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This is the first part of Exercise 1.4.2 in An Introduction to Homological Algebra by Weibel. The first part is showing that a chain complex, $C$, with boundaries $B_n$ and cycles $Z_n$ in $C_n$ is split if and only if there are $R$-module decompositions $C_n \cong Z_n \oplus C_n/Z_n$ and $Z_n \cong B_n \oplus H_n(C)$.

If $C$ is split we have maps $s_n: C_{n-1} \rightarrow C_{n}$ such that $d_{n} = d_{n}s_nd_{n}$ where $d_{n}:C_{n} \rightarrow C_{n-1}$ is our differential. Notice that then we have exact sequences $0 \rightarrow kerd_{n} \rightarrow C_{n} \rightarrow Imd_{n} \rightarrow 0$. Now $s_n|_{Imd_n}: Imd_n \rightarrow C_n$ so that if $a \in Imd_n$ we have a $b \in C_n$ so that $d_n(b)=a$. Then we have $d_ns_n|_{Imd_n}(a) = d_ns_nd_n(b) = d_n(b) = a$ so our short exact sequence splits giving us $C_n \cong kerd_n \oplus Imd_n \cong Z_n \oplus C_n/Z_n$.

Now we also have a short exact sequence $0 \rightarrow Imd_{n+1} \rightarrow kerd_n \rightarrow kerd_n/Imd_{n+1} \rightarrow 0$. I am stuck on this part of the first proof as i don't see why this is split. We would like to use the maps $s_n$ given by the assumption but these are submodules of $C_n$ so they don't seem to be of use. I also don't see why $Imd_{n+1}$ is injective or $kerd_n/Im(d_{n+1}$ is projective. I believe I am forgetting one of the ways we construct the splitting maps in this case any help would be appreciated!

noctusraid
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Michael N
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1 Answers1

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Let me tell you the answer firstly. In the notation of Weibel's book, $B^{\prime}_{n}=sd(C_{n})$, $B_{n}=ds(C_{n})$ and $H^{\prime}_{n}=(1-sd-ds)(C_{n})$. (This is the only thing I can guess after finished the "if" part)

I believe that you already showed $\mathrm{Im}\ d_{n}\cong sd(C_{n})$ in your question. Hence we have short exact sequence $$0\rightarrow Z_{n}\rightarrow C_{n} \rightarrow sd(C_{n})\rightarrow 0$$ The right split map is just the inclusion $sd(C_{n})\hookrightarrow C_{n}$.

It is also not hard to show that $ds(C_{n})=\mathrm{Im}\ d_{n+1}=B_{n}$. Then we have short exact sequence $$0\rightarrow ds(C_{n})\rightarrow Z_{n} \rightarrow Z_{n}/ds(C_{n})\rightarrow 0$$ The left split map is $ds$, so we have $Z_{n}\cong ds(C_{n})\oplus Z_{n}/ds(C_{n})$.

In summary, we have $\mathrm{Im}\ d_{n+1}=ds(C_{n})$ and $\mathrm{Im}\ d_{n}\cong sd(C_{n})$. The first splitting short exact sequence implies $\ker d_{n}=(1-sd)(C_{n})$. Then the second splitting short exact sequence implies $H^{\prime}_{n}=(1-sd-ds)(C_{n})$.

Yining Zhang
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  • I suppose that my next question may be dumb, but, I cannot understand why we need the second part of the proof. One think we know (I think) is that the sequence $0\rightarrow Im(d_{n+1})=B_n\rightarrow Z_n \rightarrow Z_n /Im(d_{n+1}) = H_n(C) \rightarrow 0$ is exact. Therefore, we know that $Z_n$ is isomorphic to $B_n \bigoplus H_n(C)$. Why do we need to work with $ds(C_n)$? Is there something that I am missing? – Rubén Ballester Jul 29 '21 at 09:15
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    Hi, if you are working over an arbitrary commutative ring instead of a field, a short exact sequence might not be split. For example, $0\rightarrow \mathbb{Z}_2 \rightarrow \mathbb{Z}_4 \rightarrow \mathbb{Z}_2 \rightarrow 0$ is exact, but $\mathbb{Z}_4$ is not isomorphic to $\mathbb{Z}_2\oplus \mathbb{Z}_2$. However if you only care about vector spaces over fields, then every complex is split. – Yining Zhang Jul 31 '21 at 18:23