A chain complex is contractible iff it is split exact

Question

Let $C.$ be a chain complex in an abelian category. Suppose that the identity $Id_{C.}$ is null homotopic. Then there are morphisms $s_n:C_n\to C_{n+1}$ such that for every $n$ the following identity holds: $$Id_{C_n}=d_{n+1}s_n+s_{n-1}d_n$$ Then we have: $$d_n=d_n Id_n=d_n(d_{n+1}s_n+s_{n-1}d_n)=d_ns_{n-1}d_n$$ So $C.$ is split. Moreover, homotopic maps have the same homology, thus the induced map $H_n(Id_{C.})$ is the zero map and $H_n(C.)$ is zero. This proves that $C.$ is acyclic, but how can I prove that it is also exact?

The converse implication is more obscure for me. Let for each $n$, $d_n=i_n\circ p_n$ be the mono-epi factorization of $d_n$. Now we are assuming $C.$ split exact and we eant to prove that the identity on $C.$ is null homotopic. Let's call $s_n:C_n\to C_{n+1}$ the splitting maps such that for each $n$, $d_n=d_ns_{n-1}d_n$. Then we have $$i_np_ns_{n-1}i_np_n=i_np_n$$ $i_n$ is mono, so we get $p_ns_{n-1}i_np_n=p_n$ so the following short sequence is split exact $$o\to Im(d_{n+1})\to C_n\to Im(d_n)\to 0$$ with splitting maps given for each $n$ by $s'_{n-1}=s_{n-1}i_n$

Now, if the abelian category were that of $R$-modules or $K$-vector spaces, I could go on by saying that $C_n$ is isomorphic to the direct sum of $Im(d_{n+1})$ and $Im(d_n)$, that each differential $d_n$ is the composite of projection onto the first coordinate, followed by inclusion, and prove the homotopy of the identity and the null morphism directly by computation. But how the same can be carried out in the more general setting of an arbitrary abelian category? Thank you

Answer 1

A couple of housekeeping things:

First, the title of your question is false; there exist chain complexes which are exact but not split. For instance, one can check that if $R=\mathbb{Z}$, then the complex of $R$-modules

$$ \cdots\xrightarrow{\cdot2}\mathbb{Z}/4\xrightarrow{\cdot2}\mathbb{Z}/4\xrightarrow{\cdot2}\mathbb{Z}/4\xrightarrow{\cdot2}\cdots $$

is exact, but not split. I believe based on the body of your question, you are asking the following:

A chain complex $(C_{\bullet},d)$ is split and exact if and only if $\mathrm{id}_{C_{\bullet}}$ is null homotopic; i.e., there exist maps $s_{n}:C_{n}\to C_{n+1}$ such that $\mathrm{id}_{C_{\bullet}}=ds+sd$.

Feel free to correct me if this is not the case.

Second, you ask "This proves that $C_{\bullet}$ is acyclic, but how can I prove that it is also exact?" but that is what acyclic means: a chain complex $C_{\bullet}$ is acyclic if it is exact as a sequence.

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Now, let us prove the claim. Your proof that "the identity is null homotopic $\Rightarrow$ $C_{\bullet}$ is split and exact" is good! Again, note that you are done because you've shown $C_{\bullet}$ is acyclic.

For the other direction, there's no need to necessarily do a monic-epic factorization, though I believe it's roughly in the spirit of what follows. Instead, assume that $C_{\bullet}$ is split and exact. We need to show that there exists $s:C_{n}\to C_{n+1}$ such that $ds+sd=\mathrm{id}$. As $C$ is split and exact, we claim that this implies $C_{n}\cong B_{n}\oplus B_{n-1}$ where $B_{n}=\mathrm{im}(d_{n+1})\subseteq C_{n}$ are the boundaries. (In fact, this is a complete characterization; $C_{n}\cong B_{n}\oplus B_{n-1}$ if and only if $C_{\bullet}$ is split and exact.)

