t'_i)\otimes \papp j(s''_j)+(-1)^i \pap i(t'_i)\otimes \papp j(s''_j)=0,$$ $$\pa{i+j-1} \pa {i+j}(t'_i\otimes s''_j)=\pa{i+j-1}\cbra{ \pap i(t'_i)\otimes s''_j + (-1)^i t'_i\otimes \papp j(s''_j) }= (-1)^{i-1} \pap i(t'_i)\otimes \papp j(s''_j)+(-1)^i \pap i(t'_i)\otimes \papp j(s''_j)=0,$$
thus $C'\otimes C''$ is a chain complex. thus $C'\otimes C''$ is a chain complex.