Showing that the mapping cone is a chain complex

Question

Let $\alpha \colon \mathcal{A} \to \mathcal{D}$ be a morphism of chain complexes. Let $$ d^{C(\alpha)}_n = \begin{bmatrix} -d^{\mathcal{A}}_{n-1} & 0 \\ \phantom{-}\alpha_{n-1} & d^{\mathcal{D}}_{n-1} \end{bmatrix} $$ I want to show that $d^{C(\alpha)}_n \circ d^{C(\alpha)}_{n+1} =0$

\begin{align*} d^{C(\alpha)}_n \circ d^{C(\alpha)}_{n+1} &= \begin{bmatrix} -d^{\mathcal{A}}_{n-1} & 0 \\ \alpha_{n-1} & d^{\mathcal{D}}_{n-1} \end{bmatrix} \begin{bmatrix} -d^{\mathcal{A}}_{n} & 0 \\ \alpha_{n} & d^{\mathcal{D}}_{n} \end{bmatrix} \\ &= \begin{bmatrix} -d^{\mathcal{A}}_{n-1} \circ d^{\mathcal{A}}_n & 0 \\ \alpha_{n-1} \circ -d^{\mathcal{A}}_n + d^{\mathcal{D}}_{n-1} \circ \alpha_n & d^{\mathcal{D}}_{n-1} \circ d^{\mathcal{D}}_n \end{bmatrix} \end{align*}

Obviously $-d^{\mathcal{A}}_{n-1} \circ d^{\mathcal{A}}_n = 0$ and $d^{\mathcal{D}}_{n-1} \circ d^{\mathcal{D}}_n = 0$. So we just need to show that $\alpha_{n-1} \circ -d^{\mathcal{A}}_n + d^{\mathcal{D}}_{n-1} \circ \alpha_n = 0$.

We have the following diagram: $$ \require{AMScd} \begin{CD} \cdots @>>> A_{n-1} @> d_{n-1}^{\mathcal{A}} > > A_n @> d_{n}^{\mathcal{A}} >> A_{n+1} @> d_{n+1}^{\mathcal{A}} >> \cdots \\ @. @V \alpha_{n-1} VV @V \alpha_n VV @V \alpha_{n+1} VV @. \\ \cdots @>>> D_{n-1} @> d_{n-1}^{\mathcal{D}} > > D_n @> d_{n}^{\mathcal{D}} >> D_{n+1} @> d_{n+1}^{\mathcal{D}} >> \cdots \end{CD} $$

Let $x \in A_{n}$. Then \begin{align*} &\, ( \alpha_{n-1} \circ -d^{\mathcal{A}}_n + d^{\mathcal{D}}_{n-1} \circ \alpha_n )(a) \\ =&\, (\alpha_{n-1} \circ -d^{\mathcal{A}}_n)(a) + (d^{\mathcal{D}}_{n-1} \circ \alpha_n)(a) \\ =&\, \alpha_{n-1}(-d^{\mathcal{A}}_n(a)) + d^{\mathcal{D}}_{n-1}(\alpha_n(a)). \end{align*}

But now I run into a problem. If we look at the diagram, $d^{\mathcal{A}}_n(a) \in A_{n+1}$, but $\alpha_{n-1} \colon A_{n-1} \to D_{n-1}$. So $\alpha_{n-1}(-d^{\mathcal{A}}_n(a))$ does not even make sense, right? We have the same problem for the second term: $\alpha_n(a) \in D_n$ but $d^{\mathcal{D}}_{n-1} \colon D_{n-1} \to D_n$, so $d^{\mathcal{D}}_{n-1}(\alpha_n(a))$ does not make sense either, right?

So I'm not really sure if I'm missing something.

score 2 · Accepted Answer · answered Oct 22 '13 at 01:37

You definition of $d^{C(\alpha)}_n$ is incorrect; only the degrees of $d^A$ and $\alpha$ should be shifted. The correct differential is $$d^{C(\alpha)}_n = \begin{pmatrix} -d^A_{n-1} & 0 \\ \alpha_{n-1} & d^B_n \end{pmatrix}.$$ Note that this only differs from yours in the $(2,2)$-entry. With this correct differential, we have that $$d^{C(\alpha)}_n \circ d^{C(\alpha)}_{n+1} = \begin{pmatrix} d_{n-1}^A \circ d_n^A & 0 \\ -\alpha_{n-1} \circ d_n^A + d_n^B \circ \alpha_n & d_n^B \circ d_{n+1}^B \end{pmatrix}.$$ This is the zero map for the same reason you attempted above: since $\alpha$ is a chain map, we have a commutative diagram $$\require{AMScd} \begin{CD} \cdots @>>> A_n @>{d^A_n}>> A_{n-1} @>>> \cdots\\ \ @V{\alpha_n}VV @V{\alpha_{n-1}}VV \ \\ \cdots @>>> B_n @>{d^B_n}>> B_{n-1} @>>> \cdots \end{CD}$$ Hence the $(2,1)$-entry is the zero map. The other entries are obviously zero maps.

Showing that the mapping cone is a chain complex

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