Splitting a short exact sequence of complexes of vector spaces It's well-known that any complex of vector spaces is isomorphic to a direct sum of two types of indecomposable complexes (a one-dimensional space concentrated in one degree, or two one dimensional spaces in adjacent degrees, connected by an isomorphism).
I have a related question: consider a short exact sequence of complexes of vector spaces. Can we tell when it is split, without actually exhibiting the splitting?
A necessary condition is for all the induced boundary maps on homology to be zero. Is this sufficient?
edit: Originally I mistakenly said "the category of complexes of vector spaces is semisimple", but that was not correct terminology (the indecomposable two-term complex is not a simple object!)
 A: I hate to answer my own question, but I think the answer is yes and here is a nice proof: compute all the first Ext groups of the indecomposable complexes. There's only one nonvanishing Ext group and it "classifies" or "records" the boundary map.
Let $P[i]$ denote the indecomposable complex $k \xrightarrow{\sim} k$ in degrees $i,i-1$. Let $S[i]$ denote the indecomposable single-term complex $k$ in degree $i$.
Lemma. Any inclusion $P[i] \hookrightarrow C$ into any complex $C$ splits. Likewise, any surjection $C \twoheadrightarrow P[i]$ splits.
Proof: First split the inclusion $k \hookrightarrow C_{i-1}$, then pull this back across the isomorphism $k \xrightarrow{\sim } k$ of $P[i]$ to obtain commuting surjections
$$\begin{array}{ccccccccc}0&& k & \xrightarrow{\sim} & k &&0 \\&& \uparrow & &\uparrow \\ \cdots & \xrightarrow{d} & C_i & \xrightarrow{d} & C_{i-1} & \xrightarrow{d} & \cdots \end{array}$$
The $i\to(i-1)$ square obviously commutes and splits the inclusion of $P[i]$; we get commutativity of the $(i+1)\to i$ square because $C_i \to k$ technically involves passing through $C_i \to C_{i-1}$ first. There's nothing to check in the $(i-1) \to (i-2)$ square or the other squares. The statement for surjections is equivalent, just dualize everything. $\blacksquare$
The Lemma shows that $$\mathrm{Ext}^1(P[i],-) = \mathrm{Ext}^1(-,P[i]) = 0,$$
so these are all split extensions.
The Ext groups of $S[i]$ are also pretty easy and by checking the possible cases, the only nonvanishing one is $Ext^1(S[i],S[i-1])$, which corresponds to
$$\begin{array}{ccccccccc} 0 & & k \\ & & \downarrow \\ k & \xrightarrow{f} & k \\ \downarrow \\ k & & 0, \end{array}$$
where the vertical arrows are isomorphisms. The Ext class is determined by the scalar multiplication $f$: it's split if and only if $f=0$, which is also if and only if the corresponding boundary map is zero.
So now we have an arbitrary short exact sequence of complexes $$0 \to A_\bullet \to B_\bullet \to C_\bullet \to 0,$$ giving an element of $$B \in \mathrm{Ext}^1(C,A).$$
By decomposing $C$ and $A$ into indecomposable summands, this becomes a sum of Ext groups. (In fact, this corresponds to picking bases for homology in order to write down the boundary maps explicitly as matrices.) If all the boundary maps are zero, then the class $B$ is the zero class and corresponds to the split extension.
