Interpreting a chain map as a double complex (Weibel Exercise 1.2.8)

Question

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.

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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 $d^{B[p]} = (-1)^p d^B$.

Answer 1

Weibel is suggesting a cohomological double complex (i.e. the arrows point up and right). The desired double complex $D$ has $D_{p,0} = C_p$ and $D_{p,1} = B_{q+1}[-1]$. The differential of the row $q=1$ is $d_B$, the vertical arrow is induced by $f$, and the differential of the row $q=0$ is $-d_C$.

Then the total complex of this has $B_{p-1}\oplus C_p$ in degree $p$, with differential $d(c,b) = (f(b)-dc,db)$. You can also change the signs of $d_B$ and $d_C$, or as you mention change interchange the sings of $f$ and leave $d_C$ and $d_B$ with the usual signs.

The exact sequence is obtained by including $C$ as a row and projecting onto $B$.

Answer 2

I have had trouble following this too. The $n$-lab has its own explanation. https://ncatlab.org/nlab/show/mapping+cone

I don't know what he means by the shifting by $[-1]$, and I don't see how it's relevant everywhere - only in a few particular parts. It could be a mistake.

Here is my own interpretation. If we regard $f: B\to C$ as a chain complex of objects in $Ch(\mathcal{A})$ concentrated in degree $-1$ and $0$ then applying the sign trick allows us to construct a (homological) double complex $D$ with $D_{p,0} = C_p$, $D_{p,1} = B_p$, $d^h_{p,1} = d_B$, $d^h_{p,0}=d_C$, and $d^v_{p,1} = (-1)^p f_p$. Let $C', B'$ be the chain complexes in $Ch(Ch(\mathcal{A}))$ associated to $C,B$, concentrated at 0, so $C'_0 = C$ and $B'_0 = B$. Then $B'[-1]_1 = B$. Apply again the sign trick to associate double complexes D(C'), D(B'[-1]) to the chain complexes $C'$ and $B'[-1]$. This fits into a (split) short exact sequence $0\to D(C')\to D\to D(B'[-1])\to 0$. The objects of the resulting total complex are, as desired, $Tot(C)_n = B_{n-1}\oplus C_n$, and the corresponding differential maps are, in matrix form, \begin{equation} d_n^{Tot(C)}: B_{n-1}\oplus C_n\to B_{n-2}\oplus C_{n-1}= \begin{bmatrix} d^B_{n-1} & 0 \\ (-1)^{n-1}f_{n-1} & d^C_n \end{bmatrix} \end{equation}

Comparing the formula for a mapping cone, we see that one can construct a chain complex $-B$ whose objects are the same as $B$ and whose morphisms are $-d_B$, together with a morphism of chain complexes $f':-B\to C$ where $f'_n = (-1)^{n+1}f_n$. $f'$ is a morphism of chain complexes as $(-1)^n f_{n-1}\circ (-d_n)= (-1)^{n+1}f_{n-1} d_n = (-1)^{n+1}d_n f_n$ by the assumption that $f$ itself is a morphism of chain complexes. And $-B$ is isomorphic to $B$ by $g: B\to -B$ with $g_n = (-1)^{n+1}id_B$ with $f'g =f$.