Constructing a chain homotopy for quasi-isomorphisms of injectives If I have left-bounded complexes $I^.$ and $J^.$ of injectives and two quasi-isomorphisms (maps which induce isomorphisms in cohomology) $f^.$ and $g^.$ between them, is it true that they are chain homotopic?
Edit I should have said here that $f$ and $g$ induce the same maps on cohomology. But even this fails as Jeremy Rickard's answer below shows. I'm now interested in the more specialized case described below.
I want to construct maps $h^k$:$I^k\rightarrow J^{k-1}$ such that $f^k-g^k=d_I\circ h^k+h^{k+1}\circ d_J$. 
This is easy for exact complexes by induction + modding out by kernels (=images) + injectivity. 
The theme here is that I want to show the concept of a 'hyper-derived functor' is well defined. These groups are constructed for a  left-bounded complex $M^.$ of objects in some abelian category with enough injectives.
The definition: Define a quasi-isomorphism, $i^.$, of $M^.$ into an injective complex $I^.$ where each $i^k$ a monomorphism and then apply a left exact functor to $I^.$ to compute cohomology in some target category. 
The point is that the resulting groups are 'canonically' isomorphic since if we choose two such quasi-isomorphisms $i^.$, $j^.$ of $M^.$ into $I^.$ and $J^.$ respectively, there will exist a chain map $f^.$  lifting the identity on $M^.$ between them and any two chain maps must be homotopic. This last point is where I'm stuck.
The reference I'm using for this is Voisin's Complex Algebraic Geometry I page ~190.
I've looked at references elsewhere (e.g. Weibel, Gelfand-Manin) and they are quite hard to follow since they prove things that are far more general or they use spectral sequence arguments which I'm not yet familiar with. Some other notes I found online indicated that the chain homotopy arises from how the $I^.$ is constructed (something called a Cartan-Eilenberg resolution) but I don't see how that is pertinent.
I feel like I'm missing some small diagram chase argument and I'm hoping someone can suggest something.
Edit 2 The situation is $M^.$ injects via quasi-isomorphisms $i^.$ and $j^.$ into $I^.$ and $J^.$ respectively. $I^.$, $J^.$ are complexes of injectives. Then we can show there exists a $f^.$ from $I^.$ to $J^.$ commuting with $i^.$ and $j^.$. I want to show any two such maps of complexes must be chain homotopic. 
 A: Chain homotopic maps induce the same map on homology, so I think you probably want to assume not just that $f^.$ and $g^.$ are quasi-isomorphisms, but that they induce the same isomorphism on homology.
But even then they need not be chain homotopic.
Let $d:X\to Y$ be any non-split epimorphism of injectives (e.g., for abelian groups, the natural epimorphism $\mathbb{Q}\to\mathbb{Q}/\mathbb{Z}$), let
$$C^.:=\dots\to0\to X\stackrel{d}{\to}Y\to0\to\dots,$$
and let
$$D^.:=\dots\to0\to 0\to Y\to0\to\dots.$$
Then the inclusion $inc^.:D^.\to C^.$ is not null-homotopic, but induces the zero map on homology.
Now let $I^.=J^.=C^.\oplus D^.$, let $f^.:I^.\to J^.$ be the identity map, and let $g^.=f^.+\begin{pmatrix}0&inc^.\\0&0\end{pmatrix}$.
Then $f^.$ and $g^.$ both induce the identity map on homology, but are not chain homotopic.
[By the way, in the application to hyper-derived functors, you know more than that $f^.$ and $g^.$ induce the same isomorphism on homology: you also know that $f^.\circ i^.=g^.\circ i^.$, and this is crucial for what you want to prove.]
Edit (in reply to "Edit 2"): Suppose $i^.:M^.\to I^.$ and $j^.:M^.\to J^.$ are inclusions of complexes that are quasi-isomorphisms, and that $f^.,g^.:I^.\to J^.$ satisfy $f^.\circ i^.=j^.=g^.\circ i^.$ Then $f^.-g^.$ restricts to zero on $M^.$ and so induces a map $I^./M^.\to J^.$. Since the inclusion of $M^.$ into $I^.$ is a quasi-isomorphism, $I^./M^.$ is acyclic, and it's straightforward to construct a null-homotopy for the map $I^./M^.\to J^.$. Composing with $I^.\to I^./M^.$, $f^.-g^.$ is null homotopic.
