Uniqueness in the proof of the comparison theorem for injective resolutions $\require{AMScd}$
I'm trying to prove the comparison theorem for injective resolutions:

Suppose $\mathcal{A}$ is an abelian category, $A\in\mathcal{A}$ and $A\xrightarrow{\varepsilon} I$ is an injective resolution. Then given $f^\prime:A\to B$ and an injective resolution $B\xrightarrow{\eta} Q$, there is a chain map $f:I\to Q$
\begin{CD}
  0 @>>> A @>\varepsilon>>   I^0 @>d>> I^1 @>d>> I^2 @>d>>   \cdots \\
  @.  @Vf^\prime VV @Vf^0VV @Vf^1VV @Vf^2VV \\
  0 @>>> B @>\eta>>  Q^0 @>d>>  Q^1 @>d>> Q^2 @>d>> \cdots
\end{CD}
such that $f^0\circ\varepsilon = \eta\circ f^\prime$ and $f$ is unique up to chain homotopy.

I've managed to prove the existence of $f$ but I'm struggling with proving it's unique up to chain homotopy.
Here is what I have so far (which isn't much). Let $g:I\to Q$ be another chain map such that $g^0\circ\varepsilon = \eta\circ f^\prime$, then set $h := f-g$. Then we aim to construct maps $s^n:I^{n+1}\to Q^n$ such that $h^n = s^n\circ d + d\circ s^{n-1}$ by induction. For $s^0$ we need $h^0 = s^0\circ d$, so I've been trying to construct a diagram
\begin{CD}
  0 @>>> ?  @>d>> I^1  \\
  @.  @Vh^0 VV \\
  @. Q^0 
\end{CD}
with the top row exact, and then use the injectivity of $Q^0$. But I can't for the life of me figure out what $?$ should be. It seems like this diagram should't be able to exist in the first place, because it requires $d$ to be a monomorphism from some object associated with $I^0$, but how can that be guaranteed? I run into a similar problem with the inductive step: Suppose that for $0\le k \le n$ we have maps $s^k:I^{k+1}\to Q^k$ such that $h^k = s^k\circ d + d\circ s^{k-1}$, then I have a diagram
\begin{CD}
  0 @>>> ?  @>d>> I^{n+2}  \\
  @.  @Vh^{n+1} - ds^n VV \\
  @. Q^{n+1} 
\end{CD}
But again I'm not sure what $?$ should be, for the same reason as before, how can we make $d$ a monomorphism from some object associated with $I^{n+1}$? I might be barking up the wrong tree, but for the time being it's the only tree I have. Any hints would be greatly appreciated.
 A: Instead of using this characterisation of injective, you can use the fact that $Hom(.,Q^n)$ is an exact functor. Indeed, from the exact sequence $0\to A\to I^0\to I^1$, you get an exact sequence :
$$
Hom(I^1,Q^0)\to Hom(I^0,Q^0)\to Hom(A,Q^0)\to 0
$$
The map $h^0\in Hom(I^0,Q^0)$ maps to $0$ in $Hom(A,Q^0)$. By exactness, there exist $s^0:I^1\to Q^0$ such that $s^0d=h^0$.
Similarly, assume you have constructed $s^n:I^{n+1}\to Q^n$ such that $h^n=s^nd+ds^{n-1}$. From the exact sequence $I^n\to I^{n+1}\to I^{n+2}$, you have an exact sequence
$$Hom(I^{n+2},Q^{n+1})\to Hom(I^{n+1},Q^{n+1})\to Hom(I^n,Q^{n+1})$$
Now the map $h^{n+1}-ds^n\in Hom(I^{n+1},Q^{n+1})$ and since
$$(h^{n+1}-ds^n)d=h^{n+1}d-ds^nd=h^{n+1}d-d(h^n-ds^{n-1})=h^{n+1}d-dh^n+0=0$$
its image is zero in $Hom(I^n,Q^{n+1})$. By exactness, there exist $s^{n+1}$ such that $h^{n+1}-ds^n=s^{n+1}d$, which concludes the proof.

You might wonder why I used this characterization instead of the one in your post. Note that the $Hom(.,I)$ is always left exact and your characterization means that $I$ is injective iff $Hom(.,I)$ preserves epimorphisms.
You have this general fact : for a left exact functor $F$, being exact is equivalent to preserving epimorphisms. Of course the direct implication is immediate, the converse, while not hard, require a bit of work. And indeed, if $A\to B\to C\to 0$ is exact, try to show that $F(A)\to F(B)\to F(C)\to 0$ is exact directly from $F$ preserves kernels and epimorphisms. You will see that you will need two short exact sequences to prove the exactness at $B$.
It follows that no objects can directly fill the "$?$" in your diagrams. You would have need two steps just to construct $s^0$.
