Split-Lemma for chain complexes

Question

Suppose $k$ is a field and $A$, $B$ and $C$ are chain complexes of $k$-vector spaces, i.e., objects in $\mathbf{Ch}(k\text{-}\mathbf{Vect})$. Is there are chain complex version of the split lemma, i.e.:

If there is an inclusion chain map $i:A \to B$ and a projection chain map $p:B \to C$ and a short exact secence

$$ 0 \to A \overset{i}\to B \overset{p}\to C \to 0 $$

of chain complexes, where '$0$' means the chain complex of the zero vector space with the trivial differential. Then we can split the sequence,i.e. there are sections of chain complexes $s:C \to B$ and retracts $r:B \to A$ such that $p\circ s = id_C$, $r\circ i=id_A$ and the chain complex $B$ is naturally isomoph to $s(C)\oplus i(A)$?

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If the answer is 'yes' I would like to get a reference.

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Edit: The following example suggest that the answer is no in general:

Consider a chain complex of vector spaces $B$. Now in degree $k\in \mathbb{Z}$ we take the quotient vector space $C_k:=B_k/kern(d_{k}(B_{k}))$. Then we have the short exact sequence

$$ 0 \to kern(d_B) \to B \to C \to 0 $$

where $kern(B)$ is the complex that in degree $k$ is just $kern(d_{B,k})$.

Clearly the differential on $C_k$ is trivial. So a section $s:C \to B$ has to satisfy $0=s(0)=s(d_C(x))=d_B(s(x))$, but $s$ can't map any $x\in C$ to the zero vector because then $p\circ s \neq id_C$ in general.

Answer 1

Actually, I can completely understand your counter-example. The following one is the simple I can imagine.

Let $k$ be a field and $\mathbf{Ch}(k\text{-}\mathbf{Vect})$ the category of chains of $k$-vector spaces.

Consider the short exact sequence of $k\text{-}\mathbf{Vect}$ $$0 \to k \to k\oplus k \to k \to 0\qquad(\star)$$ where the morphisms, given by the universal property of, resp., product and coproduct, are, resp., $\langle \text{id}_{k}, 0\rangle_\prod\colon k \to k\oplus k$ and $\langle 0, \text{id}_k\rangle_\coprod \colon k\oplus k \to k$; (i.e., resp., $x \mapsto (x, 0)$ and $(x_1, x_2)\mapsto x_2$.

Let $A_\bullet, C_\bullet$ and $B_\bullet$ be the exact complexes which have, resp., $\text{id}_k$ as $d^{A_\bullet}_1, d^{C_\bullet}_1$ morphisms and $\text{id}_{k\oplus k}$ as $d^{B_\bullet}_1$ morphism and for which $A_i \cong B_i\cong C_i \cong 0$, for every $i \in\mathbb Z\setminus\{0, 1\}$.

$(\star)$, as morphism of degree $0$ and $1$, induces a short exact sequence of exact complexes $$0_\bullet \to A_\bullet \to B_\bullet \to C_\bullet \to 0_\bullet$$

At degree $0$ and $1$ we consider, resp., the splittings $\langle 0, \text{id}_k\rangle_\prod$ and $\langle \text{id}_k, \text{id}_k\rangle_\prod$ (i.e., $x \mapsto (0, x)$ and $x\mapsto (x, x)$),both morphisms $k \to k\oplus k$.

Obviously, $$d^{B_\bullet}_1 \circ \langle 0, \text{id}_k\rangle_\prod = \text{id}_{k\oplus k}\circ\langle 0, \text{id}_k\rangle_\prod = \langle 0, \text{id}_k\rangle_\prod \neq \langle \text{id}_k, \text{id}_k\rangle_\prod = \langle \text{id}_k, \text{id}_k\rangle_\prod \circ \text{id}_k = \langle \text{id}_k, \text{id}_k\rangle_\prod \circ d^{C_\bullet}_1$$ directly by the universal property of product. As a matter of example, we have $$1_k \mapsto (0, 1_k)$$ on one hand and $$1_k \mapsto (1_k, 1_k)$$ on the other.

