I started reading Weibel’s *An introduction to homological algebra* but have trouble with the following exercise:

> 
> **Exercise 1.2.8** (Mapping cone)
>  Let $f \colon B \to C$ be a morphism of chain complexes.
>  Form a double chain complex $D$ out of $f$ by thinking of $f$ as a chain complex in $\mathbf{Ch}$ and using the sign trick, putting $B[-1]$ in the row $q = 1$ and $C$ in the row $q = 0$.
>  Thinking of $C$ and $B[-1]$ as double complexes in the obvious way, show that there exists a short exact sequence of double complexes
>  $$
>  0
>  \to C
>  \to D
>  \xrightarrow{\,\delta\,} B[-1]
>  \to 0 \,.
> $$
>  The total complex of $D$ is $\operatorname{cone}(f')$, the mapping cone (see section 1.5) of the map $f'$, which differs from $f$ only by some $\pm$ signs and is isomorphic to $f$.
> 
> 
> [I have added a short summary of Weibel’s indexing and sign conventions at the end of the question.]
> 
> 
> 

My problem is that I don’t understand how we can interpret $f$ as a double complex in such a way that the shifted complex $B[-1]$ appears.
So far I have tried the following:

* We can think of $f$ as a commutative square with $B$ sitting in the row $q = 1$ and $C$ sitting in the row $q = 0$.
We can then use the sign trick to replace $f\_p \colon B\_p \to C\_p$ by $(-1)^p f\_p$, resulting in a double complex which looks like this:
$$
 \require{AMScd}
 \begin{CD}
 B\_{p-1}
 @<d<<
 B\_{p}
 @.
 \hspace{2em}(q=1)
 \\
 @V (-1)^{p-1} f VV
 @VV (-1)^p f V
 @.
 \\
 C\_{p-1}
 @<d<<
 C\_{p}
 @.
 \hspace{2em}(q = 0)
 \end{CD}
$$
(All other rows $q \neq 0, 1$ vanish.)
But then the row $q = 1$ is not $B[-1]$ but $B$.
* We could also use the double complex which looks like this:
$$
 \require{AMScd}
 \begin{CD}
 B\_{p-1}
 @< -d <<
 B\_{p}
 @.
 (q=1)
 \\
 @V f VV
 @V f VV
 @.
 \\
 C\_{p-1}
 @<d<<
 C\_{p}
 @.
 (q = 0)
 \end{CD}
$$
But then the row $q = 1$ is still not $B[-1]$ because we only flipped the sign of the differential, but didn’t actually shift the degrees.
Also, we didn’t use the sign trick.
* We could forcefully put $B[-1]$ in the row $q = 1$ and $C$ in the row $q = 0$, which results in the following diagram:
$$
 \require{AMScd}
 \begin{CD}
 B\_{p-2}
 @< -d <<
 B\_{p-1}
 @.
 (q=1)
 \\
 @V ? VV
 @V ? VV
 @.
 \\
 C\_{p-1}
 @<d<<
 C\_{p}
 @.
 (q = 0)
 \end{CD}
$$
But then I don’t understand how the vertical arrow are supposed to look like, and we didn’t use the sign trick.
This option also doesn’t seem to give the desired mapping cone (up to sign) as its total complex.
* (What may be related to my problem is that I don’t know why Weibel labels the morphism $D \to B[-1]$ as $\delta$.
As far as I can tell, this notational choice has not been explained, so I may be missing some subtlety here.)

Any help is appreciated.

---

**Conventions**: Weibel uses for a double complex $D = D\_{\bullet,\bullet}$ the following indexing convention:
$$
 \require{AMScd}
 \begin{CD}
 D\_{p-1,q}
 @< d^h <<
 D\_{p,q}
 \\
 @V d^v VV
 @V d^v VV
 \\
 D\_{p-1,q-1}
 @< d^h <<
 D\_{p,q-1}
 \end{CD}
$$
He requires the above square to be anti-commutative, i.e. that $d^h d^v + d^v d^h = 0$.
He then defines the sign trick as associating to $d^v$ the chain morphisms $f\_{\bullet,q} \colon D\_{\bullet,q} \to D\_{\bullet,q-1}$ given by $f\_{p,q} = (-1)^p d^v\_{p,q}$ (i.e. we flip the sign columnwise).
The shift $B[p]$ of a chain complex $B = B\_{\bullet}$ is defined by $B[p]\_n = B\_{n+p}$ and $d^{B[p]} = (-1)^p d^B$.