Induced map on Cartan-Eilenberg resolutions through an exact functor

Question

Let $T: \mathcal{A} \to \mathcal{B}$ be an exact functor between abelian categories. Let $A^\bullet$ (resp. $B^\bullet$) be a bounded below cochain complex in $\mathcal{A}$ (resp. $\mathcal{B}$) and let $I^{\bullet, \bullet}$ (resp. $J^{\bullet, \bullet}$) be a Cartan-Eilenberg resolution of $A^\bullet$ (resp. $B^\bullet$). According to Weibel, An Introduction to Homological Algebra, Exercise 5.7.2; or Gelfand-Manin, Methods of Homological Algebra, Proposition III.7.11; any chain map extends to a morphism of Cartan-Eilenberg resolutions.

I have two CE resolutions which are not on the same category, but instead are related by an exact functor (in my particular case $\mathcal{A}$ and $\mathcal{B}$ are categories of sheaves and $T$ is the pullback by a continuous map). Therefore $T(I^{\bullet, \bullet})$ is not a CE resolution anymore (the objects are not injective). Is there still a way to get a map of double complexes? $$ T(I^{\bullet, \bullet}) \to J^{\bullet, \bullet} $$

My goal is to prove that a commutative square of continuous maps between topological spaces (not necessarily Cartesian) induces a map between the $E_2$ pages of the Leray spectral sequences, which eventually gives the induced contravariant map on sheaf cohomology (see Iversen, Cohomology of Sheaves, II.5.1.)

Answer 1

I have seen that it is actually possible to construct this map if you follow the proof in Gelfand-Manin carefully. In there, they say that the way to construct the map between Cartan-Eilenberg resolutions is to apply Theorem III.1.3 and statement III.7.11.B (inside the proof) repeatedly.

• Theorem III.1.3 asserts that it is possible to extend any map between objects to a map between projective resolutions of said objects. Inverting the arrows gives the statement we need for injective resolutions.
• Remark III.1.4(b) points out that if we aim to extend $f: X \to Y$, the resolution of $Y$ need not actually be projective, just an exact sequence. Dually, we don't need the resolution on the source to be injective.

This is precisely our setting: $T$ being exact guarantees that all exactness properties of $I^{\bullet, \bullet}$ pass on to $T(I^{\bullet, \bullet})$, which is all we needed to carry on with the construction.