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I can't understand an example Weibel gives in his Introduction to Homological Algebra, page 15.

So let $C.$ be any chain complex of vector spaces over a field. We use the following decomposition:

$$C_n=Z_n\oplus B_n'$$ where $B_n'\cong C_n/Z_n=d(C_n)=B_{n-1}$, and $$Z_n=B_n\oplus H_n'$$ where $H_n'\cong Z_n/B_n=H_n(C)$

Therefore we can form the composition $$C_n\longrightarrow Z_n\longrightarrow B_n\cong B_{n+1}'\subseteq C_{n+1}$$ to get splitting maps $s_n:C_n\longrightarrow C_{n+1}$ such that $d=dsd$.

What I can't understand is why $d=dsd$.

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1 Answer 1

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This is related to Exercise 1.1.3 on page 2 of Weibel :-)

OK, so we have a complex of vector spaces $$\cdots \to C_{n+1} \xrightarrow{d_{n+1}} C_n \xrightarrow{d_n} C_{n-1} \to \cdots$$ By definition, $Z_n = \ker d_n$, $B_n = \operatorname{im} d_{n+1}$, and we have short exact sequences $$0 \to B_n \to Z_n \to H_n \to 0$$ and $$0 \to Z_n \to C_n \xrightarrow{d_n} B_{n-1} \to 0$$

Since we are working over a field, there are some subspaces $H_n' \subseteq Z_n$, $H_n' \cong H_n$ and $B_{n-1}'\subseteq C_n$, $d_n\colon B_n' \xrightarrow{\cong} B_{n-1}$, such that $$Z_n = B_n \oplus H_n' \quad\text{and}\quad C_n \cong Z_n \oplus B_n'.$$ So we have a decomposition $$C_n \cong B_{n+1}' \oplus H_n' \oplus B_n',$$ and the differential $d_n\colon C_n\to C_{n-1}$ under this decomposition is given by

\begin{align*} d_n\colon B_{n+1}' \oplus H_n' \oplus B_n' & \to B_n' \oplus H_{n-1}' \oplus B_{n-1}',\\ (x,y,z) & \mapsto (z,0,0). \end{align*}

The splitting $s_n\colon C_n\to C_{n+1}$ is given under this decomposition by

\begin{align*} s_n\colon B_{n+1}' \oplus H_n' \oplus B_n' & \to B_{n+2}' \oplus H_{n+1}' \oplus B_{n+1}',\\ (x,y,z) & \mapsto (0,0,x). \end{align*}

Now $$d_n\circ s_{n-1}\circ d_n (x,y,z) = (z,0,0) = d_n (x,y,z).$$

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  • $\begingroup$ I can't get why does $d$ correspond to $(x,y,x)\to (0,0,x)$ under that iso. I mean, I can see why $x$ goes to zero, since $x\in\operatorname{Im}(d_{n+1})$ and $d_n\circ d_{n+1}=0$, but not the rest $\endgroup$
    – bateman
    Oct 17, 2016 at 18:06
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    $\begingroup$ $d$ corresponds to $(x,y,z) \mapsto (z,0,0)$. And $(x,y,z) \mapsto (0,0,x)$ is the definition of $s$. $\endgroup$
    – user144221
    Oct 17, 2016 at 18:21
  • $\begingroup$ Yes ok, because $B'_{n+1}$ is in both factorizations. But why $y$ and $z$ go to zero under $d$? $\endgroup$
    – bateman
    Oct 17, 2016 at 18:25
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    $\begingroup$ Maybe this notation is confusing. $B_n'$ denotes something that is isomorphic to $\operatorname{im} d_n = B_{n-1}$. And $H_n'$ is isomorphic to $H_n = \ker d_n / \operatorname{im} d_{n+1}$. $\endgroup$
    – user144221
    Oct 17, 2016 at 18:35
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    $\begingroup$ $x$ and $y$ go to zero, and $z \in B_n'$ goes to the corresponding component of $C_{n-1}$. $\endgroup$
    – user144221
    Oct 17, 2016 at 18:46

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