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Let us call a complex of abelian groups $C_* = \{C_n\}$ split if it admits a direct sum decomposition $$C_* = H_* \oplus B_* \oplus D_*$$ where the differential $d = \{d_n : C_n \to C_{n-1}\}$ in $C_*$ vanishes on $H_*$ and $B_*$ and maps $D_*$ isomorphically onto $B_*$ (so, in particular, $H_*$ is indeed the homology of $C_*$).

Show that a complex $C_*$ is split if and only if there exists a collection of maps $s_* = \{s_n\}$ with $s_n : C_n \to C_{n+1}$ such that $d_ns_{n-1}d_n = d_n$ for all $n$.

Could you help me,please.

David Ward
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Koam
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  • Reformatted, but I might have broken something. Let me know. – Thomas Andrews Apr 17 '13 at 13:30
  • What do you mean? – Koam Apr 17 '13 at 13:39
  • @Koam I mean I edited your question, trying to improve the formatting, but I'm not sure I got it right. – Thomas Andrews Apr 17 '13 at 13:43
  • Thomas Rot no i can not see. – Koam Apr 17 '13 at 13:53
  • First to the easy direction, $\Longrightarrow$. There are not so many candidates. Also try short complexes $0 \to C_2 \to C_1 \to C_0 \to 0$ first (splitting lemma). – Martin Brandenburg Apr 17 '13 at 13:57
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    You really shouldn't just post virtually all of a coursework assignment on a forum. People don't have any problem trying to help you with the understanding of a subject, but just posting the questions here is only likely to infuriate people. – David Ward Apr 19 '13 at 13:15
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    This is a question from a take-home exam which is in-progress. I kindly ask that all well-meaning respondents wait until Saturday April 27th to answer this question. – David Ward Apr 19 '13 at 13:41

1 Answers1

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Can you see how to define $s_n$ if you already know it splits? (Hint: use that $d_{n+1}$ maps $D_{n+1}$ isomorphically to $B_n$.)

Now, suppose such $s_n$ exists. We want to find $D_*, H_*, B_*$.

Define $f_n,g_n:C_n\to C_n$ as:

$$f_n = s_{n-1}d_n, g_n=d_{n+1}s_n$$

We easily see that $f_n^2=f_n$ and $g_n^2=g_n$. We also see that $f_ng_n=0$. We also have:

$$d_nf_n = d_n, d_ng_n=0$$

Define $h_n=(1-g_n)f_n$. Then we can show that $$h_n^2=h_n\\ h_ng_n = g_nh_n=0\\ d_nh_n = d_n$$

Define $k_n=1-h_n-g_n$. Show that:

$$k_n^2=k_n\\ k_nh_n=h_nk_n=k_ng_n=g_nk_n=0\\d_nk_n=0$$

All of these are pretty direct proofs.

Finally, show this means that $$C_n\cong g_n(C_n) \oplus h_n(C_n) \oplus k_n(C_n)$$

Map these to the appropriate sets: $B_n=g_n(C_n)$, $D_n=h_n(C_n)$ and $H_n=k_n(C_n)$.

It's pretty easy to show that $d_nk_n=d_ng_n=0$, so $d_n$ is zero on $H_n$ and $B_n$.

Finally, you have to show that $d_{n+1}$ acts as an isomorphism from $D_{n+1}$ to $B_n$.

Essentially, the image of $d_{n+1}$ is contained in $B_n$ because $d_{n+1} = d_{n+1}s_{n}d_{n+1}=g_nd_{n+1}$. $d_{n+1}$ is onto from $D_{n+1}$ to $B_n$ because $$d_{n+1}h_{n+1}s_n = g_n$$

Proving that it is $1-1$ is trickier. I'm not sure about that step.

Thomas Andrews
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