Answer 2

If you think of a complex as a functor in $Fun(\mathbb{Z},Mod_R)$ (where $R$ is a commutative ring with identity,for semplicity) then you're simply asking for calculating biproducts pointwise. In fact a section for the complex is a section pointwise, so that you have biproducts at each n-level, from which you can conclude.

Answer 3

@Mirco: I am sorry: I think I did not fully understand your question. I just wanted to say that with complexes you have to look at properties $componentwise$. If the split exists or not, and in which sense, this is another question. The tricky part is about differentials.

Let $0\rightarrow X\stackrel{f}{\rightarrow} Y \stackrel{g}{\rightarrow } Z \rightarrow 0$ be a short exact sequence of $\mathbb Z$-graded complexes of $R$-modules ($R$ is a ring). The sequence is said to be semi-split if there exists a morphism of $\mathbb Z$-graded $R$-modules $w: Z\rightarrow Y$ s.t. $g\circ w= 1_Z$. In other words, if $w$ exists, it does not need to commute with the differentials on $Y$ and $Z$, i.e. it is no morphism of complexes of $R$-modules, in general!

If you find a split $w$ like above, then you can identify $Y$ with $X \oplus Z$ as
$\mathbb Z$-graded $R$-modules using the componentwise splitting, as in the classical theory (forget about the differentials!). The differential on $X \oplus Z$ is in fact given by any $d_{X \oplus Z}=(d_X-h,d_Z)$ s.t. $h: X\rightarrow Z[1]$ is a degree $+1$ morphism of complexes of $R$-modules (so commutes with differentials: the differential on Z[1] is given by $d_{Z[1]}:=-d_Z$).

I hope that the role of differentials is a bit more clear. If not, please me know :) Avitus

Counter example added. Let $f:A\rightarrow B$ be a morphism of complexes. Then the short exact sequence $0\rightarrow B\rightarrow C(f)\rightarrow A\rightarrow 0$ is semisplit, denoting by $C(f)$ the cone of $f$, i.e. the complex $C(f)=(A[1]\oplus B,d)$, with differential $d(a,b)=(-d_A(a),f(a)+d_B(b))$. The splittings $r: C(f)\rightarrow B$ and $s: A[1]\rightarrow C(f)$ are the "usual" ones.

$C(f)$ is isomorphic to $A[1]\oplus B$ as $\mathbb Z$-graded module but not as complex. The split $r: C(f)\rightarrow B$ is not a morphism of complexes.

Answer 4

Consider the following specific counterexample. $A$ is the complex with the 1-dimensional space $k$ in degree 0 and zero in all other degrees (so of course the differential is 0 in all degrees). $B$ has the one-dimensional space $k$ in degree 1 and the 2-dimensional $k^2$ in degree 0; the unique non-trivial differential is given by $x\mapsto(x,0)$. The injection of $A$ in $B$ is nontrivial only in degree 0, where it sends $x\mapsto (x,0)$. This is a chain map; all the maps that are required to agree are in fact zero. Let $C$ be the cokernel of this injection. So we can take $C_0=C_1=k$ ; the projection from $B$ to $C$ is the identity in degree 1 and is $(x,y)\mapsto y$ in degree 0; the differential of $C$ is zero in all degrees.

Now suppose a chain map $s:C\to B$ were a splitting of this projection. For it to be a chain map, we need that the maps $C_1\to C_0\to B_0$ and $C_1\to B_1\to B_0$ agree. But the first of these maps is zero because $C_1\to C_0$ is. Therefore, so is the second. Since $B_1\to B_0$ is a monomorphism (namely $x\mapsto(x,0)$), we conclude that the degree 1 part $C_1\to B_1$ of $s$ is zero. That's impossible, as it has to split the isomorphism $B_1\to C_1$.