Splitting differentials of a splitting short exact sequence of complexes.

Question

Let $0\to F^{\bullet}\to E^{\bullet}\to G^{\bullet}\to 0$ be a short exact sequence of chain complexes of vector spaces. Denote by $v_E$, $v_F$ and $v_G$ the differentials. For every index $i$ we have an isomorphism of vector spaces $\varphi^i:E^i\to F^i\oplus G^i$.

My question is the following: Do we have a decomposition $$v_E\simeq \begin{pmatrix} v_F & 0 \\ 0 & v_G \end{pmatrix}$$ in the sense that $$\varphi^{i+1}\circ v_E\circ (\varphi^i)^{-1} = \begin{pmatrix} v_F & 0\\ \\ 0 & v_G \end{pmatrix}?$$ Any reference would be appreciated.

Answer 1

In an abelian category $\cal{A}$ a short exact sequence $$0 \rightarrow A \rightarrow B \rightarrow C$$ splits, if and only if it is isomorphic as a short exact sequence in $\cal A$ to the short exact sequence $$0 \rightarrow A \overset{i}\rightarrow A \oplus C \overset{\pi} \rightarrow C \rightarrow 0.$$ Any short exact sequence of vector spaces splits. Thus any short exact sequence $$0 \rightarrow A_\bullet \rightarrow B_\bullet \rightarrow C_\bullet \rightarrow 0$$ of complexes of vector spaces splits levelwise, i.e. when fixing $\bullet = i$. But nothing tells you that these splits are compatible with the boundary morphisms, which is what you need to have a split exact sequence of complexes.

If you do have a split exact sequence of complexes, the answer to your question is that it holds by the very definition of the comples $A_\bullet \oplus C_\bullet$, which has the differential $v_A\oplus v_B$ respectively $\begin{pmatrix}v_A &0\\0&v_B\end{pmatrix}$. The isomorphism $\phi:B_\bullet\rightarrow A_\bullet\oplus C_\bullet$ (from the isomorphism of split exact sequences of complexes) is in particular a morphism of complexes, so the square $$\begin{array}{ccc} B_i & \overset{\phi_i}\longrightarrow & A_i \oplus C_i\\ \downarrow_{v_{B}} && \downarrow_{v_A\oplus v_C}\\ B_{i+1} & \overset{\phi_{i+1}}\longrightarrow & A_{i+1} \oplus C_{i+1} \end{array}$$ commutes.

Answer 2

As an addition to the accepted answer, I would like to note that not every short exact sequence of complexes of vector spaces splits.

In an abelian category an object $X$ is projective if and only if any sequence

$$0\to Z\to Y\to X\to 0$$

splits.

In the category of complexes the only projective objects are split exact complexes of projectives. So for any inexact complex of vector spaces $G^\bullet$ there must exist a non-split sequence

$$0\to F^{\bullet}\to E^{\bullet}\to G^{\bullet}\to 0.$$