Is the mapping cone a split complex?

Question

This question arise from the comment in Corollary 1.5.4 in Weibel's An introduction to Homological Algebra: a cochain map $f\colon B\to C$ is a quasi-isomorphism if and only if the mapping cone complex is exact; hence this reduces questions about quasi-isomorphism to the study of split complexes.

The mapping cone $\operatorname{cone}(f)$ is the cochain complex which on degree $n$ is given by $B^{n+1}\oplus C^{n}$ and with differentials $$ d^{n}\colon B^{n+1}\oplus C^{n} \to B^{n+2}\oplus C^{n+1} $$ given by $(b,c)\mapsto (-d_{B}^{n+1}(b),d_{C}^{n}(c)-f^{n+1}(b))$.

If $f$ is the identity map on some cochain complex then we can define a splitting by sending $(b,c)\mapsto (-c,0)$. But in general I don't see how to define the splitting maps $s^{n}\colon \operatorname{cone}(f)^{n+1}\to \operatorname{cone}(f)^{n}$ with $d^{n}=d^{n}\circ s^{n}\circ d^{n}$. And I also couldn't find anywhere the statement that the mapping cone is always split. So this makes me believe that the answer to the question is no. But in that case, what does Weibel mean when he says that?

Answer 1

This is a mistake, as is explained in the official errata. The corrected sentence is as follows:

This device reduces questions about quasi-isomorphisms to the study of exact complexes.

Answer 2

The connecting morphism of a split short exact sequence is null, so it suffices you consider maps of complexes which are not trivial in homology, since the connecting map of the cone SEC is the map induced by the morphism in homology.