Morphisms into a homotopically injective complex need no roof

For any quasi-isomorphism $s:U\to V$, the cone and its shifts are acyclic. Apply ${\mathrm{Hom}}_K(-,I)$ to its triangle: the two adjacent cone Hom groups vanish by K-injectivity. Thus precomposition by $s$ gives a bijection ${\mathrm{Hom}}_K(V,I)\to {\mathrm{Hom}}_K(U,I)$. This includes all zero objects.

Given a left roof $X\stackrel{s}{\leftarrow }U\stackrel{f}{\to }I$, the bijection gives a unique $b:X\to I$ with $bs=f$ in $K$, so the roof equals $Q(b)$. An equality $Q(b)=Q(c)$ is witnessed by $bs=cs$ for a denominator $s$ and the same bijection gives $b=c$. Equivalently a right-roof denominator out of $I$ has a retraction and therefore also eliminates the roof.

For $s:I\to J$ the bijection gives $r:J\to I$ with $rs=1_I$ in $K$. Let $C=\mathrm{Cone}(s)$ and let $p:C\to I[1]$ be its projection. K-injectivity gives $p=0$ in $K$. Hom exactness supplies $v:C\to J$ with $iv=1_C$, where $i:J\to C$. Replace $v$ by $v-srv$, so also $rv=0$. Then $1_J-sr-vi$ is killed by $i$ and hence factors as $su$ by Hom exactness. Applying $r$ gives $u=0$. Consequently $(s,v):I\oplus C\to J$ and $(r,i)$ are inverse in $K$.

The cone signs can also be checked on matrices. Choose a representative homotopy $rs-1_I=d_{I}h+h{d}_I$. The degree-minus-one map $H:C^n\to I[1{]}^{n-1}$ given by $H(j,x)=rj+hx$ satisfies $d_{I[1]}H+H{d}_C=p$. Thus the split connecting map is zero with the specified cone convention, rather than after an unrecorded change of sign.