Let $f:B\to C$ be a map of chain complexes and let $C(f)$ denote the mapping cone of $f.$ In Weibel's book, he defines the differential of $C(f)$ by the formula \begin{align*} d_{C(f)}:C(f)_n&=B_{n-1}\oplus C_n\to C(f)_{n-1}\\ d_{C(f)}(b,c)&=(-d_B(b),d_C(c)-f(b)). \end{align*} Notice the minus sign attached to the $d_B.$ This makes bad things happen, like the fact that the canonical surjection $C(f)\to B[-1]$ is not a chain map (it anticommutes with $d$). Furthermore, wikipedia gives different signs: $d_C(f)(b,c)=(-d_B(b),d_C(c)+f(b)).$ Not only is this fundamentally different from the differential Weibel gives, it doesn't even alleviate the issue that I have!

So here's my question: what are the correct signs for the mapping cone? Furthermore, does it even matter? It's my understanding that one will run in to signs a lot while studying homological algebra, so this may just be a pedantic question altogether. However I'm worried that these signs will neverendingly trip me up if I don't figure them out early on.


1 Answer 1


It is usual for the differential in $B[s]$ to have a minus sign $(-1)^s$. I.e., if $d$ is one of the differentials of $B$, then $(-1)^sd$ will be one of the differentials of $B[s]$. With this convention, which both Weibel and Wikipedia use, the natural map $C(f)\to B[-1]$ is a chain map.

By the way, a warning. You and Weibel use the convention that $B[1]$ means "shift one place to the right" (i.e., in the direction of the arrows). I think pretty much everybody else uses the convention that $B[1]$ means "shift one place to the left".

The difference between Weibel's $d_C(c)-f(b)$ and Wikipedia's $d_C(c)+f(b)$. Doesn't really matter. The two versions of the mapping cone are naturally isomorphic via the isomorphism $B_{n-1}\oplus C_n\to B_{n-1}\oplus C_n$ given by $(b,c)\mapsto(b,-c)$

  • 1
    $\begingroup$ I see so there was a different sign that I was forgetting. That clears it up. Thanks! (Also thank you for the note on $B[1]$. That is probably something that would have taken me a while to figure out.) $\endgroup$
    – D. Brogan
    Apr 22, 2020 at 15:44

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .