In Methods of Homological Algebra Theorem III.6, it is claimed that for the distinguished triangle $$K^\bullet\rightarrow\operatorname{Cyl}(u)\rightarrow C(u)\xrightarrow{\delta}K[1]^\bullet$$ where $u:K^\bullet\rightarrow L^\bullet$ is a morphism of complexes, Cyl denotes the mapping cylinder, C(u) the mapping cone, that the connecting homomorphism $H^i(C(u))\xrightarrow{\partial}H^i(K[1]^\bullet)$ is induced by $\delta$. It appears to me, however, that it is induced by $-\delta$. That is, take an element of $H^i(C(u))$, lift it to an element of $C(u)$, e.g. $(k^{i+1},l^i)$, and then $\operatorname{Cyl}(u)$, such as $(0,k^{i+1},l^i)$, apply the differential, we get $(-k^{i+1},0,0)$, which is the image of $-k^{i+1}$ under the inclusion of $K^{i+1}$, and $-k^{i+1}$ is equal to $-\delta(k^{i+1},l^i)$, not $\delta(k^{i+1},l^i)$. Where have I erred in my computation?
There is a related statement about that the connecting homomorphism $H^{i}(K[1]^\bullet)\rightarrow H^{i+1}(L^\bullet)$ in the long exact sequence associated to the short exact sequence $$0\rightarrow L^\bullet\rightarrow C(u)\rightarrow K[1]^\bullet\rightarrow 0$$ which is in turn associated to the distinguished triangle $$K^\bullet\xrightarrow{u}L^\bullet\rightarrow C(u)\rightarrow K[1]^\bullet.$$ The statement is that this connecting homomorphism is $H^{i+1}(u)$ and this I can verify.