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If $C$ is a complex, show that there are exact sequences of complexes: $$ 0 \longrightarrow Z(C) \longrightarrow C \stackrel{d}{\longrightarrow} B(C)[-1] \longrightarrow 0; $$ $$ 0 \longrightarrow H(C) \longrightarrow C / B(C) \stackrel{d}{\longrightarrow} Z(C)[-1] \longrightarrow H(C)[-1] \longrightarrow 0. $$

$$ \cdots\xrightarrow{} C_{n+1}\xrightarrow{d_{n+1}}C_{n}\xrightarrow{d_{n}}C_{n-1}\xrightarrow{d_{n-1}}\cdots $$

I know there is an injective map $i_n: Z_n(C)\to C_n$, since $Z_n(C)=\ker(d_n)\subset C_n$. Could you please tell me how to get other maps? Thanks in advance.

Ryze
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1 Answers1

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$d$ is a morphism from $C_n$ to $C_{n-1}$. Its kernel is $Z_n(C)$ and its image is $B_{n-1}(C)$. Therefore $d$ induces an exact sequence $$0\to Z_n(C)\to C_n\to B_{n-1}(C)\to0.$$ Assembling these for all $n$ gives your first exact sequence.

Again consider $d:C_n\to C_{n-1}$. Its kernel is $Z_n(C)$ but that contains $B_n(C)$ so $d$ induces a morphism $C_n/B_n(Z)\to C_{n-1}$ which by abuse of notation we'll also call $d$. The image of our new $d$ is the same as that of our new $d$ and is $B_{n-1}(C)$, but that is contained in $Z_{n-1}(C)$. Then $d$ induces a morphism $C_n/B_n(C)\to B_{n-1}(C)$ which composed with the inclusion $B_{n-1}(C)\to Z_{n-1}(C)$ becomes a morphism $C_n/B_n(C)\to Z_{n-1}(C)$. Again, abuse notation and call that $d$. The kernel of $d$ is $H_n(C)$ and its cokernel is $H_{n-1}(C)$. That gives your second exact sequence.

Angina Seng
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