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Let $f: B_{\bullet} \rightarrow C_{\bullet}$ be a map between two chain complexes. In Weibel's H-book, he defined the mapping cone $\text{cone}(f)$ of $f$ to be the chain complex with $B_{n-1}\oplus C_{n}$ in degree $n$ and differential given by $$d(b,\,c)=(-d_{B}(b),\,d_{C}(c)-f(b)).$$ I understand that the minus sign before $d_{B}$ is because of switching the differential on $B$ with the shifting functor (both of them have degree $-1$). However, I am not quite sure why there is a minus sign before $f$. The reason above doesn't work here, because the degree of $f$ is $0$. Moreover, it seems like that if we delete the minus sign in the definition, we will still get the same properties related to the mapping cone.

Can someone tell me the reason of this extra minus sign in front of $f$. Thanks a lot.

Yining Zhang
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  • yes，you can delete the minus in front of $f$.It doesn't matter. – Jian Oct 15 '17 at 03:01
  • I believe the minus before the $f(b)$ is mainly for the following reasons : an element $(b,c)$ is a cycle iff $d_B(b)=0$ and $d_C(c)=f(b)$ (no sign here, this is very helpful for diagram chasing). In particular (and this is probably the main reason for the minus sign) when taking the long exact sequences associated to $0\rightarrow C_\bullet\rightarrow\operatorname{cone}(f)\bullet\rightarrow B[1]\bullet\rightarrow 0$, the boundary $H_{n+1}(B[1])\rightarrow H_n(C)$ is exactly the morphism $f$. And I disagree with @Sky, the choice matter. – Roland Oct 15 '17 at 08:38
  • @Roland some books don't have the minus in front of $f$ such as Neeman's book.the minus in front of $d_B$ is just make the mapping cone a complex. – Jian Oct 15 '17 at 09:08
  • @Roland, I guess the map $\text{cone}(f) \rightarrow B[1]$ is given by $(b,, c) \mapsto -b$, if we drop the minus sign in differential, we can also drop the minus in here. – Yining Zhang Oct 16 '17 at 13:23
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    Yes of course. Maybe, as @Sky said, you can choose the other convention, and everything will works fine. When I said that the choice matter, I meant that $\operatorname{cone}(f)$ is not quasi-isomorphic to $\operatorname{cone}(-f)$ in general. So we have to choose a convention, and stick with it. – Roland Oct 16 '17 at 15:38
  • @Roland Do you happen to have a reference for this? I am curious why this is the case. – ಠ_ಠ Nov 23 '23 at 08:57

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