To see this claim, assume that $C_{\bullet}$ is split. Write $Z_{n}=\ker(d_{n})\subseteq C_{n}$ for the cycles and $B_{n}\subseteq C_{n}$ for the boundaries. We have the short exact sequences $$ 0\to Z_{n}\to C_{n}\xrightarrow{d}B_{n-1}\to0 $$ and $$ 0\to B_{n}\xrightarrow{\partial}Z_{n}\to Z_{n}/B_{n}\to0. $$ Since $C_{\bullet}$ is split, there exists $s_{n}:C_{n}\to C_{n+1}$ such that $d=dsd$. Focusing on the first short exact sequence, since $B_{n-1}\subseteq C_{n-1}$, we have a map $\left.s_{n-1}\right\rvert_{B_{n-1}}:B_{n-1}\to C_{n}$. As $d_{n}s_{n-1}d_{n}=d_{n}$ by split-i-tude, we see that $d_{n}\left.s_{n-1}\right\rvert_{B_{n-1}}d_{n}=d_{n}$. As $d_{n}$ is surjective onto $B_{n-1}$, we get $d_{n}\left.s_{n-1}\right\rvert_{B_{n-1}}=\mathrm{id}_{B_{n-1}}$. Thus, invoking the splitting lemma, $C_{n}\cong Z_{n}\oplus B_{n-1}$.

Now focusing on the second short exact sequence, we're again going to use the splitting lemma once we've constructed a map $Z_{n}\to B_{n}$ that composes with $\partial$ to be $\mathrm{id}_{B_{n}}$. We will define the map $Z_{n}\to B_{n}$ to be $$ Z_{n}\subseteq C_{n}\xrightarrow{s_{n}}C_{n+1}\xrightarrow{d_{n+1}}B_{n}. $$ Thus, again by the splitting lemma, $Z_{n}\cong B_{n}\oplus Z_{n}/B_{n}$.

Finally, since $C_{\bullet}$ is split and exact, $$ B_{n}=\mathrm{im}(d_{n})=\ker(d_{n+1})=Z_{n}\cong B_{n}\oplus Z_{n}/B_{n}, $$ so $Z_{n}/B_{n}=0$ (alternatively, $Z_{n}/B_{n}=H_{n}(C_{\bullet})$ and exact means $H_{n}(C_{\bullet})=0$), and therefore $$ C_{n}\cong B_{n}\oplus B_{n-1}, $$ as claimed.

Now, how will we use this claim? Observe that if $C_{n}\cong B_{n}\oplus B_{n-1}=\mathrm{im}(d_{n+1})\oplus\mathrm{im}(d_{n})$, then $d:C_{n}\to C_{n-1}$ is projection onto the $\mathrm{im}(d_{n})$ factor followed by inclusion into the second coordinate; that is, $$ d(x,y)=(0,x). $$ We may define $s(x,y)=(y,0)$ and observe that $$ (ds+sd)(x,y)=ds(x,y)+sd(x,y)=d(y,0)+s(0,x)=(0,y)+(x,0)=(x,y)=\mathrm{id}(x,y), $$ so the identity is null homotopic, as we wished to show.

Answer 2

I'll add another proof that the identity map is null-homotopic for a split exact sequence, while avoiding the splitting lemma. Let $e=1-(ds+sd)$. Then by splitness, $de=ed=0.$ With exactness, this shows $$\mathrm{im}\ e\subseteq \ker d=\mathrm{im}\ d\subseteq \ker e,$$ so $e^2=0.$ Then $$ e^2 =(1-sd-ds)^2 =1-2sd-2ds+sdsd+sdds+dssd+dsds =1-sd-ds+dssd. $$ Therefore $$ ds+sd=1+dssd. $$ If we define $s'=s-dss$ then $$ ds'+s'd =d(s-dss)+(s-dss)d =ds+ds-dssd =1. $$