TITLE: Proof of 10.4.6 in Weibel
QUESTION [4 upvotes]: Let $I$ be a bounded below complex of injectives in some Abelian category, and $Z$ any bounded below complex. Suppose $u:I\to Z$ is a quasi isomorphism. We want to proof that $u$ is split injective up to homotopy. The prove outlined in Weibel's book An Introduction to Homological Algebra (pdf) is as follows:
The complex $\operatorname{cone}(u)$ is exact, where
$$
\operatorname{cone}(u)^i=I^{i+1}\oplus Z^i,
$$
and the differential is given by
$$
D=\begin{pmatrix}-d & 0 \\ -u & d\end{pmatrix}.
$$
The map $\phi:\operatorname{cone}(u)\to I[1]: (y,z)\mapsto -y$ is chain map, and it can be shown, using the fact that $I$ consists of injective objects, and $\operatorname{cone}(u)$ is exact, that $\phi$ is nullhomotopic, i.e.
$$\phi = fD+df$$
for some $\{f^i\}$, where
$$f^i:I^{i+1}\oplus Z^i\to I^i.$$
then we can write $f^i=(v^i,s^i)$, where $v^i:I^{i+1}\to I^i$ and $s^i:Z^i\to I^i$.
Then we get:
\begin{align*}
-y=\phi(y,z)&=fD(y,z)+df(y,z)\\
&=f(-dy,-uy+dz)+d(vy+sz)\\
&=-vdy-suy+sdz+dvy+dsz\\
&=-suy+(dvy-vdy)+(sdz+dsz).
\end{align*}
On the other hand, Weibel claims in his proof (after correcting a typo as per the errata) that
$$y=(kdy+suy+dky)+(dsz-sdz).$$
Obviously this is not the same, and my answer does lead to a proof of the lemma, so I must be doing something wrong, but I don't see what.

REPLY [5 votes]: Notice that you are looking at $I[1]$ instead of $I$, so there is an extra minus sign involving in the differential i.e. $d_{I[1]}f=-df$.
{"set_name": "stack_exchange", "score": 4, "question_id": 2869566}